Lecture 1

Some important information on the course
Lecturer: Jorge M. Mendes (jmm@novaims.unl.pt)
Lectures timetable: Tuesdays and Thursdays, 3:30pm to 4:15pm
Assessment rules: check syllabus
Jorge M. Mendes (NOVAIMS) Multivariate Data Analysis Basics 2015 1 / 122

Textbooks
Everitt, B. and Hothorn, T. (2011). An Introduction to Applied Multivariate
Analysis with R, Springer.
Johnson, R.A and Wichern (2007). D. W., Applied Multivariate Statistical
Analysis, 6th edition, Pearson Prentice Hall.
online tools and references: https://www.datacamp.com
online tools and references: http://www.statmethods.net
And...
Reis, E. (1997), Estat´ıstica Multivariada Aplicada, Edi¸c˜oes S´ılabo
Sharma, S. (1996). Applied Multivariate Techniques, John Wiley & Sons
Timm, N. H. (2002). Applied Multivariate Analysis, Springer
Other things: lecture slides, data ﬁles, etc.

Course contents
Data analysis basics: organisation of multivariate data; parameters and
descriptive statistics; matrix manipulations for sample statistics; review of
matrix deﬁnitions and operations
R tutorial; fundamentals on data manipulation on R
Graphical display of multivariate data; Linear combinations of random variables
The multivariate normal distribution
Principal components analysis
Factor analysis
Discriminant analysis
Cluster analysis
Canonical correlation analysis
Multidimensional scaling
... eventually something else!

Lecture 1
Multivariate Data Analysis Basics
Jorge M. Mendes
Nova Information Management School
Universidade Nova de Lisboa
2015

First things ﬁrst
This ﬁrst lesson will introduce the basic tools to deal with multivariate data
and multivariate techniques.
Topics covered include multivariate sample statistics and spectral
decomposition of a matrix.
After successfully completing this lesson, you should be able to:
Understand the notation used in multivariate analysis;
Understand the organization of multivariate data using vector and matrix
notation;
Perform basic operations on matrices and vectors;
Interpret measures of central tendency, dispersion, and association;
Calculate sample means, variances, covariances, and correlations using a hand
calculator;
Compute the spectral decomposition of a matrix (eigenvalues, eigenvectors,
etc.)

Multivariate data analysis basics Organisation of multivariate data
Subsection 1
Organisation of multivariate data

Introduction
Univariate statistics is concerned with random scalar variable Y .
In multivariate analysis, we are concerned with the joint analysis of
multiple dependent variables.
These variables can be represented using matrices and vectors.
This provides simpliﬁcation of notation and a format for expressing
important formulas.

Example
Suppose that we measure the variables x1 = height (cm), x2 = left forearm
length (cm) and x3 = left foot length for participants in a study of the
physical characteristics of adult humans.
These three variables can be represented in the following column vector:
x =


x1
x2
x3



Example
The observed data for a speciﬁc individual, say the i-th individual, might also
be represented in an analogous vector.
Suppose that the i-th person in the sample has height = 175 cm, forearm
length = 25.5 cm and foot length = 27 cm.
In vector notation these observed data could be written as:
xi =


xi1
xi2
xi3

 =


175
25.5
27.0


Notice the use and placement of the subscript i to represent the i-th
individual.

Deﬁnitions of matrix and vector
A matrix is two-dimensional array of numbers of formulas.
A vector is a matrix with either only one column or only one row. A column
vector has only one column. A row vector has only one row.
The dimension of a matrix is expressed as number of rows × number of
columns.
For instance, a matrix with 10 rows and 3 columns is said to be a 10 × 3
matrix. The vectors written in previous Example are 3 × 1 matrices.
A square matrix is one for which the numbers of rows and columns are the
same. For instance, a 4 × 4 matrix is a square matrix.

The data matrix in multivariate problems
Usually the observed data are represented by a matrix in which the rows are
observations and the columns are variables.
This is exactly the way the data are normally prepared for statistical software.
The usual notation is
n: the number of observed units (people, animals, companies, etc.)
p: number of variables measured on each unit.
Thus the data matrix will be an n × p matrix.

Example
Suppose that we have scores for n = 6 college students who have taken the
verbal and the science subtests of a College Qualiﬁcation test (CQT).
We have p = 2 variables: (1) the verbal score and (2) the science score for
each student.
The data matrix is the following 6 × 2 matrix:
X =








41 26
39 26
53 21
67 33
61 27
67 29









Example
In the matrix just given, the ﬁrst column gives the data for x1 = verbal score
whereas the second column gives data for x2 = science score.
Each row gives data for a student in the sample.
To repeat - the rows are observations, the columns are variables.

Notation notes
Note that we have used a small x to denote the vector of variables in first
example and a large X to represent the data matrix in second example.
It should also be noted that, in matrix terms, the i-th row in the data matrix
X is the transpose of the data vector
xi =
xi1
xi2
, as we defined data vectors in first Example.

Multivariate data analysis basics Parameters and descriptive statistics
Subsection 2
Parameters and descriptive statistics

Introduction
Three aspects of the data are of importance, the ﬁrst two of which you should
already be familiar with from univariate statistics. These are:
Central Tendency. What is a typical value for each variable?
Dispersion. How far apart are the individual observations from a central value
for a given variable?
Association. This might (or might not!) be a new measure for you. When
more than one variable are studied together how does each variable relate to
the remaining variables? How are the variables simultaneously related to one
another? Are they positively or negatively related?

