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 $||Qx||=1$