2-norm of the orthogonal projection

Question

So far, I've deduced that if the rank of A is n, then all the columns of A are linearly independent since A has n columns. As a result, m must be greater than or equal to n.

In the case that m = n, then A is a square matrix and we can just use matrix operations to distribute the inverse operation over the parenthesis; then we can apply the associative property and compute A*A-1 and AT-1 * AT first, eventually giving that the expression inside the norm is just the identity matrix, where the 2-norm of the identity matrix is obviously 1.

I'm having trouble for the case where m > n. I've identified that the expression inside the norm is actually the matrix of orthogonal projection, but I'm not sure if that will help me with the proof. Help is greatly appreciated, thanks.

This problem is from the text of "Matrix Computations" 3rd edition by Golub after covering a section on orthogonality and SVDs for some background.

Answer 1

Hint: You’re almost there. $\|M\|_2$ is the largest singular value of $M$, i.e., the square root of the largest eigenvalue of $M^TM$. You’ve figured out that you’re working with a projection. What do you know about the eigenvalues of a projection?

Answer 2

A simple approach: The matrices $XY$ and $YX$ have the same nonzero eigenvalues. Since $A(A^TA)^{-1}A^T$ is symmetric, $\|A(A^TA)^{-1}A^T\|_2$ is equal to the largest eigenvalue of the argument. Now set $X=A(A^TA)^{-1}$ and $Y=A^T$ and note that $YX=I$.