Population parameters and sample statistics
Statistics, as a subject matter, is the science and art of using sample
information to make generalizations about populations.
A population is the collection of all people, plants, animals, or objects of
interest about which we wish to make statistical inferences (generalizations).
A population parameter is a numerical characteristic of a population. In
nearly all statistical problems we do not know the value of a parameter
because we do not measure the entire population. We use sample data to
make an inference about the value of a parameter.

Population parameters and sample statistics
A sample is the subset of the population that we actually measure or observe.
A sample statistic is a numerical characteristic of a sample. A sample
statistic estimates the unknown value of a population parameter.
Information collected from sample statistic is sometimes referred to as
descriptive statistic.

Notation
Here are the notation that will be used:
Xij : Observation for variable j in subject i.
p: Number of variables
n: Number of subjects

Example: USDA Women’s Health Survey
Let us take a look at an example.
In 1985, the USDA commissioned a study of women’s nutrition.
Nutrient intake was measured for a random sample of 737 women aged 25-50
years.
The following variables were measured:
Calcium (mg)
Iron (mg)
Protein (g)
Vitamin A (µg)
Vitamin C (mg)

As a notation example let’s use the women’s nutrition data.
p = 5 nutritional outcomes
n = 737 subjects

In multivariate statistics we will always be working with vectors of
observations.
So in this case we are going to arrange the data for the p variables on each
subject into a vector.
In the expression below, Xi is the vector of observations for the i-th subject,
i = 1, ..., n(= 737).
Therefore, the data for the j-th variable will be located in the j-th element of
this subject’s vector, j = 1, ..., p(= 5).
Xi =





Xi1
Xi2
...
Xip






The mean
Throughout this course, we’ll use the ordinary notations for the mean of a
variable.
That is, the symbol µ is used to represent a (theoretical) population mean
and the symbol ¯x is used to represent a sample mean computed from
observed data.
In the multivariate setting, we add subscripts to these symbols to indicate the
speciﬁc variable for which the mean is being given.
For instance, µ1 represents the population mean for variable x1 and ¯x1
denotes a sample mean based on observed data for variable x1.

The mean
The population mean is the measure of central tendency for the population.
Here, the population mean for variable j is
µj = E(Xij )
The notation E stands for statistical expectation; here E(Xij ) is the mean of
Xij over all members of the population, or equivalently, over all random draws
from a stochastic model.
For example, µj = E(Xij ) may be the mean of a normal variable.

The mean
The population mean µj for variable j can be estimated by the sample mean
¯xj =
1
n
n
i=1
Xij
Note: the sample mean ¯xj , because it is a function of our random data is
also going to have a mean itself.
In fact, the population mean of the sample mean is equal to population mean
µj ; i.e.,
E(¯xj ) = µj
Therefore, the ¯xj is unbiased for µj .

The mean
Another way of saying this is that the mean of the ¯xj ’s over all possible
samples of size n is equal to µj .
Recall that the population mean vector is µ which is a collection of the
means for each of the population means for each of the diﬀerent variables.
µ =





µ1
µ2
...
µp






The mean
We can estimate this population mean vector, µ, by ¯x. This is obtained by
collecting the sample means from each of the variables in a single vector.
This is shown below.
¯x =





¯x1
¯x2
...
¯xp





=





1
n
n
i=1 Xi1
1
n
n
i=1 Xi2
...
1
n
n
i=1 Xip





=
1
n
n
i=1
Xi

The mean
Just as the sample means, ¯xj , for the individual variables are unbiased for
their respective population means, note that the sample mean vector is
unbiased for the population mean vector.
E(¯x) = E





¯x1
¯x2
...
¯xp





=





E(¯x1)
E(¯x2)
...
E(¯xp)





=





µ1
µ2
...
µp





= µ

Dispersion: variance and standard deviation
A variance measures the degree of spread (dispersion) in a variable’s values.
Theoretically, a population variance is the average squared diﬀerence between
a variable’s values and the mean for that variable. The population variance
for variable xj is
σ2
j = E(xj − µj )2
Note that the squared residual (xj − µj )2
is a function of the random variable
Xij .
Therefore, the squared residual itself is random, and has a population mean.

The population variance is thus the population mean of the squared residual.
We see that if the data tend to be far away from the mean, the squared
residual will tend to be large, and hence the population variance will also be
large.
Conversely, if the the data tend to be close to the mean, the squared residual
will tend to be small, and hence the population variance will also be small.

The population variance σ2
j can be estimated by the sample variance
s2
j =
1
n − 1
n
i=1
(Xij − ¯xj )2
=
1
n − 1
n
i=1
X2
ij −
n
n − 1
¯x2
j
The ﬁrst expression in this formula is most suitable for interpreting the
sample variance.
We see that it is a function of the squared residuals; that is, take diﬀerence
between the individual observations and their sample mean, and then square
the result.

Here, we may observe that if tend to be far away from their sample means,
then the squared residuals and hence the sample variance will also tend to be
large.
If on the other hand, if the observations tend to be close to their respective
sample means, then the squared diﬀerences between the data and their means
will be small, resulting is a small sample variance value for that variable.
The last part of the expression above, gives the formula that is most
suitable for computation, either by hand or by a computer!

Since the sample variance is a function of the random data, the sample
variance itself is a random quantity, and so has a population mean.
In fact, the population mean of the sample variance is equal to the
population variance:
E(s2
j ) = σ2
j
That is, the sample variance s2
j is unbiased for the population variance σ2
j .

