LinearAlgebra_2016updatedFromwiki.ppt

Editor's Notes

• #28: We’ll see implications of this feature shortly… (motivation for “rank 1”/1-dimensional matrix) and also underlying components of SVD
• #46: Such rank 1 matrices can be written as an outer product, e.g. in this case as (1 2)’ * (1 -2)
• #58: Note: this way of matrix multiplying, i.e. breaking into outer products is key to Singular Value Decomposition (SVD) which I’ll discuss at end of lecture; note that each outer product is rank 1 since each column (or row) is a multiple of every other column (or row)
• #62: For scalars, there is no inverse if the scalar’s value is 0. For matrices, the corresponding quantity that has to be nonzero for the inverse to exist is called the determinant, which we’ll show next corresponds geometrically (up to a sign) to an area (or, in higher dimensions, volume).
• #65: Note, potentially a bit confusing: the vector (1,0) got mapped into (0,2) and (0,1) got mapped into (3,1): this relates to fact that determinant is negative (i.e. in some sense, orientation of parallelogram flipped)
• #69: If above determinant is zero, then inverse doesn’t exist and can’t conclude that x=0 by simply multiplying both sides of the equation by the inverse.
• #73: c
• #95: Step 4: (solve for c and) transform back to original coordinates.
• #100: Notes: 1) sometimes left eigenvectors are called adjoint eigenvectors 2) later we’ll show that in some cases the right and left eigenvectors are the same (Normal matrices). It is easily shown that they obey the equation: e_left M = lambda * e_left, i.e. they are analogous to the usual (right) eigenvectors except that they multiply M from the *left* (and they must be written as row vectors to do this). Proof (can do on board): M = E*Lambda*E^(-1) E^(-1) M = Lambda*E^(-1) 3) the left eigenvectors of M transpose are the same as the right eigenvectors of M Proof: M*E = E * Lambda E’ * M’ = Lambda * E’ (i.e. rows of E’ are the left eigenvectors of M’, and are identical to columns of E) 4) Even though eigenvectors are not necessarily orthogonal, left & right eigenvectors corresponding to distinct eigenvalues are orthogonal Proof: follows directly from E^(-1) * E = Identity .
• #103: Key feature: if decomposing a general vector into components in an orthonormal coordinate system, then do this through simple orthogonal projection along the orthonormal axes of the coordinate system. This can be done through a dot product (as noted near the beginning of this tutorial…). Since transforming into the eigenvector coordinate system is accomplished by a dot product with the rows of E^(-1), we see that this operation is simply a dot product with the eigenvectors when we have E^(-1) = E transpose. Note: reflections could technically be through any axis, because one can then rotate more to make up for which axis one reflected through. Note that performing reflection changes the sign of the determinant (e.g. from =1 to -1). [Aside/not relevant here but for cultural interest: Unitary or orthonormal matrices that are restricted to have positive determinant (= +1) are called “special unitary” or “special orthonormal” matrices.] Also note that the absolute value of the determinant equaling 1 for unitary/orthonormal matrices reflects the earlier comment that determinants give the area of the transformation of a unit square, and a rotation matrix is area-preserving.
• #106: Proof that eigvals real: Suppose Av = (lambda)v, where lambda potentially complex and A = A’ (‘ denotes transpose) and A real-valued Consider v’*Av = v’*(Av) = lambda v’*v = lambda |v|^2, where * denotes complex conjugate (NOT multiplication) but also v’*Av = (v’*A)v = (A’v*)’v = (Av*)’v = lambda* |v|^2 Thus, lambda = lambda*  lambda real Proof that eigvectors orthogonal: now consider two eigenvectors v and w with distinct eigenvalues Consider v’Aw two ways above to see that v’w must equal zero Note: covariance matrices have further restriction that eigenvalues are positive, this results from the fact that covariance matrices are of the form XX’
• #110: Note 1: Each neural mode has an associated temporal mode, and that M can be written as a linear sum of outer products. Note 2: Alternatively, the time course of each spatial mode (i.e. the strength of the spatial mode at each point in time) is given by the associated temporal eigenvector times the singular value. Sometimes people refer to each temporal eigenvector scaled by its singular value as the temporal ‘component’ to distinguish it from the spatial ‘mode’. Note 3: remember that outer products are 1d (i.e. rank 1) Note 4: if you are centering your data, you can subtract off the mean of *either* your rows *or* your columns, but you cannot do both (i.e. first subtract off mean of rows, then subtract off means of resulting columns) or you will lose a dimension. More generally, if you think of your individual data points as occupying either the rows or columns of M, then you want to do the operation that corresponds to subtracting the means of the data points, i.e. “centering the data”; if you do the other mean subtraction, you will see that your data loses a dimension. Try this with a 2x3 matrix, plotting it with the interpretation of its being 3 two-dimensional data points to see what happens if you subtract off the “wrong mean”.