91影视

Let A be a symmetric matrix. Hence$A=A^t.$

Now let $位$be a eigenvalue of A. Then there exists a nonzero vector $v$such that$Av=位v$ . Thus we have

$\begin{array}{rcl}\left(A^{t}A\right)v& =& A^t\left(Av\right)\\ & =& A^t\left(位v\right)=位\left(A^{t}v\right)\\ & =& 位\left(Av\right)\left[SinceA=A^t\right]\\ & =& 位\left(位v\right)={位}^{2}v\\ & & \end{array}$

This shows that ${位}^2$is an eigenvalue of $A^{t}A$. This implies that $\left|位\right|$is a singular value of A.

Conversely, let$蟽$be a singular value of A. Then ${蟽}^2$a eigenvalue of$A^{t}A$. Then there exists a nonzero vector $u$such that$\left(A^{t}A\right)u={蟽}^{2}u$

Then we have$A^{2}u={蟽}^{2}usinceA=A^t.$

$Hencedet\left(A^2-{蟽}^2{I}_n\right)=0$This implies that

$0=det\left(A^2-{蟽}^2{I}_n\right)=det\left(A-蟽I_n\right)det\left(A+蟽I_n\right)$

This means either$det\left(A-蟽I_n\right)=0ordet\left(A+蟽I_n\right)=0$ which shows that$蟽or-蟽isaneigenvalueofA.$

Hence in a nutshell we have the following relationship between the singular value and eigenvalue of A:

If$位$ is an eigenvalue of A, then$\left|位\right|$ is a singular value of A and if $蟽$ is a singular value of A, then $蟽or-蟽$ is an eigenvalue of A.