91影视

Given that,$Ais2脳2matrixanduisaunitvector{R}^2.$

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

$Thus,wehave\left|\right|Bu\left|\right|=\left|\right|u\left|\right|foranyvectoru.$

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

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

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

$危=\left[\begin{array}{cc}{蟽}_1& 0\\ 0& {蟽}_2\end{array}\right]$

Since ${蟽}_1$is the largest and ${蟽}_2$is the smallest singular value, hence we have

$||\left[\begin{array}{cc}{蟽}_2& 0\\ 0& {蟽}_2\end{array}\right]V^{t}u||鈮�||危V^{t}u||鈮�||\left[\begin{array}{cc}{蟽}_1& 0\\ 0& {蟽}_1\end{array}\right]V^{t}u||$

which implies

$\left|{蟽}_2\right|||V^{t}u||鈮�||危V^{t}u||鈮�\left|{蟽}_1\right|||V^{t}u||$

Using this we get that

$\left|{蟽}_2||V^{t}u||\right|鈮�||Au||=||危V^{t}u||鈮�\left|{蟽}_1\right|||V^{t}u||$

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

$\left|{蟽}_2\right|||u||鈮�||Au||鈮�\left|{蟽}_1\right|鈮�||u||$

which implies that

$\left|{蟽}_2\right|鈮�||Au||鈮�\left|{蟽}_1\right|$

since $u$is a unit vector. As ${蟽}_1,{蟽}_2猢�0,$therefore we have

${蟽}_2猢�||A\mathbf{u}||猢�{蟽}_1$

Now by using Theorem 8.3 .2 we get that, the image of the unit circle under A is an ellipse whose semi-major and semi-minor axes have lengths and respectively. Also, as$u$is a unit vector therefore, $u$lies on the unit circle. Hence, the vector lies on the ellipse whose semi-major and semi-minor axes have lengths respectively. Therefore, clearly $||Au||$is greater than or equal to ${蟽}_2$and less than or equal to ${蟽}_1$.

The following image will justify this geometrically;