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There are many discussions about the singular values and eigenvalues, such as What is the difference between Singular Value and Eigenvalue?. I want to ask the particular one in title.

Usually, for a general square matrix, singular valure are not equal to eigenvalues. But singular values are alwyas nonnegative. My claim is

$A\in \mathbb{R}^{n\times n}$. $|\lambda_i(A)|\leq \sigma_\max(A)$, for all $i$.

The reason is the definition of the maximal singular value of $A$, which is $$\sigma_\max(A) = \|A\|_2 = \max_{\|x\|=1} \|Ax\|.$$ It reflects the maximal gain of $A$. And this $x$ does not have to be the eigenvector of $A$. However, the eigenvalue of $A$ is $$Av = \lambda v, \ \ \ \ \|v\|=1.$$ To get $\lambda_\max$, we have to have the extra constraint $\|Av\| = \|\lambda_\max v\|$. So $|\lambda(A)|\leq \sigma_\max (A)$.

Here $A$ is any real matrix not necessarily symmetric. I am not sure if I am correct.

sleeve chen
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1 Answers1

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I think is odd that the maximal singular value is definite in this way, but it coincides with the maximal eigenvector of $A^{*}A$.

Let $\sigma_1^2 > \dots > \sigma_n^2 $ be the eigenvalues of $A^{*}A$ and let $v_1, \dots v_n$ be the corresponding eigenvectors, which can also be a ortonormal basis. Take $u$ a vector and write as $u = \alpha_1 v_1 + \dots + \alpha_n v_n$ so

$\|Au\|^2 = \langle Au,Au\rangle = \langle A^{*}Au,u \rangle = \langle \sum_{i} \sigma_i^{2}\alpha_iv_i, \sum_{i}\alpha_iv_i \rangle = \sum_{i} \sigma_i^{2}|\alpha_iv_i|^2 \leq \sum_{i} \sigma_1^{2}|\alpha_iv_i|^2 = \sigma_1^2 |u|$

So $\|Au\| \leq \sigma_1$ for every $u$ unitary. So your claim is true.

bz0
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Non
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