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: 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).
A: Problem
Detailed instructions for computing the SVD abound on Math Stack Exchange. For example: SVD and the columns, SVD -obligation of normalization, how SVD is calculated in reality, Pseudo-inverse of a matrix that is neither fat nor tall?.
This problem is straightforward. Let the target matrix be the covector
$$
\mathbf{A} = \left[ \begin{array}{ccr}
 2 & 1 & -2
\end{array} \right]
$$
Find the singular value decomposition
$$ 
 \mathbf{A} = \mathbf{U} \, \Sigma \, \mathbf{V}^{*}
$$
Methods
The first link presents a table showing two paths for finding the SVD for a matrix $\mathbf{A} \in \mathbb{C}^{m \times n}_{\rho}$:
$$
\begin{array}{lll}
%
 \text{Operation} &  
\text{Row space first} & \text{Column space first} \\\hline
%
  \text{1. Construct product matrix}  &
  \mathbf{W} = \mathbf{A}^{*} \mathbf{A} & 
  \mathbf{W} = \mathbf{A} \, \mathbf{A}^{*} \\
%
  \text{2. Solve for eigenvalues}  &
  \sigma = \tilde{\lambda} \left( \mathbf{W} \right) & 
  \sigma = \tilde{\lambda} \left( \mathbf{W} \right) \\
%
  \color{blue}{\text{3. Solve for eigenvectors }} w_{k},\  k=1, \rho  &
  \left( \mathbf{W} - \lambda_{k} \mathbf{I}_{n} \right) w_{k} = \mathbf{0} &
  \left( \mathbf{W} - \lambda_{k} \mathbf{I}_{m} \right) w_{k} = \mathbf{0} \\
%
  \text{4. Assemble domain matrix}  &
  \mathbf{V}_{k} = \frac{w_{k}}{\lVert w_{k} \rVert_{2}} &
  \mathbf{U}_{k} = \frac{w_{k}}{\lVert w_{k} \rVert_{2}} &
\\
%
  \text{5. Compute complementary domain matrix}  &
  \mathbf{U}_{k} = \sigma_{k}^{-1} \mathbf{A} \mathbf{V}_{k} &
  \mathbf{V}_{k} = \sigma_{k}^{-1} \mathbf{A}^{*} \mathbf{U}_{k} &
\\
%
\end{array}
$$
The two product matrices to wrok with are
$$
\mathbf{A} \mathbf{A}^{*} =
\left[
\begin{array}{c}
  9 \\
\end{array}
\right]
\qquad
\mathbf{A}^{*} \mathbf{A} =
\left[
\begin{array}{rrr}
  4 &  2 & -4 \\
  2 &  1 & -2 \\
 -4 & -2 &  4 \\
\end{array}
\right].
$$
Solution
The product matrix $\mathbf{A} \mathbf{A}^{*}$ is much easier to work with!
Singular values
The eigenvalue is
$$
 \lambda \left( \mathbf{A}\mathbf{A}^{*} \right) = 9
$$
which implies the singular value is
$$
 \sigma = \sqrt{\lambda \left( \mathbf{A}\mathbf{A}^{*} \right)} = 3
$$
The matrix of singular values is
$$
 \mathbf{S} = 
\left[
\begin{array}{c}
  3 
\end{array}
\right]
$$
There is no need for $0$ padding, so
$$ 
 \Sigma = \mathbf{S}
$$
Matrix $\mathbf{U}$
The eigenvector is already normalized and represents the first column 
$$
\mathbf{U} =
\color{blue}{\left[
\begin{array}{c}
  1 
\end{array}
\right]}
$$
Coloring distinguishes $\color{blue}{range}$ space vectors from $\color{red}{null}$ space vectors.
Column space
$$
\begin{align}
 \color{blue}{\mathcal{R}\left(\mathbf{A}\right)} 
&=
\text{span} \left\{ \,
\color{blue}{\left[
\begin{array}{c}
  1 
\end{array}
\right]}
\, \right\} \\
%
 \color{red}{\mathcal{N}\left(\mathbf{A}^{*}\right)}  &= 
\left\{ 0 \right\}
\end{align}
$$
Matrix $\mathbf{V}$
$$
  \mathbf{V}_{1} 
= \frac{1}{\sigma} \mathbf{A}^{*} \mathbf{U}_{k}
= \frac{1}{3}
\color{blue}{\left[
\begin{array}{r}
  2 \\ 1 \\ -2 
\end{array}
\right]}
$$
For the $\color{red}{null}$ space vectors, we have options. We could call upon the process of Gram and Schmidt. Or we could eyeball the problem to find any $\color{red}{null}$ space vector, and use the cross product to find the missing vector.
Choosing the latter, we select 
$$
\color{red}{\mathbf{V}_{2}} = \frac{1}{\sqrt{2}}
\color{red}{\left[
\begin{array}{r}
  1 \\ 0 \\ 1 
\end{array}
\right]}
$$
The missing vector is then
$$
\color{blue}{\left[
\begin{array}{r}
  2 \\ 1 \\ -2 
\end{array}
\right]}
\times
\color{red}{\left[
\begin{array}{r}
  1 \\ 0 \\ 1 
\end{array}
\right]}
=
\color{red}{\left[
\begin{array}{r}
  1 \\ -4 \\ -1 
\end{array}
\right]}
$$
Final assembly
$$
\begin{align}
 \mathbf{A} &= \mathbf{U} \, \Sigma \mathbf{V}^{*}
% U
\\ &=
\color{blue}{\left[
\begin{array}{c}
  1 
\end{array}
\right]}
% Sigma
\left[
\begin{array}{c}
  3 
\end{array}
\right]
% V*
\left[
\begin{array}{ccc}
  % c1
\frac{1}{3}
\color{blue}{\left[
\begin{array}{r}
  2 \\ 1 \\ -2 
\end{array}
\right]}
  % c2
\frac{1}{\sqrt{2}}
\color{red}{\left[
\begin{array}{r}
  1 \\ 0 \\ 1 
\end{array}
\right]}
  % c3
\frac{1}{3\sqrt{2}}
\color{red}{\left[
\begin{array}{r}
  1 \\ -4 \\ -1 
\end{array}
\right]}
\end{array}
\right]
%
\\ &=
 \left[ \begin{array}{ccr}
 2 & 1 & -2
\end{array} \right]
%
\end{align}
$$
