<s>[INST] Finding the singular value(s) of a given matrix without SVD. I am struggling on a problem that asks to find the singular value(s) that are unequal to 0 in the following matrix:
$M = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \end{bmatrix}$ 
I am not supposed to do a singular value decomposition on the matrix.
The given answer shows how you should rewrite the matrix as 
$
\sqrt{20}
\begin{bmatrix} \frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{2} \end{bmatrix}
\begin{bmatrix}
\frac{1}{\sqrt{5}} & \frac{1}{\sqrt{5}} & \frac{1}{\sqrt{5}} & \frac{1}{\sqrt{5}} & \frac{1}{\sqrt{5}}
\end{bmatrix}
$, and that therefore the singular value must be $\sqrt{20}$.
I see how the two vectors are orthonormal but other than that I don't really see why $\sqrt{20}$ must be a singular value here. It is also unclear to me why this is the only singular value unequal to zero.
Any help would be much appreciated.
 [/INST] For any $n$, let $e=(1,...,1)^T$, and let $x_2,...,x_n$ be an orthonormal basis of $\operatorname{sp}\{e\}^\bot$. Let $x_1 = \frac{1}{\|e\|} e$. Then $x_1,...,x_n$
is an orthonormal basis for $\mathbb{R}^n$. Let $U_n = \begin{bmatrix} x_1 & \cdots & x_n \end{bmatrix}$.
Note that $U_4^* M U_5 = \begin{bmatrix} \begin{matrix} \sqrt{4} & \cdots & \sqrt{4} \end{matrix} \\ 0 \end{bmatrix} U_5 = \Sigma = \begin{bmatrix} \sqrt{20} & 0 & \cdots & 0 \\
0 & 0 & \cdots & 0 \\
\vdots & \vdots & & \vdots\\
0 & 0 & \cdots & 0
\end{bmatrix} $
and so $M = U_4 \Sigma U_5^*$.
$U_4,U_5$ are unitary, $\Sigma$ is diagonal with non-negative entries, hence this is a singular value decomposition.
</s>