minimizing vectors Could please check whether my solution is right? 
Q. $$A=\begin{pmatrix}
        1 & 1 \\
        1 & 1 \\
        0 & 0 \\
        \end{pmatrix}$$Find the set of vectors $x$ that minimize the value $$||
        A
x-
        \begin{pmatrix}
        1  \\
        2 \\
        3  \\
        \end{pmatrix}||$$
My solution. 
$Ax-(1,2,3)^T=0$ (inconsistent) → least square method 
$A^TAx=A^Tb$,                $b=(1,2,3)^T$ 
$$
\begin{pmatrix}
        2 & 2 \\
        2 & 2 \
  \end{pmatrix}\begin{pmatrix}
        x_1 \\
        x_2 \\
        \end{pmatrix}=\begin{pmatrix}
        3 \\
        3 \\
        \end{pmatrix}$$
$2x_1+2x_2=3$ → $x_2=-x_1+1.5$ 
Therefore the set of $x$ is {$ 
\begin{pmatrix}
        k \\
        -k+1.5 \\
        \end{pmatrix}$| k∈ℝ }
 A: I would use a more "basic" method.
Writing the vector x as $\begin{pmatrix}x \\ y \end{pmatrix}$,  $Ax= \begin{pmatrix}1 & 1 \\ 1 & 1 \\ 0 & 0\end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} x+ y \\ x+ y \\ 0 \end{pmatrix}$ so that $Ax- \begin{pmatrix}1 \\ 2 \\ 3 \end{pmatrix}= \begin{pmatrix}x+ y- 1 \\ x+ y- 2 \\ -3\end{pmatrix}$.
The norm of that is $\sqrt{(x+ y- 1)^2+ (x+ y- 2)^2+ 9}$.
Writing that as f(x,y), we have $f_x= (1/2)((x+ y- 1)^2+ (x+ y- 2)^2+ 9)^{-1/2}[(2(x+y-1)+ 2(x+ y- 2)]= 0$ and $f_y= (1/2)((x+ y- 1)^2+ (x+ y- 2)^2+ 9)^{-1/2}[(2(x+y-1)+ 2(x+ y- 2)]= 0$. Those both reduce to $2[(x+ y- 1)+ 2(x+ y- 2)]= 4x+ 4y- 6= 0$.  The function will be minimized at any point along the line $x+ y= \frac{3}{2}$.  That is, $y= \frac{3}{2}- x$ which is exactly what you have.
A: Problem statment
Given the matrix
$$
\mathbf{A} = 
\left[
\begin{array}{cc}
 1 & 1 \\
 1 & 1 \\
 0 & 0 \\
\end{array}
\right] 
\in \mathbb{C}^{m\times n}_{\rho}
$$
and the data vector 
$$
b = 
\left[
\begin{array}{cc}
 1 \\
 2 \\
 3 \\
\end{array}
\right]
$$
find the least squares solution
$$
 x_{LS} = \left\{
 x\in\mathbb{C}^{n} \colon
\lVert
 \mathbf{A} x - b
\rVert_{2}^{2}
\text{ is minimized}
\right\}
$$
Solution
The general solution to the least squares problem is
$$
 x_{LS} = 
\color{blue}{\mathbf{A}^{+} b} +
\color{red}{ 
\left(
\mathbf{I}_{n} - \mathbf{A}^{+} \mathbf{A}
\right) y}, \quad y \in \mathbb{C}^{n}
\tag{1}
$$
where colors distinguish $\color{blue}{range}$ and $\color{red}{null}$ spaces.
The simple structure of the matrix $\mathbf{A}$ encourages a solution using the pseudoinverse built from the singular value decomposition (SVD).
Singular value decomposition
The SVD is straightforward:
$$
  \mathbf{A} =
  \mathbf{U} \, \Sigma \, \mathbf{V}^{*}
\tag{2}
$$
Resolve the eigensystem for
$$
 \mathbf{W} = \mathbf{A}^{*} \mathbf{A} =
\left[
\begin{array}{cc}
 2 & 2 \\
 2 & 2 \\
\end{array}
\right]
$$
Singular values
The eigenvalues are
$$
 \lambda \left( \mathbf{W} \right) =
\left\{ 4, 0 \right\}
$$
The matrix $\mathbf{A}$ has rank $\rho=1$, and the lone singular value is
$$
 \sigma_{1} = \sqrt{\lambda_{1}} = 2
$$
The matrix of singular values is 
$$
\mathbf{S} = 
\left[
\begin{array}{c}
 2
\end{array}
\right]
$$
and the sabot matrix is
$$
 \Sigma =
\left[
\begin{array}{c}
 \mathbf{S} & 0 \\
 \mathbf{0} & \mathbf{0}
\end{array}
\right]
=
\left[
\begin{array}{c}
 2 & 0 \\
 0 & 0 \\
 0 & 0 \\
\end{array}
\right]
$$
Domain matrix $\mathbf{V}$
The normalized eigenvectors of $\mathbf{W}$ are the column vectors of $\mathbf{V}$:
$$
 \mathbf{V} = \frac{1}{\sqrt{2}}
\left[
\begin{array}{cr}
 \color{blue}{1} & \color{red}{-1} \\
 \color{blue}{1} & \color{red}{1} \\
\end{array}
\right]
$$
Domain matrix $\mathbf{U}$
Equation $(2)$ can be rewritten to provide the $\color{blue}{range}$ space vector for $\mathbf{U}$:
$$
 \mathbf{U}_{1} =
\frac{1}{\sqrt{2}}
\color{blue}{\left[
\begin{array}{r}
 1 \\ 0 \\ -1
\end{array}
\right]}
$$
For this simple problem, we can eyeball the $\color{red}{null}$ space vectors.
$$
 \mathbf{U} =
\left[
\begin{array}{ccc}
%
\frac{1}{\sqrt{2}}
\color{blue}{\left[
\begin{array}{r}
 1 \\ 0 \\ -1
\end{array}
\right]} &
%
\frac{1}{\sqrt{2}}
\color{red}{\left[
\begin{array}{r}
 1 \\ 0 \\ 1
\end{array}
\right]} &
%
\color{red}{\left[
\begin{array}{c}
 0 \\ 1 \\ 0
\end{array}
\right]}
%
\end{array}
\right]
$$
Pseudoinverse
The pseudoinverse matrix is
$$
 \mathbf{A}^{+} = \mathbf{V} \, \Sigma^{+} \mathbf{U}^{*} =
\frac{1}{4}
\left[
\begin{array}{ccc}
 1 & 1 & 0 \\
 1 & 1 & 0 \\
\end{array}
\right]
$$
Least squares solution
Equation $(1)$ provides
$$
 x_{LS} = 
\color{blue}{\mathbf{A}^{+} b} +
\color{red}{ 
\left(
\mathbf{I}_{n} - \mathbf{A}^{+} \mathbf{A}
\right) y} =
\color{blue}{
\frac{3}{4}
\left[
\begin{array}{ccc}
 1 \\
 1 \\
\end{array}
\right]
}
+
\color{red}{
\left[
\begin{array}{rr}
  1 & -1 \\
 -1 &  1 \\
\end{array}
\right] y
}
, \quad y \in \mathbb{C}^{2}
$$
Plot
The following plot shows the least squares merit function $\lVert
 \mathbf{A} x - b
\rVert_{2}^{2}
$ as a function of $x$. The white dot represents $\color{blue}{x_{LS}}$, the dashed, yellow line $\color{red}{x_{LS}}$.

