SVD and linear least squares problem Edit:
I've actually found an error: 
Instead of full SVD I had to use, "economy size" SVD, where $U$ has only first $n$ columns, and $\Sigma$ becomes a square matrix. I also forgot to take the transpose of $V$, that's why I was getting wrong numbers. SO, the primary question is solved :)
But it would be great if someone could answer the bonus question.
I need to solve the linear least squares problem $\min_{x}\|Ax - b\|_{2}^{2}$ given the matrix
$A = \begin{pmatrix}
1&1 \\
0&1 \\
-1&0 
\end{pmatrix} $ and vector $b = \begin{pmatrix}
1 \\
1 \\
1 
\end{pmatrix} $
The normal equations method $A^{T}Ax = A^{T}b$ and the QR-decomposition $Rx = Q^{T}b$
give the same result: $x = \begin{pmatrix}
-0.667 \\
1.333
\end{pmatrix}$
However, I have some questions about the SVD method.
My lecture notes say that the solution to the LLS is
$$x = V\Sigma^{-1}U^{T}b$$
But $$A_{32} = U_{33}\Sigma_{32}V_{22}^{T}$$
Therefore, $\Sigma$ is an $3\times2$ matrix, and the inverse is not defined for it.
I though $\Sigma^{-1}$ could mean $\Sigma^{+}$ (pseudo inverse), then the dimensions agree.
But the solution is $x = \begin{pmatrix}
1.333 \\
0.667
\end{pmatrix}$ which looks quite close to the first solution, but it doesn't give the least value. 
Thank you for your time! I really hope someone could help me figure it out.
As a bonus question, why is it written $\min_{x}\|Ax - b\|_{2}^{2}$ ?
Isn't $\min_{x}\|Ax - b\|_{2}^{2} \Leftarrow\Rightarrow \min_{x}\|Ax - b\|_{2}$, because norm is always $\geq0$?

Problem definition
$$
\begin{align}
  \mathbf{A} x &= b\\
\left(
\begin{array}{rr}
 1 & 1 \\
 0 & 1 \\
 -1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{c}
 x_{1} \\
 x_{2}
\end{array}
\right)
&=
\left(
\begin{array}{c}
 1 \\
 1 \\
 1 \\
\end{array}
\right)
\end{align}
$$
Normal equations
$$
\begin{align}
  \mathbf{A}^{*} \mathbf{A} x &= \mathbf{A}^{*} b\\
\left(
\begin{array}{cc}
 2 & 1 \\
 1 & 2 \\
\end{array}
\right)
\left(
\begin{array}{c}
 x_{1} \\
 x_{2}
\end{array}
\right)
&=
\left(
\begin{array}{c}
 0 \\
 2
\end{array}
\right)
\end{align}
$$
Given the inverse matrix $\left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} = \frac{1}{3}\left(
\begin{array}{rr}
 2 & -1 \\
 -1 & 2 \\
\end{array}
\right)$, the solution is
$$
  x_{LS} = \left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} \mathbf{A}^{*}b = \frac{1}{3}
\left(
\begin{array}{r}
 -2 \\
  4
\end{array}
\right)
$$
SVD
The singular value decomposition is
$$
\begin{align}
  \mathbf{A} &= \mathbf{U}\ \Sigma \ \mathbf{V}^{*} \\
&=
\left(
\begin{array}{ccc}
 \frac{1}{\sqrt{6}} 
\left[
\begin{array}{r}
  2 \\
  1 \\
 -1 \\
\end{array}
\right] &
 \frac{1}{\sqrt{2}} 
\left[
\begin{array}{r}
  0 \\
  1 \\
  1 \\
\end{array}
\right] &
 \frac{1}{\sqrt{3}} 
\left[
\begin{array}{r}
  1 \\
  1 \\
  1 \\
\end{array}
\right] &
\end{array}
\right)
%
\left(
\begin{array}{cc}
 \sqrt{3} & 0 \\
 0 & 1 \\
 0 & 0 \\
\end{array}
\right)
%
\frac{1}{\sqrt{2}}
\left(
\begin{array}{rr}
  1 & 1 \\
 -1 & 1
\end{array}
\right)
\end{align}
$$
The pseudoinverse solution provides the same answer
$$
  x_{LS} = \mathbf{A}^{\dagger} b = \mathbf{V}\ \Sigma^{\dagger} \mathbf{U}^{*} b = 
\frac{1}{3}
\left(
\begin{array}{rrr}
 1 & -1 & -2 \\
 1 & 2 & 1 \\
\end{array}
\right)
\left(
\begin{array}{c}
 1 \\
 1 \\
 1 \\
\end{array}
\right)
=
\frac{1}{3}
\left(
\begin{array}{r}
 -2 \\
  4
\end{array}
\right)
$$
Bonus answer: You can show that when the inverse to the product matrix exists, it must be the pseudoinverse. Start with the basic definitions
$$
  \mathbf{A}= \mathbf{U} \ \Sigma \ \mathbf{V}^{*}, \quad
  \mathbf{A}^{*}= \mathbf{V} \ \Sigma^{\mathrm{T}}\ \mathbf{U}^{*}, \quad
  \mathbf{A}^{\dagger} = \mathbf{V} \ \Sigma^{\dagger} \mathbf{U}^{*}.
$$ 
To get there, use the following identities
$$
\Sigma = \left(
\begin{array}{cc}
 \sqrt{3} & 0 \\
 0 & 1 \\
 0 & 0 \\
\end{array}
\right), \quad
%
\Sigma^{\mathrm{T}} = \left(
\begin{array}{ccc}
 \sqrt{3} & 0 & 0\\
 0 & 1 & 0
\end{array}
\right), \quad
%
\Sigma^{\dagger} = 
\left(
\begin{array}{ccc}
 \frac{1}{\sqrt{3}} & 0 & 0\\
 0 & 1 & 0
\end{array}
\right).
$$
Given $\mathbf{S} = 
\left(
\begin{array}{cc}
 \sqrt{3} & 0 \\
 0 & 1 
\end{array}
\right),$
$$
  \Sigma = 
\left(
\begin{array}{c}
 \mathbf{S} \\
 0 
\end{array}
\right)_{3\times 2}, \quad
%
  \Sigma^{\mathrm{T}} = 
\left(
\begin{array}{cc}
 \mathbf{S} & 0 
\end{array}
\right)_{2\times 3}, \quad
%
  \Sigma^{\dagger} = 
\left(
\begin{array}{cc}
 \mathbf{S}^{-1} & 0 
\end{array}
\right)_{2\times 3}
$$
When the normal equations matrix product exists, it must be the same as the pseudoinverse:
$$
\left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} \mathbf{A}^{*} =
\left( \mathbf{V}\ \mathbf{S}^{2}\ \mathbf{V}^{*} \right)^{-1} \mathbf{V} \ \Sigma^{\mathrm{T}}\ \mathbf{U}^{*} =
\left( \mathbf{V}\ \mathbf{S}^{-2}\ \mathbf{V}^{*} \right) \mathbf{V}\ \Sigma^{\mathrm{T}}\ \mathbf{U}^{*} =
\mathbf{V}\ \Sigma^{\dagger}\ \mathbf{U}^{*} = \mathbf{A}^{\dagger}
$$