Some textbooks use a sample variance formula derived using maximum
likelihood estimation principles. In this formula, the division is by n rather
than n − 1.
s2
j =
n
i=1(xij − ¯xj )2
n

Example
Suppose that we have observed the following n = 5 resting pulse rates: 64,
68, 74, 76, 78. The sample mean is ¯x = 64+68+74+76+78
5 = 72.
The maximum likelihood estimate of the variance is
s2
=
(64 − 72)2
+ (68 − 72)2
+ (74 − 72)2
+
5
+(76 − 72)2
+ (78 − 72)2
5
=
136
5
= 27.2
The standard deviation based in this method is s =
√
27.2 = 5.215.
The more commonly used variance estimate, the one given by statistical
software, would be 136
5−1 = 34. The standard deviation would be
s =
√
34 = 5.83.

Measures of association: covariance and correlation
Association is concerned with how each variable is related to the other
variable(s). In this case the ﬁrst measure that we will consider is the
covariance between two variables j and k.
The population covariance is a measure of the association between pairs of
variables in a population. Here, the population covariance between variables j
and k is
σjk = E{(Xij − µj )(Xik − µk )}
Note that the product of the residuals (Xij − µj ) and (Xik − µk ) for variables
j and k, respectively, is a function of the random variables Xij and Xik .
Therefore, (Xij − µj )(Xik − µk ) is itself random, and has a population mean.

The population covariance is deﬁned to be the population mean of this
product of residuals. We see that if either both variables are greater than
their respective means, or if they are both less than their respective means,
then the product of the residuals will be positive.
Thus, if the value of variable j tends to be greater than its mean when the
value of variable k is larger than its mean, and if the value of variable j tends
to be less than its mean when the value of variable k is smaller than its
mean, then the covariance will be positive.

Positive population covariances mean that the two variables are positively
associated; variable j tends to increase with increasing values of variable k
and vice-versa.
Negative association can also occur. If one variable tends to be greater than
its mean when the other variable is less than its mean, the product of the
residuals will be negative, and you will obtain a negative population
covariance. Variable j will tend to decrease with increasing values of variable
k.

The population covariance σjk between variables j and k can be estimated by
the sample covariance. This can be calculated using the formula below:
sjk =
1
n − 1
n
i=1
(Xij − ¯xj )(Xik − ¯xk ) =
1
n − 1
n
i=1
Xij Xik −
n
n − 1
¯x2
j ¯x2
k
Just like in the formula for variance we have two expressions that make up
this formula. The ﬁrst half of the formula is most suitable for understanding
the interpretation of the sample covariance, and the second half of the
formula is what is used for calculation.

Looking at the first half of the expression what you see inside the sum is the
product of the residual differences between variable j and its mean times the
residual differences between variable k and its mean.
We can see that if either both variables tend to be greater than their
respective means or less than their respective means, then the product of the
residuals will tend to be positive leading to a positive sample covariance.
Conversely if one variable takes values that are greater than its mean when
the opposite variable takes a value less than its mean, then the product will
take a negative value. In the end, when you add up this product over all of
the observations, you will end up with a negative covariance.

So, in effect, a positive covariance would indicate a positive association
between the variables j and k. And a negative association is when the
covariance is negative.
For computational purposes we will use the second half of the formula. For
each subject, the product of the two variables is obtained, and then the
products are summed to obtain the first term in the numerator.
The second term in the numerator is obtained by taking the product of the
sums of variable over the n subjects, then dividing the results by the sample
size n. The difference between the first and second terms is then divided by
n − 1 to obtain the covariance value.

Again, sample covariance is a function of the random data, and hence, is
random itself.
As before, the population mean of the sample covariance sjk is equal the
population covariance σjk ; i.e.,
E(sjk ) = σjk
That is, the sample covariance sjk is unbiased for the population covariance
σjk .

The sample covariance is a measure of the association between a pair of
variables:
sjk = 0 implies that the two variables are uncorrelated. (Note that this does
not necessarily imply independence, we’ll get back to this later.)
sjk > 0 implies that the two variables are positively correlated; i.e., values of
variable j tend to increase with increasing values of variable k. The larger the
covariance, the stronger the positive association between the two variables.
sjk < 0 implies that the two variables are negatively correlated; i.e., values of
variable j tend to decrease with increasing values of variable k. The smaller the
covariance, the stronger the negative association between the two variables.

Recall, that we had collected all of the population means of the p variables
into a mean vector.
Likewise, the population variances and covariances can be collected into the
population variance-covariance matrix. This is also known by the name of
population dispersion matrix.
Σ =





σ2
1 σ12 . . . σ1p
σ21 σ2
2 . . . σ2p
...
...
...
...
σp1 σp2 . . . σ2
p





Note that the population variances appear along the diagonal of this matrix,
and the covariance appear in the oﬀ-diagonal elements. So, the covariance
between variables j and k will appear in row j and column k of this matrix.

The population variance-covariance matrix may be estimated by the sample
variance-covariance matrix.
The population variances and covariances in the above population
variance-covariance matrix are replaced by the corresponding sample
variances and covariances to obtain the sample variance-covariance matrix:
S =





s2
1 s12 . . . s1p
s21 s2
2 . . . s2p
...
...
...
...
sp1 sp2 . . . s2
p






Note that the sample variances appear along diagonal of this matrix and the
covariances appear in the oﬀ-diagonal elements. So the covariance between
variables j and k will appear in the jk-th element of this matrix.
Notes:
S (the sample variance-covariance matrix) is symmetric; i.e., sjk = skj .
S is unbiased for the population variance covariance matrix Σ ; i.e.,
E(S) =





E(s2
1 ) E(s12) . . . E(s1p)
E(s21) E(s2
2 ) . . . E(s2p)
...
...
...
...
E(sp1) E(sp2) . . . E(s2
p )





= Σ

Since this matrix is a function of our random data, this means that the
elements of this matrix are also going to be random, and the matrix on the
whole is random as well.
The statement “σ2
is unbiased”means that the mean of each element of that
matrix is equal to the corresponding elements of the population.

