91影视

Given that, A is anmatrix and is a real eigenvalue of A. Then there exists a nonzero vector v such that $Av=位v$

Now we will show that for any orthogonal matrix $B,\left|\right|Bv\left|\right|=\left|\right|v\left|\right|.$

Thus we have $||Bv||=||v||$for any vector v

Nest, consider the singular value decomposition of $AasA=U危V^t,$ where U, V are orthogonal matrix. Then we have

$\begin{array}{rcl}\left|位\right|v& =& 位=Av=||\left(U危V^t\right)v||=||U\left(危V^{t}v\right)||\\ & =& ||危V^{t}v||\left[SinceUisanorthogonalmatrix\right]\\ & & \end{array}$

Now let, ${蟽}_1,{蟽}_2,...,{蟽}_n$are singular values of A, where ${蟽}_1$ is the largest and${蟽}_n$is the smallest singular value.

Then we have

role="math" localid="1660708558952" $危=\left[\begin{array}{cccc}{蟽}_1& 0& 鈰�& 0\\ & {蟽}_2& 鈰�& 0\\ 鈰�& M& 鈰�& 鈰�\\ 0& 0& 鈰�& {蟽}_n\end{array}\right]$

Since ${蟽}_1$is the largest and ${蟽}_n$ is the smallest singular value, hence we have

$||\left[\begin{array}{cccc}{蟽}_1& 0& 鈰�& 0\\ & {蟽}_2& 鈰�& 0\\ 鈰�& M& 鈰�& 鈰�\\ 0& 0& 鈰�& {蟽}_n\end{array}\right]V^{t}v||鈮�||危V^{t}v||鈮�||\left[\begin{array}{cccc}{蟽}_1& 0& 鈰�& 0\\ & {蟽}_2& 鈰�& 0\\ 鈰�& M& 鈰�& 鈰�\\ 0& 0& 鈰�& {蟽}_n\end{array}\right]V^{t}v||$

which implies

$\left|{蟽}_n\right|||V^{t}v||猢�||危V^{t}v||猢�\left|{蟽}_1\right|||V^{t}v||$

Using this we get that

$\left|{蟽}_n\right|||V^{t}v||猢�\left|位\right|v=||危V^{t}v||猢�\left|{蟽}_1\right|||V^{t}v||$

Now since V is an orthogonal matrix, therefore $V^t$ is also orthogonal. Hence we have

$\left|{蟽}_n\right|v猢�\left|位\right|v猢�\left|{蟽}_1\right|v$

As V is a nonzero vector, therefore dividing by $||v||$ the above equation implies that

$\left|{蟽}_n\right|猢�\left|位\right|猢�\left|{蟽}_1\right|$

$Since{蟽}_1,{蟽}_n猢�0,thereforewehave$

${蟽}_n猢�\left|位\right|猢�{蟽}_1$

$\left|{蟽}_n\right|||v||猢�\left|位\right|||v||猢�\left|{蟽}_1\right|||v||$