upperbound on eigenvalue of $A^TA$, where A is a real square matrix

Question

I have a real square $n\times n$ matrix $A$, and I would like to give an upperbound on $|\lambda|_{\max}$ on $A^TA$. I wonder is there a good upperbound on $|\lambda|_{\max}$ of $A^TA$ using only eigenvalues of $A$?

Answer 1

An upper bound on $|\lambda_{\max}(A^TA)|$ can be provided in terms of the singular values of $A$, call them $\sigma_1, ..., \sigma_n$ (not eigenvalues as requested, but doable with singular values).

Because $A^TA$ is symmetric positive semidefinite, all its eigenvalues are nonnegative, therefore, $\lambda_{\max}(A^TA) \le trace(A^TA)$.

Applying Von_Neumann's_trace_inequality , and because the singular values of $A$ and $A^T$ are the same, $trace(A^TA) \le \Sigma_{i=1}^n\sigma_i^2$, which actually holds with equality in this case.

So $\lambda_{\max}(A^TA) \le \Sigma_{i=1}^n\sigma_i^2$

I make no claim that this is the best possible bound.

Answer 2

No such upper bound exists (in terms of the eigenvalues of $A$)

Note, for instance, that the matrix $$ A = \pmatrix{0&0\\t&0}, \quad t > 0 $$ has only $0$ as an eigenvalue. However, $$ A^TA = \pmatrix{t^2&0\\0&0} $$ has $t^2$ as its maximal eigenvalue, which can be made arbitrarily large.