In matrix notation sample variance-covariance matrix may be computed used
the following expressions:
S =
1
n − 1
n
i=1
(Xi − ¯x)(Xi − ¯x) =
1
n − 1
n
i=1
Xi Xi −
n
n − 1
¯x¯x
Just as we have seen in the previous formulas, the ﬁrst half of the formula is
used in interpretation, and the second half of the formula is what is used for
calculation purposes.

Looking at the second term you can see that the first term in the numerator
involves taking the data vector for each subject and multiplying by its
transpose. The resulting matrices are then added over the n subjects.
To obtain the second term in numerator, first compute the sum of the data
vectors over the n subjects, then take the resulting vector and multiply by its
transpose; then divide the resulting matrix by the squared number of subjects
n2
. Take the difference between the two terms in the numerator and divide
by n − 1.

Example
Suppose that we have observed the following n = 4 observations for variables
x1 and x2.
x1 x2
6 3
10 4
12 7
12 6
The sample means are x1 = 10 and x2 = 5.

Example
The maximum likelihood estimate of the covariance is the average product of
deviations from the mean:
s12 =
(6 − 10)(3 − 5) + (10 − 10)(4 − 5)+
4
+(12 − 10)(7 − 5) + (12 − 10)(6 − 5)
4
=
8 + 0 + 4 + 2
4
= 3.5
The positive value reﬂects the fact that as x1 increases, x2 also tends to
increase.

Problem
Problem: The magnitude of the covariance value is not particularly helpful
as it is a function of the magnitudes (scales) of the two variables.
It is hard to distinguish the eﬀects of the association between the two
variables from the eﬀects of their dispersions.
Note, however, that the covariance between variables j and k must lie
between the product of the two component standard deviations of variables j
and k, and negative of that same product:
−sj sk ≤ sjk ≤ sj sk .

Example
In an undergraduate statistics class, n = 30 females reported their heights
(inches), and also measured their left forearm length (cm), left foot length
(cm), and head circumference (cm).
The sample variance-covariance matrix is the following:
Notice that the matrix has four rows and four columns because there are four
variables being considered. Also notice that the matrix is symmetric.

Example
Here are a few examples of the information in the matrix:
The variance of the height variable is 8.74. Thus the standard deviation is√
8.74 = 2.956.
The variance of the left foot measurement is 1.908 (in the 3rd diagonal
element). Thus the standard deviation for this variable is
√
1.908 = 1.381.
The covariance between height and left arm is 3.022, found in the 1st row,
2nd column and also in the 2nd row, 1st column.
The covariance between left foot and left arm is 1.234, found in the 3rd row,
2nd column and also in the 2nd row, 3rd column.
All covariance values are positive so all pairwise associations are positive.
But, the magnitudes do not tell us about the strength of the associations. To
assess the strength of an association, we use correlation values.

Correlation
This suggests an alternative measure of association. The population
correlation is deﬁned to be equal to the population covariance divided by the
product of the population standard deviations:
ρjk =
σjk
σj σk
.
The population correlation may be estimated by substituting into the formula
the sample covariances and standard deviations:
rjk =
sij
si sj
=
n
i=1 Xij Xik − (
n
i=1 Xij )(
n
i=1 Xik )/n2
(
n
i=1 X2
ij − (
n
i=1 Xij /n)2)(
n
i=1 X2
ik − (
n
i=1 Xik /n)2)

Correlation
It is very important to note that the population as well as the sample
correlation must lie between -1 and 1.
−1 ≤ ρjk ≤ 1
−1 ≤ rjk ≤ 1
Therefore:
ρjk = 0 indicates, as you might expect, that the two variables are uncorrelated .
ρjk close to +1 will indicate a strong positive dependence
ρjk close to -1 indicates a strong negative dependence.
Sample correlation coeﬃcients also have similar interpretation.

Correlation matrix
For a collection of p variables, the correlation matrix is a p × p matrix that
displays the correlations between pairs of variables.
For instance, the value in the j-th row and k-th column gives the correlation
between variables xj and xk .
The correlation matrix is symmetric so that the value in the k-th row and
j-th column is also the correlation between variables xj and xk .
The diagonal elements of the correlation matrix are all identically equal to 1.

Correlation matrix
The sample correlation matrix is denoted as R.
R =





1 r12 . . . r1p
r21 1 . . . r2p
...
...
...
...
rp1 rp2 . . . 1






Example
The following covariance matrix show the pairwise covariances for the height,
left forearm, left foot and head circumference measurements of n = 30
female college students.
Here are two examples of calculating a correlation coeﬃcient:
The correlation between height and left forearm is 3.022√
8.74
√
2.402
= 0.66.
The correlation between head circumference and left foot is
0.118√
3.434
√
1.908
= 0.046.

