The eigenvalue of $A^TA$

Question

If $\lambda$ is the eigenvalue of matrix $A$,what is the eigenvalue of $A^TA$?I have no clue about it. Can anyone help with that?

real matrix...complex matrix...? Not that it matters, but it makes the question more neatly posted. — DonAntonio, Apr 19 '14 at 16:51
Do you mean $\lambda$ is one eigenvalue or the only eigenvalue? In general, the eigenvalues of $A^\top A$ have little to do with the eigenvalues of $A$. See singular values. — Ted Shifrin, Apr 19 '14 at 16:59
@TedShifrin what about $\lambda_{\mbox{max}}A^TA$ and $\lambda_{\mbox{max}}(A)$ — 89085731, Apr 19 '14 at 17:27

score 2 · Accepted Answer · answered Apr 19 '14 at 18:18

2

The good exercise for you is to prove, in general, that $$\lambda_{\text{max}}(A)\le \max_{\|x\|=1} \|Ax\|=\sqrt{\lambda_{\text{max}}(A^\top A)}.$$

answered Apr 19 '14 at 18:18

Ted Shifrin

115,160

Is $\frac{(\frac{1}{2}\lambda_{\mbox{min}}(A+A^T))^2}{\lambda_{\mbox(max)}A^TA}<1$ if A is positive definite – 89085731 Apr 19 '14 at 19:26

score 0 · Answer 2 · answered Apr 19 '14 at 17:44

Generally, you won't be able to say much about them.

However, if $A$ is for instance real symmetric, it is diagonalizable with real eigenvalues, meaning there is an orthogonal matrix B (that is with $B^{-1}=B^T$) and a diagonal matrix D such that: $$A = B D B^T$$ $$A^TA = (B D B^T)^T B D B^T = B D^T B^T B D B^T = BD^2B^T$$ In other words, the eigenvalues of $A^TA$ are the squares of the eigenvalues of $A$.

The eigenvalue of $A^TA$

2 Answers2