Linear Algebra.pptx
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The document provides solutions for various linear algebra problems, detailing steps to derive results for angle calculations, vector dependencies, singular matrices, and convergence to steady-state vectors. It includes examples involving Markov matrices, intersection of planes, and permuting equations in matrix multiplication. Additionally, it covers problem-specific MATLAB code implementations and their implications in linear systems.
![� � � �
combination of left sides that gives zero. What combination of b1, b2, b3, b4, b5 must
be zero?
The 5 by 5 centered difference matrix is
⎛
0
⎞
⎜
⎜
⎜
⎝
C = 0 ⎟
⎠
0 1 0 0
−1 0 1 0 0⎟
−1 0 1 0 .
0 0 −1 0 1
0 0 0 −1 0
The five equations C x = bare
x2 = b1, −x1 + x3 = b2, −x2 + x4 = b3, −x3 + x5 = b4, −x4 = b5.
Observe that the sum of the first, third, and fifth equations is zero. Similarly,
b1 + b3 + b5 = 0.
Section 2.1. Problem 29: Start with the vector u0 = (1, 0). Multiply again and
again by the same “Markov matrix” A = [.8 .3; .2 .7]. The next three vectors are
u1, u2, u3:
� �
� � � �
1
u = 2 1 3 2
= u = Au = u = Au = .
.8 .3 1 .8
.2 .7 0 .2
What property do you notice for all four vectors u0, u1, u2, u3.
Computing, we get
.7
.3
.65
.35
2
u = 3
u = .
All four vectors have components that sum to one.
Section 2.1. Problem 30: Continue Problem 29 from u0 = (1, 0) to u7, and also
from v0 = (0, 1) to v7. What do you notice about u7 and v7? Here are two MATLAB
codes, with while and for. They plot u0 to u7 and v0 to v7.
The u’s and the v’s are approaching a steady state vector s. Guess that vector and
check that As = s. If you start with s, then you stay with s.
Solution
Solution](https://image.slidesharecdn.com/linearalgebra-230606104015-aa5f9b2a/85/Linear-Algebra-pptx-3-320.jpg)
![Solution
Here is the code entered into MATLAB.
u= [1; 0];
A= [.8 .3; .2 .7];
x= u;
k= [0 : 7];
while size (x, 2)<=7
u= A*u; x= [x, u];
end
plot (k,x)
v= [0;1];
A= [.8 .3; .2 .7];
x= v;
k= [0:7];
for j=1:7
v= A*v; x=[x, v];
end
plot (k, x)
In this graph, we see that the sequence u1, u2, . . . , u7 is approaching (.6, .4).](https://image.slidesharecdn.com/linearalgebra-230606104015-aa5f9b2a/85/Linear-Algebra-pptx-4-320.jpg)
Linear Algebra.pptx
- 1. Linear Algebra Problem Set 1 Solutions For any Assignment related queries, Call us at:- +1 (315) 557-6473 You can mail us at info@mathsassignmenthelp.com or reach us at: https://www.mathsassignmenthelp.com/ https://mathsassignmenthelp.com/
- 2. Section 1.2. Problem 23: The figure shows that cos(α) = v1/ǁvǁ and sin(α) = v2/ǁvǁ. Similarly cos(β) is and sin(β) is . The angle θ is β −α. Substitute into the trigonometry formula cos(α) cos(β) +sin(β) sin(α) for cos(β −α) to find cos(θ) = v · w/ǁvǁǁwǁ. First blank: w1/ǁwǁ. Second blank: w2/ǁwǁ. Substituting into the trigonome try formula yields cos(β −α) = (w1/ǁwǁ)(v1/ǁvǁ) + (w2/ǁwǁ)(v2/ǁvǁ) = v · w/ǁvǁǁwǁ. Section 1.2. Problem 28: Can three vectors in the xy plane have u · v < 0 and v · w < 0 and u · w < 0? √ 1 3 √ 1 3 Yes. For instance take u = (1, 0), v = ( 2 2 2 2 —, ), w = (− , − ). Notice 1 2 u · v = v · w = u · w = − . Section 1.3. Problem 4: Find a combination x1w1 + x2w2 + x3w3 that gives the zero vector: ⎡ 1 ⎤ ⎡ 4 ⎤ ⎡ 7 ⎤ 1 w = 2 ; w ⎣ ⎦ ⎣ ⎦ 2 3 ⎣ ⎦ = 5 ; w = 8 . 