What is the singular value decomposition for [2 -1 2] Somehow its easier to find SVD for higher order matrices but i cant get my head around a 1*n matrix.
 A: 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
$$
\begin{bmatrix} 2&-1&2\end{bmatrix}
=\begin{bmatrix} 1\end{bmatrix} \,
\begin{bmatrix} 3&0&0\end{bmatrix}\,
\begin{bmatrix}2/3&-1/3&2/3 \\
1/\sqrt2&0&-1/\sqrt2\\
1/3\sqrt2& 2\sqrt2/3&1/3\sqrt2
 \end{bmatrix}.
$$
