# Is $\sqrt{\lambda_{\max}(A^2)} = \lambda_{\max}(A)$?

I'm trying to prove a statement on the condition number of a matrix in the 2-norm for a symmetric positive definite matrix $A$. I nearly have the proof completed, if $\sqrt{\lambda_{\max}(A^2)} = \lambda_{\max}(A)$ then the proof is finished. But I'm not sure if this is the case? Is $\sqrt{\lambda_{\max}(A^2)} = \lambda_{\max}(A)$ for a symmetric positive definite matrix?

• Hint: What is the relationship between the eigenvalues of $A$ and those of $A^2$? – ekkilop Aug 28 '16 at 14:49
• The eigenvalues of $A^2$ are the eigenvalues of $A$ squared. – ManUtdBloke Aug 28 '16 at 15:17
• That's right. So if $\mu$ is an eigenvalue of $A$ then $\mu^2$ is an eigenvalue of $A^2$. Further, $\mu$ is real and positive so we can safely say that $\sqrt{\mu^2}$ is an eigenvalue of $A$. Can you proceed from here? – ekkilop Aug 28 '16 at 15:37
• Yes thats fine thanks! – ManUtdBloke Aug 29 '16 at 8:00

For a symmetric positive definite matrix $A$, singular values and eigenvalues are identical.
So, $\sqrt{\lambda_{max}(A^2)}=\lambda_{max}(A)$.
Let $A$ be any square matrix and $(\lambda,v)$ such that $Av = \lambda v$. Then, $A^2v = \lambda Av = \lambda^2 v$ and so $\lambda_{\max}(A^2)= \lambda_{\max}(A)^2$.
So, if $\lambda_{\max}(A)\geq 0$ we are done, i.e. $\sqrt{\lambda_{\max}(A^2)}= \lambda_{\max}(A)$.
This is clearly the case when $A$ is positive semi-definite.
It is also the case if $A$ has nonnegative entries (by the Perron-Frobenius theorem).