3 6 9 Those vectors are (independent)(dependent). T he three vectors lie in a . The matrix W with those columns is not invertible. Observe w1 −2w2 + w3 = 0. The vectors are dependent. They lie in a plane. Section 1.3. Problem 13: The very last words say that the 5 by 5 centered difference matrix is not invertible. Write down the 5 equations C x = b. Find a Solution Solution Solution
- 3. � � � � combination of left sides that gives zero. What combination of b1, b2, b3, b4, b5 must be zero? The 5 by 5 centered difference matrix is ⎛ 0 ⎞ ⎜ ⎜ ⎜ ⎝ C = 0 ⎟ ⎠ 0 1 0 0 −1 0 1 0 0⎟ −1 0 1 0 . 0 0 −1 0 1 0 0 0 −1 0 The five equations C x = bare x2 = b1, −x1 + x3 = b2, −x2 + x4 = b3, −x3 + x5 = b4, −x4 = b5. Observe that the sum of the first, third, and fifth equations is zero. Similarly, b1 + b3 + b5 = 0. Section 2.1. Problem 29: Start with the vector u0 = (1, 0). Multiply again and again by the same “Markov matrix” A = [.8 .3; .2 .7]. The next three vectors are u1, u2, u3: � � � � � � 1 u = 2 1 3 2 = u = Au = u = Au = . .8 .3 1 .8 .2 .7 0 .2 What property do you notice for all four vectors u0, u1, u2, u3. Computing, we get .7 .3 .65 .35 2 u = 3 u = . All four vectors have components that sum to one. Section 2.1. Problem 30: Continue Problem 29 from u0 = (1, 0) to u7, and also from v0 = (0, 1) to v7. What do you notice about u7 and v7? Here are two MATLAB codes, with while and for. They plot u0 to u7 and v0 to v7. The u’s and the v’s are approaching a steady state vector s. Guess that vector and check that As = s. If you start with s, then you stay with s. Solution Solution
- 4. Solution Here is the code entered into MATLAB. u= [1; 0]; A= [.8 .3; .2 .7]; x= u; k= [0 : 7]; while size (x, 2)<=7 u= A*u; x= [x, u]; end plot (k,x) v= [0;1]; A= [.8 .3; .2 .7]; x= v; k= [0:7]; for j=1:7 v= A*v; x=[x, v]; end plot (k, x) In this graph, we see that the sequence u1, u2, . . . , u7 is approaching (.6, .4).
- 5. In this graph, we see that the sequence v1, v2, . . . , v7 is approaching (.6, .4).
- 6. From the graphs, we guess that s = (.6, .4) is a steady state vector. We verify this with the computation � � � � � � .8 .3 .6 .2 .7 .4 .6 .4 As = = . Section 2.2. Problem 20: Three planes can fail to have an intersection point, even if no planes are parallel. The system is singular if row 3 of A is a of the first two rows. Find a third equation that can’t be solved together with x + y + z = 0 and x −2y −z = 1. (a) Zero. (b) Zero. (c) There are many possible answers. For instance, the matrix for which every row is (1 2 3 · · · 100). (d) The row picture is 100 copies of the hyperplane in 100-space defined by the equation x1 + 2x2 + 3x3 + ··· + 100x100 = 0. The column picture is the 100 vectors proportional to (1 1 1 ··· 1) of lengths 10, 20, . . . , 1000. Solution The system is singular if row 3 of A is a linear combination of the first two rows. There are many possible choices of a third equation that cannot be solved together with the ones given. An example is 2x + 5y + 4z = 1. Note that the left hand side of