2-Norm of PD Matrix

Quote:

Originally Posted by leoemil

Hi all,

In most books i've seen, it says that:

$<br /> ||A||_2 = \textnormal{ Root of largest eigenvalue of } A^T A<br />$

If $A$ is symmetric and positive definite, this just becomes
equal to the largest eigenvalue. However, my teacher says that this even hold for positive semidefinite. Can anybody tell me if they agree/disagree, and also why?

Teacher is correct, the result still holds in the positive semidefinite case.

One way to see it, thinking of A as a linear transformation rather than a matrix, is to let L be the kernel of A (equivalently, the eigenspace corresponding to the eigenvalue 0). Let B denote the restriction of A to the orthogonal complement $L^\perp$ . Then B is positive definite on $L^\perp$ , and $B^{\textsc t}B$ is the restriction of $A^{\textsc t}A$ to $L^\perp$ . So the nonzero eigenvalues of $A^{\textsc t}A$ are exactly the eigenvalues of $B^{\textsc t}B$ . The square root of the largest one is equal to the largest eigenvalue of B, which is equal to the largest eigenvalue of A.