Why the null space of pseudo inverse equals the null space of the matrix transpose? The pseudoinverse $A^+$ of A is the matrix for which $x = A^+Ax$ for all x in the row space of A. The nullspace of $A^+$ is the nullspace of $A^T$.
I don't understand this cause the above seems to imply that $A^+=A^T$ which doesn't make sense as $x = A^+Ax$ while $A^TA$ gives a matrix which is not an identity matrix.
Here is the source-Page 2 on "Finding the pseudo Inverse"
 A: Define matrix
Start with a matrix 
$$
 \mathbf{A} \in\mathbb{C}^{m\times n}_{\rho}
$$
Fundamental Theorem of Linear Algebra
The Fundamental Theorem of Linear Algebra can be expressed as
$$
\begin{align}
%
  \mathbf{C}^{n} = 
    \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus
    \color{red}{\mathcal{N} \left( \mathbf{A} \right)} \\
%
  \mathbf{C}^{m} = 
    \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus
    \color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)}
%
\end{align}
$$
Singular value decomposition
The singular value decomposition of the matrix is
$$
\begin{align}
  \mathbf{A} &=
  \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\
%
 &=
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cccc|cc}
     \sigma_{1} & 0 & \dots &  &   & \dots &  0 \\
     0 & \sigma_{2}  \\
     \vdots && \ddots \\
       & & & \sigma_{\rho} \\\hline
       & & & & 0 & \\
     \vdots &&&&&\ddots \\
     0 & & &   &   &  & 0 \\
  \end{array} \right]
% V 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
  \end{array} \right]  \\
%
  & =
% U
   \left[ \begin{array}{cccccccc}
    \color{blue}{u_{1}} & \dots & \color{blue}{u_{\rho}} & \color{red}{u_{\rho+1}} & \dots & \color{red}{u_{n}}
  \end{array} \right]
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S}_{\rho\times \rho} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
% V
   \left[ \begin{array}{c}
    \color{blue}{v_{1}^{*}} \\ 
    \vdots \\
    \color{blue}{v_{\rho}^{*}} \\
    \color{red}{v_{\rho+1}^{*}} \\
    \vdots \\ 
    \color{red}{v_{n}^{*}}
  \end{array} \right]
%
\end{align}
$$
The connection to the Fundamental Theorem is intimate:
$$
\begin{array}{ll}
%
     column \ vectors & span \\\hline
%
     \color{blue}{u_{1}}  \dots  \color{blue}{u_{\rho}} & 
     \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \\
%
     \color{blue}{v_{1}}  \dots  \color{blue}{v_{\rho}} & 
     \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \\
%
     \color{red}{u_{\rho+1}}  \dots  \color{red}{u_{m}} & 
     \color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)} \\
%
     \color{red}{v_{\rho+1}}  \dots  \color{red}{v_{n}} & 
     \color{red}{\mathcal{N} \left( \mathbf{A} \right)} \\
%
  \end{array}
$$
Pseudoinverse matrix
The least squares solution with the SVD produces the pseudoinverse:
$$
\begin{align}
  \mathbf{A}^{+} &= \mathbf{V} \, \Sigma^{+} \mathbf{U}^{*} \\
%
&=
% V
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{V}_{\mathcal{R}}} & 
     \color{red}{\mathbf{V}_{\mathcal{N}}}
  \end{array} \right] 
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S}^{-1} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
%
% U
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{U}_{\mathcal{N}}}^{*}
  \end{array} \right]  \\
%
\end{align}
%
$$
$$
 \mathbf{A}^{+} \in\mathbb{C}^{n\times m}_{\rho}
$$
The subspace decomposition in terms of the pseudoinverse is now explicit.
Adjoint matrix
For comparison, the adjoint matrix is
$$
\begin{align}
  \mathbf{A}^{*} &= \mathbf{V} \, \Sigma^{T} \mathbf{U}^{*} \\
%
&=
% V
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{V}_{\mathcal{R}}} & 
     \color{red}{\mathbf{V}_{\mathcal{N}}}
  \end{array} \right] 
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
%
% U
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{U}_{\mathcal{N}}}^{*}
  \end{array} \right]  \\
%
\end{align}
$$
$$
 \mathbf{A}^{*} \in\mathbb{C}^{n\times m}_{\rho}
$$

Further reading
How the pseudoinverse solution arises in least squares: How does the SVD solve the least squares problem?; Singular value decomposition proof
Variant forms of the pseudoinverse are presented in What forms does the Moore-Penrose inverse take under systems with full rank, full column rank, and full row rank?; generalized inverse of a matrix and convergence for singular matrix Note the null space relationships with the pseudoinverse.
Fundamental projectors and the pseudoinverse: Least squares solutions and the orthogonal projector onto the column space