Example
The complete sample correlation matrix for this example is the following:
Overall, we see moderately strong linear associations among the variables
height, left arm and left foot and quite weak (almost 0) associations between
head circumference and the other three variables.
In practice, use scatter plots of the variables to fully understand the
associations between variables. It is not a good idea to rely on correlations
without seeing the plots. Correlation values are aﬀected by outliers and
curvilinearity.

Overall measures of dispersion
Sometimes it is also useful to have an overall measure of dispersion in the
data.
In this measure it would be good to include all of the variables
simultaneously, rather than one at a time.
In the past we looked at the individual variables and their variances to
measure the individual variances.
Here we are going to look at measures of dispersion of all variables together,
particularly we are going to look at such measures that look at total variation.

The variance σ2
j measures the dispersion of an individual variable Xj .
The following two are used to measure the dispersion of all variables
together.
Total Variation
Generalized Variance
To understand total variation we ﬁrst must ﬁnd the trace of a square matrix.
A square matrix is a matrix that has an equal number of columns and rows.
Important examples of square matrices include the variance-covariance and
correlation matrices.

The trace of an n × n matrix A is
trace(A) =
n
i=1
aii .
For instance, in a 10 × 10 matrix, the trace is the sum of the diagonal
elements.
The total variation, therefore, of a random vector X is simply the trace of the
population variance-covariance matrix.
trace(Σ) = σ2
1 + σ2
2 + . . . σ2
p.

Thus, the total variation is equal to the sum of the population variances.
The total variation can be estimated by:
trace(S) = s2
1 + s2
2 + · · · + s2
p .
The total variation is of interest for principal components analysis and factor
analysis and we will look at these concepts later in this course.

Example
Let us use the data from the USDA women’s health survey again to illustrate
this.
We have taken the variances for each of the variables from software output
and have placed them in the table below.
Variable Variance
Calcium 157,829.4
Iron 35.8
Protein 934.9
Vitamin A 2,668,452.4
Vitamin C 5,416.3
Total 2,832,668.8

Example
The total variation for the nutrient intake data is determined by simply
adding up all of the variances for each of the individual variables.
The total variation equals 2,832,668.8. This is a very large number.

Example
Problem: The problem with total variation is that it does not take into
account correlations among the variables.
These plots show simulated data for pairs of variables with diﬀerent levels of
correlation. In each case, the variances for both variables are equal to 1, so
that the total variation is 2.
The corresponding variance-covariance matrix is:
Σ =
1 ρjk σj σk
ρjk σj σk 1
where ρjk is the correaltion coeﬃcient between the two variables.

Example
When the correlation r = 0, then we see a shotgun-blast pattern of points, widely
dispersed over the entire range of the plot.
Σ =
1 0
0 1
, being the Total Variation trace(Σ) = 2

Example
Increasing the correlation to r = 0.7, we see an oval-shaped pattern. Note that
the points are not as widely dispersed.
Σ =
1 0.7
0.7 1

Example
Increasing the correlation to r = 0.9, we see that the points fall along a 45 degree
line, and are even less dispersed.
Σ =
1 0.9
0.9 1

Generalized variance
To take into account the correlations among pairs of variables an alternative
measure of overall variance is suggested.
This measure takes a large value when the various variables show very little
correlation among themselves.
In contrast, this measure takes a small value if the variables show very strong
correlation among themselves, either positive or negative.
This particular measure of dispersion is the generalized variance. In order to
define the generalized variance we first define the determinant of the matrix.

We will start simple with a 2 × 2 matrix and then we will move on to more
general deﬁnitions for larger matrices.
Let us consider the determinant of a 2 × 2 matrix B as shown below.
Here we can see that it is the product of the two diagonal elements minus the
product of the oﬀ-diagonal elements.
|B| =
b11 b12
b21 b22
= b11b22 − b12b21.

Here is an example of a simple matrix that has the elements 5, 1, 2 and 4.
You will get the determinant 18.
The product of the diagonal 5 × 4 subtracting the elements of the
oﬀ-diagonal 1 × 2 yields an answer of 18:
5 1
2 4
= 5 × 4 − 1 × 2 = 20 − 2 = 18.

More generally the determinant of a general p × p matrix B is given by the
expression shown below:
|B| =
p
j=1
(−1)j+1
b1j |B1j |.
The expression involves the sum over all of the ﬁrst row of B. Note that
these elements are noted by b1j .
These are pre-multiplied by −1 raised to the (j + 1)-th power, so that
basically we are going to have alternating plus and minus signs in our sum.
The matrix B1j is obtained by deleting row i and column j from the matrix B.

For a 3 × 3 matrix the determinant is:
|B| =
b11 b12 b13
b21 b22 b23
b31 b32 b33
= b11b22b33 + b12b23b31 + b13b21b32
−b13b22b31 − b32b23b11 − b33b21b12.

By deﬁnition the generalized variance of a random vector X is equal to |Σ|,
the determinant of the variance-covariance matrix.
The generalized variance can be estimated by calculating |S|, the
determinant of the sample variance-covariance matrix.

Multivariate data analysis basics Matrix manipulations for sample statistics
Subsection 3
Matrix manipulations for sample statistics

The Mean and mean corrected data
Suppose we have a n × p data matrix X and a 1 × n vector of ones, 1 . The
mean vector is given by:
¯x =
1
n
1 X,
and the mean corrected data are given by
Xm = X − 1¯x
where Xm gives the matrix containing the mean-corrected data.

