Does the spectral norm of a square matrix equal its largest eigenvalue in absolute value?

I have one simple question.

Given the spectral norm $\left \| \cdot \right \| _2$ of a matrix $A$, which is equal to the square root of the largest eigenvalue of $A^{^*}A$

$$\left \| A \right \| _2=\sqrt{\lambda_{\text{max}}(A^{^*}A)}=\sigma_{\text{max}}(A)$$

for a square matrix $A$, is $\left \| A \right \| _2$ equal to the largest eigenvalue of $A$ in absolute value?

I know it is true for a symmetric matrix but I don't know for a random square one.

For any square $A$, $\rho(A)\leq\|A\|_2$, where $\rho(A)$ is the spectral radius of $A$, with the equality (but not necessarily) if $A$ is normal. Besides the general inequality, $\rho(A)$ and $\|A\|_2$ can be completely unrelated. Consider, e.g., $$A_\alpha:=\pmatrix{0&\alpha\\0&0}$$ with $\rho(A_\alpha)=0$ but $\|A_\alpha\|_2=|\alpha|$. All the eigenvalues are zero but the 2-norm can be an arbitrary non-negative number (depending on $\alpha$).

• What is $\rho(A)$? – gen Mar 6 '18 at 16:00
• @gen Spectral radius of $A$. – Algebraic Pavel Mar 7 '18 at 10:15

For a symmetric matrix you want to maximise $||Ax||$ for $||x||=1$, which is the same as maximising $||Ax||^2 = xA^TAx$

Since we know that if $A$ is symmetric, it is diagnolaisable, so we write $$A = Q^T\Lambda Q$$, with $Q^{-1}=Q^T$then $$xA^TAx= (Qx)^T\Lambda^2(Qx)$$ where $Q$ is orthonormal so $||Qx||=1$

It is straightforward to see what you have written must follow from the argument given above in the symmetric case, but this cannot possibly be expected to hold in general. You could have probably realised this fact does not hold by finding a counter-example. (and use Wolfram alpha or matlab to speed up calculation if necessary)

• The only part of your answer that addresses the question is your brief statement this cannot possibly be expected to hold in general, without any semblance of a justification. This answer is not useful. – Omnomnomnom Jul 8 '15 at 14:50