Calculate singular value decomposition

Question

I need to find the reduced singular value decomposition of this matrix.

$$\begin{pmatrix} -2 & -1 & 2 \\ 2 & 1 & -2 \\ \end{pmatrix} $$

I formed $$ A^TA= \begin{pmatrix} 8 & 4 & -8 \\ 4 & 2 & -4 \\ -8 & -4 & 8 \\ \end{pmatrix} $$

Found the eigenvalue 18 and eigenvector $$\begin{pmatrix} -1 \\ -1/2 \\ 1 \end{pmatrix}$$

$$ u_1=\begin{pmatrix} \frac{3\sqrt2}{4} \\ \frac{-3\sqrt2}{4} \end{pmatrix} $$

So $$ A=\begin{pmatrix} \frac{3\sqrt2}{4} \\ \frac{-3\sqrt2}{4} \end{pmatrix}\begin{pmatrix}3\sqrt2\end{pmatrix}\begin{pmatrix} -1 & -1/2 & 1 \end{pmatrix} $$

But this is clearly not correct

Answer 1

Guide:

This is a rank $2$ matrix, your computation of $A^TA$ is wrong.

octave:1> A = [-2 -1 1; 2 1 -2]
A =

  -2  -1   1
   2   1  -2

octave:2> A'*A
ans =

   8   4  -6
   4   2  -3
  -6  -3   5

Also, rather than dealing with $A^TA$, perhaps working with $AA^T$ is simpler.

Edit:

\begin{align} \begin{bmatrix} -2 & -1 & 2 \\2 & 1 & -2\end{bmatrix} &= \begin{bmatrix} 1 \\ -1\end{bmatrix}\begin{bmatrix} -2 & -1 & 2 \end{bmatrix} \\ &=\begin{bmatrix} \frac{1}{\sqrt2} \\ -\frac1{\sqrt{2}}\end{bmatrix}3\sqrt{2}\begin{bmatrix} -\frac23 & -\frac13 & \frac23 \end{bmatrix} \end{align}

Singular vectors are of unit length.