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Singular Valued Decomposition: Nearly SVD

LINALG-IDAQ9X

If $A=\begin{bmatrix} -5 & 0 & 0 \\\ 0 & 3 & 0\end{bmatrix}$ then $A=U\Sigma V^{T}$ for $U=\begin{bmatrix} 0 & 1 \\\ 1 & 0\end{bmatrix}$, $\Sigma=\begin{bmatrix} 3 & 0 & 0 \\\ 0 & 5 & 0\end{bmatrix}$, and $V=\begin{bmatrix} 0 & 1& 0 \\\ -1 & 0 & 0 \\\ 0 & 0 & 1\end{bmatrix}$.

Why is the equation $A=U\Sigma V^{T}$ not a singular value decomposition of $A$?

A

The entries of $A$ must be non-zero.

B

The eigenvalues of $A^{T}A$ are 9 and 25.

C

The diagonal entries of $\Sigma$ must be arranged in descending order starting with the upper left entry.

D

The matrix $A$ is not a square matrix.

E

The columns of $U$ are not eigenvectors of $A^{T}A$