Maximizing symmetric matrices v.s. non-symmetric matrices Quick clarification on the following will be appreciated.
I know that for a real symmetric matrix $M$, the maximum of $x^TMx$ over all unit vectors $x$ gives the largest eigenvalue of $M$. Why is the "symmetry" condition necessary? What if my matrix is not symmetric? Isn't the maximum of $x^TMx=$ still largest eigenvalue of $M$?
Thanks.
 A: You can decompose any asymmetric matrix $M$ into its symmetric and antisymmetric parts, $M=M_S+M_A$, where
$$\begin{align}
M_S&=\frac12(M+M^T),\\
M_A&=\frac12(M-M^T).
\end{align}$$
Observe that $x^TM_Ax=0$ because $M_A=-M_A^T$. Then
$$x^TMx=x^T(M_S+M_A)x=x^TM_Sx+x^TM_Ax=x^TM_Sx.$$
Therefore, when dealing with something of the form $x^TMx$, we may as well assume $M$ to be symmetric; if it wasn't, we could replace it with its symmetric part $M_S$ and nothing would change.
A: The reason why the maximum of $x^T M x$ (I think you meant "over the set of $x$'s such that $\|x\|_2 = 1$", because if you don't add that condition your maximum is not defined) is the largest eigenvalue of $M$ is because any real square matrix can be factored as
$$
M = Q_1^T D Q_2
$$
where $Q_1$ and $Q_2$ are orthogonal and $D$ is a diagonal matrix containing the eigenvalues of $A$ on its diagonal. When $M$ is symmetric, we have $Q_1 = Q_2 = Q$, thus
$$
x^T M x = x^T Q^T D Q x = (Qx)^T D (Qx)
$$
and since $Q$ is orthogonal, not only is it invertible but it also preserves distances, so that
$$
\max_{\|x\| = 1} x^T M x = \max_{\| x \| = 1} (Qx)^T D (Qx) = \max_{\|x\| = 1} x^T D x.
$$
Now I believe you said that you knew the largest eigenvalue of $A$ was this maximum, so I assume you understand how to compute this last maximum to be the biggest eigenvalue of $M$ (it is a simple Lagrange multiplier argument that gives you that extremums are attained where $x$ is an eigenvector of $D$.)
Now if $M$ is not symmetric, none of these things apply because things started to look better when I assumed $Q_1 = Q_2 = Q$. But you can show that 
$$
\|M\|_2 = \sqrt{ \rho(M^T M) } 
$$
where $\rho( M^T M)$ is the spectral norm of $M$, i.e. $\rho(M^TM)$ is the largest eigenvalue (in absolute value) of $M^TM$. When $M$ is symmetric, $M^T = M$ and the spectral norm is just the square of the largest eigenvalue of $M$, so we know it still works. But in general the eigen values of $M^T M$ are not the square of the eigenvalues of $M$ because $M^T$ and $M$ do not have the same eigenvectors if $M$ is not symmetric, so the same arguments don't apply in general.
As an example, take 
$$
M = 
\begin{bmatrix}
0 & 2 \\
1 & 0 \\
\end{bmatrix}.
$$
You can readily see that its characteristic polynomial is just $\lambda^2 - 2$, thus that its eigenvalues are $\pm \sqrt 2$. But if you use the above formula, 
$$
M^T M = 
\begin{bmatrix}
4 & 0 \\
0 & 1 \\
\end{bmatrix}.
$$
Therefore the largest eigenvalue of $M^T M$ is $4$, hence $\|M \| = 2 > \sqrt 2, -\sqrt 2$. Perhaps that was more convincing.
Hope that helps,
