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The 2nd largest eigenvalue of a symmetric $n \times n$ matrix is min of max Rayleigh quotient over the subspace of dimension $1$. From this definition, how do we get that it is equal to the max Rayleigh quotient over a $\vec{x}$ such that $\vec{x} \perp \vec{1}$?

Rodrigo de Azevedo
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sudhin s
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  • It sounds like you're dealing with a Laplacian (or Laplacian-like) matrix. In general, the eigenspace corresponding to the smallest eigenvalue will not contain $\vec{1}$. – Batman Mar 25 '15 at 19:27

2 Answers2

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The answer for 2nd smallest eigenvalues and courant-fisher actually answers this as well.

Suppose that $\lambda_1,\geq \cdots,\geq \lambda_n$ are the eigenvalues of $A$.

In order for the statement to be correct, you need to assume that $\bar 1$ is an eigenvector corresponding to $\lambda_1$.

Let $(\bar 1, x_2, \ldots, x_n)$ is an orthogonal basis of $\Bbb{R}^n$ using eigenvectors corresponding to the eigenvalues $\lambda_1,\ldots,\lambda_n$ . Then every $x\in\Bbb{R}^n$, $x\perp\bar 1$ can be represented as $\alpha_2x_2+\cdots+\alpha_nx_n$. Substituting in the Rayleigh quotient: $$ \frac{x^tAx}{x^tx} = \frac{(\alpha_2x_2+\cdots+\alpha_nx_n)^tA(\alpha_2x_2+\cdots+\alpha_nx_n)}{(\alpha_2x_2+\cdots+\alpha_nx_n)^t(\alpha_2x_2+\cdots+\alpha_nx_n)} = \frac{\lambda_2\alpha_2\|x_2\|^2+\cdots\lambda_n\alpha_n\|x_n\|^2}{\alpha_2\|x_2\|^2+\cdots\alpha_n\|x_n\|^2} \leq \frac{\lambda_2 (\alpha_2\|x_2\|^2+\cdots\alpha_n\|x_n\|^2)}{\alpha_2\|x_2\|^2+\cdots\alpha_n\|x_n\|^2}=\lambda_2 $$

So $\max_{x\perp \bar 1}\frac{x^tAx}{x^tx}\leq\lambda_2$. On the other hand, if $x\in\Bbb{R}^n$ is an eigenvector belonging to the eigenvalue $\lambda_2$ we have $\frac{x^tAx}{x^tx} = \lambda_2$, so actually $$ \max_{x\perp \bar 1}\frac{x^tAx}{x^tx}=\lambda_2 $$

Community
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Guy
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  • i think the answer has a typo. $\alpha_2$ and $\alpha_n$ in the last two fractions should be $\alpha_2^2$ and $\alpha_n^2$. In words, the Rayleigh quotient is a weighted average of the eigenvalues, but the weights are $\alpha^2$s which are positive, instead of $\alpha$s, which can be negative. –  Jan 18 '15 at 10:55
  • @purewater01: I assume you wanted your post to be a comment to Guy's answer. If you meant something else, or if the problem has been fixed, then post again. – Jyrki Lahtonen Jan 18 '15 at 12:29
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You got the Courant-Fischer minimax principle wrongly. Denote by $\lambda_2^\downarrow(A)$ the second largest eigenvalue of an $n\times n$ real symmetric matrix $A$. The correct statement should be: $$ \lambda_2^\downarrow(A) = \min_{\dim(V)=n-1}\ \max_{x\in V,\,x\neq0} \frac{\|x^TAx\|}{\|x\|^2}.\tag{1} $$ That is, $\lambda_2^\downarrow(A)$ is the minimum of the maximum Rayleigh quotient over all subspaces of codimension $1$ (rather than over subspces of dimension $1$). If $V$ is an $(n-1)$-dimensional subspace of $\mathbb{R}^n$, it has a unit normal vector $y$. So $x\in V$ if and only if $x\perp y$. Therefore $(1)$ can be rewritten as $$ \lambda_2^\downarrow(A) = \min_{y\neq0}\ \max_{x\perp y,\,x\neq0} \frac{\|x^TAx\|}{\|x\|^2}. $$

user1551
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