# 2-Norm of PD Matrix

• Sep 19th 2010, 02:05 PM
leoemil
2-Norm of PD Matrix
Hi all,

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

\$\displaystyle
||A||_2 = \textnormal{ Root of largest eigenvalue of } A^T A
\$

If \$\displaystyle 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?
• Sep 20th 2010, 12:46 PM
Opalg
Quote:

Originally Posted by leoemil
Hi all,

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

\$\displaystyle
||A||_2 = \textnormal{ Root of largest eigenvalue of } A^T A
\$

If \$\displaystyle 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 \$\displaystyle L^\perp\$. Then B is positive definite on \$\displaystyle L^\perp\$, and \$\displaystyle B^{\textsc t}B\$ is the restriction of \$\displaystyle A^{\textsc t}A\$ to \$\displaystyle L^\perp\$. So the nonzero eigenvalues of \$\displaystyle A^{\textsc t}A\$ are exactly the eigenvalues of \$\displaystyle 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.