the third equation is the three times the left hand side of thte first minus the left hand side of the second. However, the right hand side does not satisfy this relation. Section 2.2. Problem 32: Start with 100 equations Ax = 0 for 100 unknowns x = (x1, . . . , x100). Suppose elimination reduces the 100th equation to 0 = 0, so the system is “singular”. (a)Elimination takes linear combinations of the rows. So this singular system has the singular property: Some linear combination of the 100 rows is . (b)Singular systems Ax = 0 have infinitely many solutions. This means that some linear combination of the 100 columns is . (c) Invent a 100 by 100 singular matrix with no zero entries. (d) For your matrix, describe in words the row picture and the column picture of Ax = 0. Not necessary to draw 100-dimensional space. Solution
- 7. Σ 1j j 11 1 1n n 21 subtracts row 1 from row component of Ax is a x = a x + ···+ a x . If E Section 2.3. Problem 22: The entries of A and x are aij and xj . So the first 2, write a formula for (a) the third component of Ax (b) the (2, 1) entry of E 21A (c) the (2, 1) entry of E 21(E 21A ) (d) the first component of E21Ax. Solution (a) Σ a3jxj . (b) a21 −a11. (c) a21 −2a11. (d) Σ a1jxj . Section 2.3. Problem 29: Find the triangular matrix E that reduces “Pascal’s matrix ” to a smaller Pascal: ⎡ 0 ⎤ ⎡ 1 0 ⎤ E ⎢ ⎢ ⎣ = ⎥ ⎢ ⎦ ⎣ 0⎥ ⎦ 1 0 0 1 1 0 0 0 1 2 1 0 1 3 3 1 0 0 1 0 1 1 1 2 0 0 0 1 . Which matrix M (multiplying several E’s) reduces Pascal all the way to I? Solution ⎡ 0 ⎤ ⎢ ⎢ ⎣ ⎥ ⎦ 1 0 0 −1 1 0 0 0 −1 1 0 0 0 −1 1 E = . One can eliminate the second column with the matrix 1 0 0 0 ⎡ ⎢0 1 0 0 ⎤ ⎥ ⎣0 0 −1 0 1 −1 0 ⎦ 1 and the third column with the matrix ⎤ ⎡ 0 0 1 ⎢ ⎣ 0 0 −1 0 ⎥ ⎦ 1 1 0 0 0 0 1 0 0 .
- 8. Multiplying these together, we get ⎥ ⎦ ⎡ 0 ⎤ ⎡ 1 0 ⎤ ⎡ 1 0 ⎤ ⎡ 1 0 ⎤ M = ⎢ ⎢ ⎣ ⎥ ⎢ ⎦ ⎣ ⎥ ⎢ ⎦ ⎣ ⎥ ⎢ ⎦ ⎣ = 0 0 −1 1 0 0 1 −2 1 0 . 0 −1 1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 −1 1 0 0 0 0 1 0 0 −1 1 0 −1 0 0 −1 1 0 1 0 0 −1 1 −1 3 −3 1 Section 2.4. Problem 32: Suppose you solve Ax = bfor three special right sides b: ⎡ 1 ⎤ ⎡ 0 ⎤ ⎡ 0 ⎤ 3 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 1 2 Ax = 0 ; Ax = 1 ; Ax = 0 . 0 0 1 If the three solutions x1, x2, x3 are the columns of a matrix X , what is A times X ? Solution The matrix A X has columns Ax1, Ax2, and Ax3. Therefore, A X = I. Section 2.4. Problem 36: Suppose A is m by n, B is n by p, and C is p by q. Then the multiplication count for (AB)C is mnp + mpq. The multiplication count from A times B C with mnq + npq separate multiplications. (a) If A is 2 by 4, B is 4 by 7, and C is 7 by 10, do you prefer (AB)C or A(BC)? (b) With N -component vectors, would you choose (uT v)wT or uT (vwT )? (c) Divide by mnpq to show that (AB)C is faster when n−1 + q−1 < m−1 + p−1. Solution (a)Note that (AB)C has 2 · 4 · 7 + 2 · 7 · 10 = 196 multiplications and A ( B C ) has 2 · 4 · 10 + 4 · 7 · 10 = 360. Hence, we prefer (A B )C . (b)We prefer (uT v)wT ; it requires 2N multiplications. On the other hand, the multiplication count for uT (vwT ) is 2N 2. (c)Note (AB)C is faster when mnp + mpq < mnq + npq. Dividing by mnpq, we get q−1 + n−1 < p−1 + m−1.