<s>[INST] how SVD is calculated in reality let us suppose that we have following matrix
$
       A=
      \left[ {\begin{array}{cc}
       4 & 0 \\
       3 & -5 \\
      \end{array} } \right]
    $
for calculation of  SVD,first i have calculated $A'*A$   which is equal to
$
       A'*A=
      \left[ {\begin{array}{cc}
       25 & -15 \\
       -15 & 25 \\
      \end{array} } \right]
    $
eigenvalues of this  matrix is equal to   $40$ and $10$,  eigenvector of  following matrix
$    
      \left[ {\begin{array}{cc}
       -15 & -15 \\
       -15 & -15 \\
      \end{array} } \right]
    $
i got this matrix after subtraction of $40$ from diagonal elements, eigenvector is equal to
\begin{bmatrix}-1 \\ 1\\ \end{bmatrix}
after inserting of eigenvalues of $10$, i got  following matrix
$    
      \left[ {\begin{array}{cc}
       15 & -15 \\
       -15 & 15 \\
      \end{array} } \right]
    $
eigenvector of this matrix is equal to 
\begin{bmatrix}1 \\ 1\\ \end{bmatrix}
so normalization of these  vectors and putting in  one matrix  $V$ will  have  the following  form
$    
      \left[ {\begin{array}{cc}
       -1/\sqrt{2} & 1/\sqrt{2} \\
       1/\sqrt{2} & 1/\sqrt{2} \\
      \end{array} } \right]
    $
now i know that 
$A*V=U*E$ where  $E$ is equal to
$    
      \left[ {\begin{array}{cc}
       \sqrt{40} & 0 \\
       0 & \sqrt{10} \\
      \end{array} } \right]
    $
we know that 
$A*v1=u1*\sigma$
let us try to multiply
$
       \left[ {\begin{array}{cc}
       4 & 0 \\
       3 & -5 \\
      \end{array} } \right] 
    $ 
by 
\begin{bmatrix}-1/\sqrt{2} \\ 1/\sqrt{2} \\ \end{bmatrix}
i got  the following result
\begin{bmatrix}-4/\sqrt{2} \\ -8/\sqrt{2} \\ \end{bmatrix}
but i can't get equation for  $\sigma$ and $u$  please help me
 [/INST] You did everything correctly. You already found $\sigma_1,\sigma_2$ - they are the square roots of the eigenvalues of $A^TA$ so $\sigma_1 = \sqrt{40},\sigma_2 = \sqrt{10}$. After you found $V$ with columns $v_1,v_2$, you must have
$$ Av_1 = \sigma_1 u_1, Av_2 = \sigma_2 u_2 $$
so $u_1$ is just $\frac{Av_1}{\sigma_1}$ and $u_2 = \frac{Av_2}{\sigma_2}$. In your case,
$$ u_1 = \begin{pmatrix} -\frac{4}{\sqrt{2}} \\ -\frac{8}{\sqrt{2}} \end{pmatrix} \frac{1}{\sqrt{40}} = \begin{pmatrix} -\frac{1}{\sqrt{5}} \\ -\frac{2}{\sqrt{5}} \end{pmatrix}, \\
u_2 = \begin{pmatrix} \frac{4}{\sqrt{2}} \\ -\frac{2}{\sqrt{2}} \end{pmatrix} \frac{1}{\sqrt{10}} = \begin{pmatrix} \frac{2}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}} \end{pmatrix} $$
and indeed
$$ \begin{pmatrix} -\frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} & -\frac{1}{\sqrt{5}} \end{pmatrix} \begin{pmatrix} 4 & 0 \\ 3 & -5\end{pmatrix}  \begin{pmatrix} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}  \end{pmatrix} = \begin{pmatrix} \sqrt{40} & 0 \\ 0 & \sqrt{10} \end{pmatrix}. $$
</s>