91影视

Given that:

$A=\left[\begin{array}{cc}p& -q\\ q& p\end{array}\right]$

Now the singular values of A are found by first computing the eigenvalues of the square matrix

$A^{t}A=\left[\begin{array}{cc}p& q\\ -q& p\end{array}\right]\left[\begin{array}{cc}p& -q\\ q& p\end{array}\right]=\left[\begin{array}{cc}{p}^2+q^2& 0\\ 0& p^2+q^2\end{array}\right]$

Then

$(x{l}_2-A^{t}A)=\left[\begin{array}{cc}x-(p^2+q^2)& 0\\ 0& x-(p^2+q^2)\end{array}\right]$

Let characteristic polynomial of$A^{t}Abep\left(x\right)$.

Then

$$\begin{gathered} p\left(x\right)=\det \left(x{I}_2-A^{t}A\right) \\ ={\left[x-\left(p^2+q^2\right)\right]}^2-0 \\ ={\left[x-\left(p^2+q^2\right)\right]}^2 \end{gathered}$$

Thus, the characteristic polynomial is ${\left[x-\left(p^2+q^2\right)\right]}^2$.

This implies that the roots of p(x) are $p^2+q^2,p^2+q^2$.

Hence the eigenvalues of A^tA are ${位}_1=p^2+q^2$and${位}_2=p^2+q^2$.
Thus the singular values of A are ${蟽}_1=\sqrt{p^2+q^2}$and ${蟽}_2=\sqrt{p^2+q^2}$.

Clearly,${蟽}_1={蟽}_2$, which means the singular values of A are same. Now by using Theorem 8.3.2 we get that, if $p^2+q^2鈮�0$, equivalently if L(x) = Ax is an invertible linear transformation from $R^{2}to{R}^2$, then the image of the unit circle under L is a circle [Since ${蟽}_1={蟽}_2$implies that the lengths of the semi-major and semi-minor axes are equal].