SVD of (2,1,-2) not ok I'm trying to find the SVD of 
$$
\begin{pmatrix}
2&1&-2\\
\end{pmatrix}
$$
I found $$\Sigma , u$$ 
But on the V matrix I got
$$
\begin{pmatrix}
-\frac{2}{3}&&\frac{1}{\sqrt{2}}&&\frac{-1}{\sqrt{5}} \\
-\frac{1}{3}&&0&&\frac{2}{\sqrt{5}} \\
\frac{2}{3}&&\frac{1}{\sqrt{2}}&&0 \\
\end{pmatrix}
$$
While Wolfram Alpha gives : 
result
I checked the eigenvectors on wolfram alpha as well and they are correct:
$$
\begin{pmatrix}
-2&&1&&-1\\
-1&&0&&2\\
2&&1&&0\\
\end{pmatrix}
$$
I realized that the 3rd collumn of V (the one from wolfram alpha) is obtained by doing the cross product of the first two eigenvectors , why is that ?

A singular value decomposition of a $1\times 3$ matrix can be written almost "by inspection":
$$ \begin{pmatrix} 2 & 1 & -2 \end{pmatrix} = U \Sigma V^* $$
where $U,V$ are unitary (in this case, orthogonal) matrices and $\Sigma$ is a rectangular diagonal matrix with the required singular values on the diagonal:
$$ \Sigma = \begin{pmatrix} \lambda & 0 & 0 \end{pmatrix} $$
Since $U$ is a $1\times 1$ unitary matrix, actually $U = (1)$ is just the $1\times 1$ identity.  $\Sigma$ is a $1\times 3$ matrix of equal Euclidean norm to $\begin{pmatrix} 2 & 1 & -2 \end{pmatrix}$, so $\Sigma = \begin{pmatrix} 3 & 0 & 0 \end{pmatrix}$.
The requirement that $V$ be unitary (or orthogonal) does not uniquely specify its entries.  So long as:
$$ \begin{pmatrix} 3 & 0 & 0 \end{pmatrix} V^* = \begin{pmatrix} 2 & 1 & -2 \end{pmatrix} $$
it will meet the definition of a singular value decomposition, and from this we see that only the first column of $V$ (top row of $V^*$) is determined:
$$ V^* = \begin{pmatrix} \frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\
{-} & {-} & {-} \\ {-} & {-} & {-} \end{pmatrix} $$
The bottom two rows of $V^*$ can be filled in with any pair of unit vectors that make up an orthonormal basis for $\mathbb{R}^3$ together with the top row shown.  Therefore it is not surprising that the third row might be the cross-product of the first two rows (since the third row will be determined, up to sign, by the first two rows).
