Let A ∈Mn. Use the singular value decomposition to prove...

Key Concepts

Singular Value Decomposition

Singular Value Decomposition (SVD) is a factorization of a matrix into three components: A = U?V*, where U and V are unitary matrices and ? is a diagonal matrix with nonnegative real numbers arranged in descending order. This decomposition is a powerful tool in linear algebra because it provides insight into the intrinsic geometric properties of the matrix, such as its range, null space, and conditioning. It is especially useful in proving various properties of matrices, including norm-related inequalities and identities.

Spectral Norm

The spectral norm of a matrix, often defined as the largest singular value, is a measure of the maximum stretching factor of the matrix when acting on a vector. This norm is widely used because it captures the worst-case amplification of the input and has a natural interpretation in terms of eigenvalues when the matrix is symmetric or normal. In proofs involving the spectral norm, one leverages the diagonal form obtained by the SVD to relate the norm to the entries of ?, the diagonal matrix of singular values.

Unitary Invariance

Unitary invariance is a property of certain matrix norms, including the spectral norm, which states that the norm of a matrix remains unchanged under multiplication by unitary matrices. Since unitary matrices preserve lengths and angles, applying U or V* in the singular value decomposition does not affect the norm. This invariance simplifies many proofs by reducing the analysis to the diagonal matrix of singular values, allowing one to work directly with nonnegative numbers rather than dealing with the full complexity of the original matrix.

Transcript

00:02 In this question, we're asked to show that u orthogonally diagonalizes aa transpose, given that u sigma v transpose is a singular value decomposition of a.

00:21 So recall that u orthogonally diagonalizes a matrix b when u transpose bu is a diagonal matrix, and u is an orthogonal matrix.

00:40 So we already know that u is an orthogonal matrix, because by definition of singular value decomposition, u and v are both orthogonal.

00:52 So all we have to show is that u transpose aa transpose u is a diagonal matrix.

01:12 So first of all, let's substitute the singular value decomposition in for a.

01:25 So that would look like this.

01:30 Now, because matrix multiplication is associative, we don't need these brackets.

01:38 I was just using them to clarify the substitution, but we have it here.

01:51 Now, when a is transposed, when this product of three matrices is transposed, that's the same thing as v transpose, or it's the last matrix transposed times this matrix transposed times this matrix transposed.

02:11 In other words, when you transpose a product of matrices, the result is the backwards order of the matrices transposed.

02:24 Each transposed, i should say.

02:31 And we also know that v transpose transpose is just v itself.

02:43 Okay, so how do we simplify this product of eight matrices? well, recall that u and v are orthogonal matrices, and orthogonal matrices have a special property that their transposes are their inverses.

03:06 So u transpose u is just u inverse u, or the identity matrix.

03:14 And the same thing applies for v, and u transpose u also occurs here.

03:27 Now, since multiplying by the identity, multiplying a matrix by the identity does not change that matrix, the result is just sigma sigma transpose.

03:44 So now what do we do? how do we show that this is a diagonal? well, we know that in a singular value decomposition, sigma is a matrix that has the singular values of a as its terms along its main diagonal.

04:08 That is, the terms starting in the top left and then going diagonally down.

04:21 So those are the only non -zero terms, or potentially non -zero terms, in sigma, and the rest are just zeros.

04:37 Now, you might be tempted to think that sigma is a diagonal matrix, and so therefore sigma transpose is just the same thing as sigma, and so their product is also diagonal.

04:52 However, recall that in a singular value decomposition, sigma is the same size as a, and a is not necessarily square.

05:06 So sigma is not necessarily square.

05:14 However, even with a non -square matrix, sigma has a useful form in that the only non -zero terms are along the main diagonal.

05:31 So how can we represent sigma sigma transpose and show that it is diagonal? well, remember, by definition of matrix multiplication, that multiplying two matrices together involves taking dot products of the rows of the first matrix with the columns of the second...

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