Is the second norm of matrix $A$ equal to its maximum absolute eigenvalue if $A^TA$ is symmetric?

Question

The second norm of matrix $A$ whose elements are real numbers is the square root of the spectral radius of $A^TA$:

$$ ||A||_2= \sqrt{\rho (A^T*A)} \tag1$$

If matrix $A$ is symmetric, its second norm is equal to its maximum eigenvalue:

$$ ||A||_2= \sqrt{\rho (A*A)} = \sqrt{\rho (A^2)} = \sqrt{\rho ^2(A)}=\rho(A)=\max(|\lambda_i|) \tag2$$

Suppose now that $A$ is not symmetric, but the elements of any of its diagonals are the same. For example:

$$A = \begin{bmatrix} a & b & 0 & c & 0 \\ d & a & b & 0 & c \\ 0 & d & a & b & 0 \\ e & 0 & d & a & b \\ 0 & e & 0 & d & a \end{bmatrix} \tag 3$$

Is the second norm of matrix $A$ equal to its maximum absolute eigenvalue in this case? I am guessing that it is because $A^TA$ is a symmetric matrix.

Answer 1

This is a counterexample: $$A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ $\lVert A \rVert_2 = 1$, $\max(|\lambda_i|) = 0$.

This is a diagonalizable counterexample: $$A = \begin{pmatrix} 1 & 1 \\ 4 & 1 \end{pmatrix}$$ $\lVert A \rVert_2 = \frac12 (19 - 5 \sqrt{13})$, $\max(|\lambda_i|) = 3$.

It is true when $A$ is circulant, meaning that the diagonals loop around: $$A = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{pmatrix} \,.$$ Such real-valued matrices are normal, and hence diagonalizable with orthogonal matrices, so that their eigenvalues and singular values coincide (up to signs).