SVD -obligation of normalization let us suppose we have following  matrix
 $
       A=
      \left[ {\begin{array}{cc}
       2 & 2 \\
       -1 & 1 \\
      \end{array} } \right]
    $
and  i  want to compute  SVD  of this matrix, i have calculated first of all
$A*A'$ , which is equal to
$
      \left[ {\begin{array}{cc}
       8 & 0 \\
       0 & 2 \\
      \end{array} } \right]
    $
because it is  symmetric matrix, i can  decompose it using  eigenvalue decomposition, its corresponding  eigenvalues are
$8$ and  $2$,  while Eigenvectors are
$
    U=  \left[ {\begin{array}{cc}
       1 & 0 \\
       0 & 1 \\
      \end{array} } \right]
    $
for computing right  singular vectors  i used  $A'*A $ which is equal to 
$
      \left[ {\begin{array}{cc}
       5 & 3 \\
       3 & 5 \\
      \end{array} } \right]
    $
eigenvectors of this matrix is equal to
$
      V=\left[ {\begin{array}{cc}
       1 & -1 \\
       1 & 1 \\
      \end{array} } \right]
    $
and diagonal matrix will be
$
      E=\left[ {\begin{array}{cc}
       \sqrt{8}  & 0 \\
       0 & \sqrt{2} \\
      \end{array} } \right]
    $
so
original matrix  equal to
$A=U*E*V'$
but it gives me different result and  why?  where i am making mistake?
 A: Compute the singular value decomposition of a matrix $\mathbf{A}\in\mathbb{C}^{m\times n}_{\rho}$
$$
\mathbf{A} =
\mathbf{U} \, \Sigma \, \mathbf{V}^{*}
=
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S}_{\rho\times \rho} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
% V 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
  \end{array} \right]  \\
 \\
\tag{1}
$$
The beauty of the SVD is that it provides an orthonormal basis for the four fundamental subspace for a matrix $\mathbf{A}\in\mathbb{C}^{m\times n}$
$$
\begin{align}
%
  \mathbf{C}^{n} = 
    \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus
    \color{red}{\mathcal{N} \left( \mathbf{A} \right)} \\
%
  \mathbf{C}^{m} = 
    \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus
    \color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)}
%
\end{align}
$$
To compute the SVD,


*

*Resolve the domain by finding eigenvectors of $\mathbf{A}^{*}\mathbf{A}$. Outputs: matrix of singular values $\mathbf{S}$, $\color{blue}{\mathbf{V}_{\mathcal{R}}}$.

*Compute $\color{blue}{\mathbf{U}_{\mathcal{R}}}$ using $\mathbf{S}$ and  $\color{blue}{\mathbf{V}_{\mathcal{R}}}$



