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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.

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As counterexample you can consider $M = \left( \begin{array}{c c} 0 & 1 \\ 0 & 0 \end{array}\right)$ –  leonbloy Dec 1 '11 at 0:31

2 Answers 2

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.

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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,

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More compactly, the singular values and the eigenvalues of a matrix have a nice relationship only for symmetric matrices. –  J. M. Dec 1 '11 at 0:36
    
I believed OP would not have been satisfied with such compactness... sometimes explaining stuff is nice =) –  Patrick Da Silva Dec 1 '11 at 0:47
    
That's why I left that severe summary as a comment. :) –  J. M. Dec 1 '11 at 0:49
    
You are indeed a very severe man. Haha :P Good you did that in a comment –  Patrick Da Silva Dec 1 '11 at 0:56
    
This is a very nice answer to a slightly different question. You are considering the maximum of $\lVert Mx\rVert^2 = x^TM^TMx$, while the question asks about the maximum of $x^TMx$ instead. –  Rahul Jan 27 '13 at 0:53

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