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.
Best Answer
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.