1. Resolve  $\ \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)}$

Step 1: 
Compute product matrix
$$
%
\begin{align}
%
 \mathbf{W} = \mathbf{A}^{T}\mathbf{A} = 
%
\left[
\begin{array}{cr}
 2 & -1 \\
 2 & 1 \\
\end{array}
\right]
\left[
\begin{array}{rr}
 2 & 2 \\
 -1 & 1 \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{cc}
 5 & 3 \\
 3 & 5 \\
\end{array}
\right]
%
\end{align}
%
$$
Step 2: 
Compute eigenvalue spectrum $\lambda \left(\mathbf{W}\right)$
$$
 \det \mathbf{W} = 16, \qquad \text{trace } \mathbf{W} = 10
$$
The characteristic polynomial is
$$
 p(\lambda) = \lambda^{2} - \lambda \text{ trace } \mathbf{W} + \det \mathbf{W}
= \lambda ^2-10 \lambda +16 = 
\left( \lambda - 8 \right) \left( \lambda - 2 \right)
$$
The roots of the $p(\lambda)$ are the eigenvalues of $\mathbf{W}$:
$$
 \lambda \left(\mathbf{W}\right) = \left\{ 8, 2 \right\}
$$
Step 3: 
Compute singular value spectrum $\sigma$
To obtain the singular values: form $\tilde{\lambda}$, a list arranged in decreasing order with $0$ values culled:
$$
 \sigma = \sqrt{\tilde{\lambda}} = \left\{ 2\sqrt{2}, \sqrt{2} \right\}
$$
The singular values are the diagonal entries of the $\mathbf{S}$:
$$
\boxed{
 \mathbf{S} = \sqrt{2}\left[
\begin{array}{cc}
 2  & 0 \\
 0 & 1 \\
\end{array}
\right]
}
$$
Step 4: 
Compute eigenvectors of $\mathbf{W}$
Fundamental tool: eigenvalue equation
$$
  \mathbf{W} v_{k} = \lambda_{k} v_{k}, \qquad k = 1, 2
$$
$k=1$:
$$
%
\begin{align}
%
 \mathbf{W} v_{1} &= \lambda_{1} v_{1} \\
%
\left[
\begin{array}{cc}
 5 & 3 \\
 3 & 5 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
 x \\ y \\
\end{array}
\right]
%
&=
%
8
%
\left[
\begin{array}{c}
 x \\ y \\
\end{array}
\right] \\[3pt]
% % %
\left[
\begin{array}{c}
 5 x + 3 y \\ 3 x + 5 y \\
\end{array}
\right]
&=
\left[
\begin{array}{c}
 8x \\ 8y \\
\end{array}
\right]\\[3pt]
% % %
\left[
\begin{array}{c}
 x \\ y \\
\end{array}
\right]
&=
\left[
\begin{array}{c}
 1 \\ 1 \\
\end{array}
\right]
%
\end{align}
%
$$
The normalized vector is the first column vector in $\color{blue}{\mathbf{V}_{\mathcal{R}}}$.
$$
  \hat{v}_{1} = \frac{1}{\sqrt{2}}
\left[
\begin{array}{r}
 1 \\ 1 \\
\end{array}
\right]
$$
$k=2$:
$$
%
\begin{align}
%
 \mathbf{W} v_{2} &= \lambda_{2} v_{2} \\
%
\left[
\begin{array}{cc}
 5 & 3 \\
 3 & 5 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
 x \\ y \\
\end{array}
\right]
%
&=
%
2
%
\left[
\begin{array}{c}
 x \\ y \\
\end{array}
\right] \\[3pt]
% % %
\left[
\begin{array}{c}
 5 x + 3 y \\ 3 x + 5 y \\
\end{array}
\right]
&=
\left[
\begin{array}{c}
 2x \\ 2y \\
\end{array}
\right]\\[3pt]
% % %
\left[
\begin{array}{c}
 x \\ y \\
\end{array}
\right]
&=
\left[
\begin{array}{c}
 -1 \\ 1 \\
\end{array}
\right]
%
\end{align}
%
$$
The normalized vector is the second column vector in $\color{blue}{\mathbf{V}_{\mathcal{R}}}$.
$$
  \hat{v}_{2} = \frac{1}{\sqrt{2}}
\left[
\begin{array}{r}
 -1 \\ 1 \\
\end{array}
\right]
$$
Assemble:
$$
\boxed{
\color{blue}{\mathbf{V}_{\mathcal{R}}} =  \frac{1}{\sqrt{2}}
%
\left[
\begin{array}{cr}
 1 & -1 \\
 1 &  1 \\
\end{array}
\right]
}
$$
2. Resolve  $\ \color{blue}{\mathcal{R} \left( \mathbf{A} \right)}$
Rearrange (1) to recover
$$
 \color{blue}{\mathbf{U}_{\mathcal{R}}} = \mathbf{A} \color{blue}{\mathbf{V}_{\mathcal{R}}} \mathbf{S}^{-1}
$$
The power of the SVD is that is aligns the $\color{blue}{range}$
 spaces and accounts for scale differences. This allows direct computation using equation (1):
$$
\begin{align}
  \color{blue}{\mathbf{U}_{\mathcal{R}}} =
\mathbf{A} \color{blue}{\mathbf{V}_{\mathcal{R}}} \mathbf{S}^{-1}
%
&=
\left[
\begin{array}{rc}
  2 & 2 \\
 -1 & 1 \\
\end{array}
\right]
%
\frac{1}{\sqrt{2}}
\left[
\begin{array}{cr}
 1 & -1 \\
 1 &  1 \\
\end{array}
\right]
% Sinv
\left[
\begin{array}{cc}
 \frac{1}{2 \sqrt{2}} & 0 \\
 0 & \frac{1}{\sqrt{2}} \\
\end{array}
\right]
%
\end{align}
%
$$
At last,
$$
\boxed{
\color{blue}{\mathbf{U}_{\mathcal{R}}} = 
\left[
\begin{array}{cc}
 1 & 0 \\
 0 & 1 \\
\end{array}
\right]
}
$$

Final answer
$$
  \mathbf{A} = 
\color{blue}{\mathbf{U}_{\mathcal{R}}} 
\mathbf{S} 
\color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} =
%
\left[
\begin{array}{cc}
 1 & 0 \\
 0 & 1 \\
\end{array}
\right]
%
\sqrt{2}
\left[
\begin{array}{cc}
 2 & 0 \\
 0 & 1 \\
\end{array}
\right]
 \frac{1}{\sqrt{2}}
%
\left[
\begin{array}{cr}
 1 & -1 \\
 1 &  1 \\
\end{array}
\right]
%
%
$$
A: First problem: selecting your $U$ determines the matrix $V$.  After you find $U$, all that remains is
$$
A = U \Sigma V' \implies V = \Sigma^{-1}U'A
$$
The second problem is that when you found $V$, you didn't normalize the columns.  This is not strictly relevant, however, since your method of finding $V$ in the first place was incorrect.
See my answer here for a more thorough explanation.  My answer is about how one can compute the SVD by finding $V$ first, but the idea is the same.
