What is the singular value decomposition for [2 -1 2]

Question

Somehow its easier to find SVD for higher order matrices but i cant get my head around a 1*n matrix.

Answer 1

Let $A=\begin{bmatrix} 2&-1&2\end{bmatrix}$. The singular value decomposition looks like $ A=UDV$, with $U$ a $1\times 1$ unitary, $D$ a $1\times 3$ "diagonal", and $V$ a $3\times 3$ unitary.

So $U=\begin{bmatrix} 1\end{bmatrix}$, and $D=\begin{bmatrix} 3&0&0\end{bmatrix}$, since $$ AA^*=\begin{bmatrix} 2&-1&2\end{bmatrix} \,\begin{bmatrix} 2\\-1\\2\end{bmatrix} =\begin{bmatrix} 9 \end{bmatrix} $$ so the only singular value is $3$. Then we have $$ \begin{bmatrix} 3&0&0\end{bmatrix}\,V = A = \begin{bmatrix} 2&-1&2\end{bmatrix}, $$ which tells us that the first row of $V$ is $\frac23,-\frac13,\frac23$. To complete $V$ we need the three rows to be orthonormal. For instance $$ V=\begin{bmatrix}2/3&-1/3&2/3 \\ 1/\sqrt2&0&-1/\sqrt2\\ 1/3\sqrt2& 2\sqrt2/3&1/3\sqrt2 \end{bmatrix}. $$ Thus $$ \begin{bmatrix} 2&-1&2\end{bmatrix} =\begin{bmatrix} 1\end{bmatrix} \, \begin{bmatrix} 3&0&0\end{bmatrix}\, \begin{bmatrix}2/3&-1/3&2/3 \\ 1/\sqrt2&0&-1/\sqrt2\\ 1/3\sqrt2& 2\sqrt2/3&1/3\sqrt2 \end{bmatrix}. $$