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Let $\mathrm{A}=\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]$and role="math" localid="1664181842990" $\mathrm{B}=\left[\begin{array}{cc}0& 0\\ 1& 1\end{array}\right]$

The characteristic polynomial of$\mathrm{B}$is$位\left(位-1\right)$. This implies that 0 and 3 are eigenvalues of$\mathrm{B}$.

Now, find $\mathrm{P}$such that

$\mathrm{B}={\mathrm{P}}^{-1}\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\mathrm{P}$

$={\mathrm{P}}^{-1}\mathrm{AP}$

Thus, $\mathrm{A}$and $\mathrm{B}$are similar matrices.

Here, $\mathrm{A}$ is already in the diagonal form with non-negative diagonal entries. Hence, its singular values are its diagonal entries which are 0 and 3.

Find ${\mathrm{B}}^{\mathrm{T}}\mathrm{B}$as follows:

$$\begin{gathered} {\mathrm{B}}^{\mathrm{T}}\mathrm{B}=\left[\begin{array}{cc}0& 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}0& 1\\ 0& 1\end{array}\right] \\ =\left[\begin{array}{cc}0& 0\\ 0& 2\end{array}\right] \end{gathered}$$

To find the singular value, substitute

$\left({\mathrm{B}}^{\mathrm{T}}\mathrm{B}-\mathrm{位滨}\right)=0$

$$\begin{gathered} \left[\begin{array}{cc}0& 0\\ 0& 2\end{array}\right]-位\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=0 \\ \left[\begin{array}{cc}0& 0\\ 0& 2\end{array}\right]-\left[\begin{array}{cc}位& 0\\ 0& 位\end{array}\right]=0 \\ \left[\begin{array}{cc}-位& 0\\ 0& 2-位\end{array}\right]=0 \end{gathered}$$

Simplify further,

$$\begin{gathered} \left(-位\right)\left(2-位\right)=0 \\ -2位+{位}^2=0 \\ 位=0;-2+位=0 \\ 位=0;位=2 \end{gathered}$$

The singular values are found from the non-zero eigenvalues of ${\mathrm{B}}^{\mathrm{T}}\mathrm{B}$.

Hence, singular value $\mathrm{蟽}=\sqrt{\mathrm{位}}$.

The singular values of $\mathrm{B}$are 0 and $\sqrt{2}$.

Similar matrices do not always have the same singular values. The singular value of the matrix is determined by the determinant's value, which can be zero. As a result, the value for two similar matrices is not always the same.

The given statement is FALSE.