91影视

Given that, Amatrix and V is a vector $R^m$

Now we will show that for any orthogonal matrix role="math" localid="1660705370616" $B,||\mathrm{Bv}||=||\mathrm{V}||$

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

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

$Nowlet,{蟽}_1,{蟽}_2,L,{蟽}_{m}aresingularvaluesofA,where{蟽}_{1}isthelargestand{蟽}_m$is the smallest singular value.

Then we have

$危=\left[\begin{array}{cccc}{蟽}_1& 0& 鈰�& 0\\ 0& {蟽}_2& 鈰�& 0\\ 鈰�& 鈰�& 鈰�& 鈰�\\ 0& & 鈰�& {蟽}_m\\ 0& 0& 鈰�& 0\\ 鈰�& 鈰�& 鈰�& 鈰�\\ 0& 0& 鈰�& 0\end{array}\right]$

Since is the largest and is the smallest singular value, hence we have

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

which implies

$\left|{蟽}_m\right|||V^{t}v||鈮�||危V^{t}v||鈮�\left|{蟽}_1\right|||V^{t}v||$

Using this we get that

$\left|{蟽}_m\right|||V^{t}v||鈮�||Av||=||危V^{t}v||鈮�\left|{蟽}_1\right|||V^{t}v||$

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

$\left|{蟽}_m\right|||v||鈮�||Av||鈮�\left|{蟽}_1\right|||v||$

Since${蟽}_1,{蟽}_m猢�0,$ therefore we have

${蟽}_m||v||鈮�||Av||鈮�{蟽}_1||v||$