Sum of squares of cross products and covariance
The matrix of sum of squares and cross products (SSCP) SSCPm is given by
SSCPm = XmXm
and the sample variance-covariance matrix is given by
S =
1
n − 1
SSCPm =
1
n − 1
XmXm

Standardized data
If we deﬁne a p × p diagonal matrix D, which has variances of the variables
in the diagonal, then standardized data are given by
Xs = XmD−1/2
.
The SSCPs of standardized data is given by
SSCPs = XsXs

Correlation matrix
The correlation matrix is given by
R =
1
n − 1
SSCPs =
1
n − 1
XsXs
The generalized variance is given by |S|
All these computations can be easily performed in R using the matrix
operations...
The following table presents and hypothetical data set

Firm Original Mean-corrected Standardized
data data data
X1 X2 X1 X2 X1 X2
1 13.000 4.000 7.917 3.833 1.619 1.108
2 10.000 6.000 4.917 5.833 1.006 1.686
3 10.000 2.000 4.917 1.833 1.006 0.530
4 8.000 -2.000 2.917 -2.167 0.597 -0.626
5 7.000 4.000 1.917 3.833 0.392 1.108
6 6.000 -3.000 0.917 -3.167 0.187 -0.915
7 5.000 0.000 -0.083 -0.167 -0.017 -0.048
8 4.000 2.000 -1.083 1.833 -0.222 0.530
9 2.000 -1.000 -3.083 -1.167 -0.631 -0.337
10 0.000 -5.000 -5.083 -5.167 -1.040 -1.493
11 -1.000 -1.000 -6.083 -1.167 -1.244 -0.337
12 -3.000 -4.000 -8.083 -4.167 -1.653 -1.204
Mean 5.083 0.167 0.000 0.000 0.000 0.000
SS 262.917 131.667 11.000 11.000
Var 23.902 11.970 23.902 11.970 1.000 1.000

Example
The SSCP matrix of this data set is:
SSCP =
262.917 136.375
136.375 131.667
The S matrix is:
S =
1
n − 1
SSCP =
23.902 12.398
12.398 11.970

Example
The correlation matrix R of these data is:
R =
s2
1
s1s1
s12
s1s2
s21
s2s1
s2
2
s2s2
=
1 0.733
0.733 1

Multivariate data analysis basics A review of matrix deﬁnitions and operations
Subsection 4
A review of matrix deﬁnitions and operations

Just a review!
The following material reviews basic matrix definitions and operations.
You should have previous knowledge on matrix algebra.
This review does not relieve you from responsibility of being sufficiently
proficient in these matters!
Therefore be cautious and if you do not feel comfortable with these issues
please get back again to these matters!

Transpose of a matrix
The transpose of a matrix A is a matrix in which the rows of the transpose
are the columns of A (so the columns of the transpose are the rows of A).
The transpose of A is denoted as A .
Example: for the 4 × 2 matrix,
A =




3 6
2 5
3 7
1 8




the transpose is the 2 × 4 matrix
A =
3 2 3 1
6 5 7 8
Notice that the columns of A are the rows of A

Symmetric matrices
A square matrix A is symmetric if aij = aji , for all i and j.
That is throughout the matrix, the element in the i-th row and j-th column
equals the element in the j-th row and i-th column.
Said another way, a matrix is symmetric when, A = A.
Important examples of symmetric matrices in multivariate statistics include
the variance-covariance matrix and the correlation matrix.
These were deﬁned when we considered descriptive statistics.

Symmetric matrices
Example: the following is an example of a symmetric matrix.
Notice that the ﬁrst row and the ﬁrst column are identical, the second row
and the second column are the same, and that the third row is the same as
the third column. 

4 2 1
2 3 0
1 0 5



Adding two matrices
Two matrices may be added if and only if they have the same dimensions
(same number of rows and also same number of columns as each other. To
add two matrices, add corresponding elements (in terms of location).
Example:
2 4
6 8
+
1 3
5 7
=
2 + 1 4 + 3
6 + 5 8 + 7
=
3 7
11 15
Note: one matrix is subtracted from another in the same way that matrices
are added. To do the subtraction, subtract each element of the ﬁrst matrix
from the corresponding (in location) element of the ﬁrst matrix

Multiplying a matrix by a scalar
The word scalar is a synonym for a numerical constant.
In matrix terms, a scalar is a matrix with one row and one column.
To multiply a matrix by a scalar, multiply each element in the matrix by the
scalar.
Example: in this example, we multiply a matrix by the value 3.
3 ×
2 4
6 8
=
3 × 2 3 × 4
3 × 6 3 × 8
=
6 12
18 24

Multiplication of matrices
To be able to perform the matrix multiplication C = A × B, the number of
columns in the matrix A must equal the number of rows in the matrix B.
The element in the i-th row and j-th column of the answer matrix C, is the
cross product sum of the i-th row of the matrix A and the j-th column of the
matrix B.
This is done of all combinations of rows of A and columns of B.

Multiplication of matrices
Example: we’ll carry out the multiplication A×B =


1 2
2 3
4 3

×
4 2
1 3
.
The answer is
C =


1 × 4 + 2 × 1 1 × 2 + 2 × 3
2 × 4 + 3 × 1 2 × 2 + 3 × 3
4 × 4 + 3 × 1 4 × 2 + 3 × 3

 =


6 8
11 13
19 17

 .
Notice, for instance, that the element in the 1st row and 1st column of the
answer C is the cross product sum of the 1st row of A and the 1st row of B.
As another example, the 2nd row, 1st column element of C is the cross
product sum of the 2nd row of A and the 1st column of B.

The Identity Matrix
An identity matrix is a square matrix that has the value one in each main
diagonal position (from upper left to bottom right) and has the value 0 in all
other locations.
As an example of an identity matrix, the 3 × 3 identity matrix is
I =


1 0 0
0 1 0
0 0 1

 .
I is called the identity matrix because multiplication of any square matrix A
by the identity matrix yields the original matrix A as the answer. That is
AI = IA = A.

Matrix inverse
An inverse (in the traditional sense) can be found only for square matrices.
The inverse of a square matrix A is the matrix A−1
such that
A−1
A = AA−1
= I.
The calculation of an inverse for large matrices is a laborious process that
we’ll leave to the computer.
For 2 × 2 matrices, however, the formula is relatively simple.

Matrix inverse
For
A =
a11 a12
a21 a22
,
the inverse is
A−1
=
1
a11a22 − a12a21
a22 −a12
−a21 a11
.

Matrix inverse
Example: we’ll determine the inverse of
A =
10 6
8 5
.
The inverse is
A−1
=
1
10 × 5 − 6 × 8
5 −6
−8 10
=
1
2
5 −6
−8 10
=
2.5 −3
−4 5
.
You might want to check that A−1
A = I. (It does!)

Eigenvalues, eigenvectors and spectral decomposition
Let A be a p × p square matrix and I be the p × p identity matrix.
The scalars λ1, λ2, ..., λp satisfying the characteristic equation
|A − λI| = 0
are called the eigenvalues (or characteristic roots) of matrix A.

On the left-hand side, we have the matrix A minus λ times the Identity
matrix.
When we calculate the determinant of the resulting matrix, we end up with a
polynomial of order p.
Setting this polynomial equal to zero, and solving for λ we obtain the desired
eigenvalues.
In general, we will have p solutions and so there are p eigenvalues, not
necessarily all unique.

Example
Let
A =
1 0
1 3
.
Then
|A − λI| =
1 0
1 3
− λ
1 0
0 1
=
1 − λ 0
1 3 − λ
= (1 − λ)(3 − λ)
implies that there are two roots, λ1 = 1 and λ2 = 3. The eigenvalues of A
are 3 and 1 (the eigenvalues are in generally listed in descending order).

The corresponding eigenvectors e1, e2, ..., ep are obtained by solving the
expression below:
(A − λj I)ej = 0
Here, we have the diﬀerence between the matrix A minus the j-th eigenvalue
times the Identity matrix, this quantity is then multiplied by the j-th
eigenvector and set it all equal to zero.
This will obtain the eigenvector ej associated with eigenvalue λj .

Note: This does not have a generally have a unique solution.
So, to obtain a unique solution we will often require that ej transposed ej is
equal to 1.
Or, if you like, the sum of the square elements of ej is equal to 1, which
means the length (norm) of the eigenvector is the unity.
ej ej = 1
Also note that eigenvectors corresponding to diﬀerent eigenvalues are
orthogonal. In situations, where two (or more) eigenvalues are equal,
corresponding eigenvectors will still be orthogonal.

Example (continued)
The eigenvectors of matrix A associated to those eigenvalues (1 and 3) can
be determined by solving the following equations:
Ax = λ1x
1 0
1 3
x1
x2
= 1 ×
x1
x2
x1
x1 + 3x2
=
x1
x2

Example (continued)
There are many solutions! Therefore, setting x2 = 1 (arbitrarily) gives
x1 = −2, and hence
−2
1
is an eigenvector corresponding to the eigenvalue 1.
After transforming for unit norm:
e = (−2/
√
5 1/
√
5).

Example (continued)
Ax = λ2x
1 0
1 3
x1
x2
= 3 ×
x1
x2
x1
x1 + 3x2
=
3x1
3x2
Again, setting x2 = 1 and x1 = 0
0
1
is an eigenvector corresponding to the eigenvalue 3.

Any symmetric square matrix can be reconstructed from its eigenvalues and
eigenvectors.
Let A be a p × p symmetric matrix. Then A can be expressed in terms of its
p eigenvalue-eigenvector pairs (λi , ei ) as
A =
p
j=1
λj ej ej
Or, if you dispose the eigenvalues in a diagonal matrix Λ and the eigenvectors
in a matrix E, then
A = EΛE .

Example
For example, let
A =
2.2 0.4
0.4 2.8
Then
|A − λI| = λ2
− 5λ + 6.16 − 0.16 = (λ − 3)(λ − 2)
so A has eigenvalues λ1 = 3 and λ2 = 2.

Example
The corresponding eigenvectors are e1 = ( 1/
√
5 2/
√
5 ) and
e2 = ( −2/
√
5 1/
√
5 ), respectively.
Therefore
A =
2.2 0.4
0.4 2.8
= 3
1/
√
5
2/
√
5
1/
√
5 2/
√
5 +
+2
−2/
√
5
1/
√
5
−2/
√
5 1/
√
5
=
0.6 1.2
1.2 2.4
+
1.6 −0.8
−0.8 0.4

Example
Or,
A =
2.2 0.4
0.4 2.8
=
=
1/
√
5 −2/
√
5
2/
√
5 1/
√
5
3 0
0 2
1/
√
5 2/
√
5
−2/
√
5 1/
√
5
=
3/
√
5 −4/
√
5
6/
√
5 2/
√
5
1/
√
5 −2/
√
5
2/
√
5 1/
√
5
=
3/5 + 8/5 6/5 − 4/5
6/5 − 4/5 12/5 + 2/5/

Usually the matrix A is taken to be either the variance-covariance matrix Σ,
or the correlation matrix, or their estimates S and R, respectively.
Eigenvalues and eigenvectors are used for:
Computing prediction and conﬁdence ellipses
Principal Components Analysis (later in the course)
Factor Analysis (also later in this course)
Discriminant analysis (also later in this course)
Multidimensional scaling (also later in this course)
...

Eigenvalues, eigenvectors and spectral decomposition: a
special case
Consider the correlation matrix R as shown below:
R =
1 ρ
ρ 1
.
Then, using the deﬁnition of the eigenvalues, we must calculate the
determinant of R − λ times the Identity matrix.
|R − λI| =
1 ρ
ρ 1
− λ
1 0
0 1

special case
Carrying out the math we end up with the matrix with 1 − λ on the diagonal
and ρ on the oﬀ-diagonal. Then calculating this determinant we obtain 1 − λ
squared minus ρ2
:
1 − λ ρ
ρ 1 − λ
= (1 − λ)2
− ρ2
= λ2
− 2λ + 1 − ρ2
.
Setting this expression equal to zero we end up with the following...
λ2
− 2λ + 1 − ρ2
= 0.

special case
To solve for λ we use the general result that any solution to the second order
polynomial below:
ay2
+ by + c = 0
is given by the following expression:
y =
−b ±
√
b2 − 4ac
2a
.
Here, a = 1, b = −2 (the term that precedes λ and c is equal to 1 − ρ2
.
Substituting these terms in the equation above, we obtain that λ must be
equal to 1 plus or minus the correlation ρ.
λ =
2 ± 22 − 4(1 − ρ2)
2
= 1 ± 1 − (1 − ρ2) = 1 ± ρ

special case
We will take the following solutions:
λ1 = 1 + ρ
λ2 = 1 − ρ
Next, to obtain the corresponding eigenvectors, we must solve a system of
equations below:
(R − λ1I)e1 = 0
and
(R − λ2I)e2 = 0

special case
Or in other words, this is translated for this speciﬁc problem in the expression
below:
1 ρ
ρ 1
− λ1
1 0
0 1
e11
e12
=
0
0
This simpliﬁes as follows:
1 − λ1 ρ
ρ 1 − λ1
e11
e12
=
0
0

special case
Yielding a system of two equations with two unknowns:
(1 − λ1)e11 + ρe12 = 0
ρe11 + (1 − λ1)e12 = 0
Note that this does not have a unique solution. If (e11, e12) is one solution,
then a second solution can be obtained by multiplying the ﬁrst solution by
any non-zero constant c, i.e., (ce11, ce12). Therefore, we will require the
additional condition that the sum of the squared values of e11 and e12 are
equal to 1.
e2
11 + e2
12 = 1

special case
Consider the ﬁrst equation:
(1 − λ1)e11 + ρe12 = 0
Solving this equation for e12 and we obtain the following:
e12 = −
(1 − λ1)
ρ
e11

special case
Substituting this into e2
11 + e2
12 = 1 we get the following:
e2
11 +
(1 − λ1)2
ρ2
e2
11 = 1
Recall that λ = 1 ± ρ. In either case we end up ﬁnding that (1 − λ1)2
= ρ2
,
so that the expression above simpliﬁes to:
2e2
11 = 1
Or, in other words:
e11 =
1
√
2

special case
Using the expression for e12 which we obtained above,
e12 = −
1 − λ1
ρ
e11 =
1
√
2
we get
e11 =
1
√
2
and e12 =
1
√
2
for λ1 = 1 + ρ
Using the same reasoning we get
e21 =
1
√
2
and e22 = −
1
√
2
for λ2 = 1 − ρ
Therefore, the two eigenvectors are given by the two vectors as shown below:
1√
2
1√
2
for λ = 1 + ρ and
1√
2
− 1√
2
for λ = 1 − ρ

special case
Some properties of the eigenvalues of the variance-covariance matrix are to
be considered at this point. Suppose that λ1, λ2, ..., λp are the eigenvalues of
the variance-covariance matrix Σ.
By deﬁnition, the total variation is given by the sum of the variances. It turns
out that this is also equal to the sum of the eigenvalues of the
variance-covariance matrix. Thus, the total variation is:
p
j=1
s2
j = s2
1 + s2
2 + · · · + s2
p = λ1 + λ2 + · · · + λp =
p
j=1
λj

special case
Some properties of the eigenvalues of the variance-covariance matrix are to
be considered at this point. Suppose that λ1, λ2, ..., λp are the eigenvalues of
the variance-covariance matrix Σ.
The generalized variance is equal to the product of the eigenvalues:
|Σ| =
p
j=1
λj = λ1 × λ2 × · · · × λp

Quadratic forms
A quadratic form Q(x) in the p variables x1, x2, ...xp is Q(x) = x Ax, where
x = [x1, x2, ..., xp] and A is a p × p symmetric matrix.
Note that a quadratic form can be written as Q(x) =
p
i=1
p
j=1
aij xi xj . For
example,
Q(x) = x1 x2
1 1
1 1
x1
x2
= x2
1 + x1x2 + x2
2 .
Q(x) = x1 x2 x3


1 3 0
3 −1 −2
0 −2 2




x1
x2
x3


= x2
1 + 6x1x2 − x2
2 − 4x2x3 + 2x2
3 .

Lecture 1

More Related Content

What's hot (20)

Viewers also liked (15)

Similar to Lecture 1 (20)

Recently uploaded (20)

Lecture 1