Numerical linear algebra (pseudoinverse of a matrix) Let $A$ be the matrix: 
$$\left(\begin{matrix}
\alpha I_{n}  \\
\beta I_{n}   
\end{matrix}\right)$$
where $\alpha,\beta\in\Bbb C$ are not both zero. Derive (a) the (reduced) QR factorization of $A$ and (b) the pseudoinverse of $A$.
Any help for the second question about pseudoinverse ?
Thanks in advance
NEW
I know that if 1) rank(A)=n then $A^{+} = (A^{T} A)^{-1} A^{T}$
and if 2) rank(A)=n=m then 
$A^{+} = A^{-1}$.
I use the 1) and I found :
$A^{+} = (a^{2} I_{n} + b^{2} I_{n})^{-1} $$\left(\begin{matrix}
\alpha I_{n}  \
\beta I_{n}   
\end{matrix}\right)$
note the second brackets is a matrix (1x2).
How could I solve this ?
Any help 
 A: Well, you know, when $A$ has linearly independent columns, $A^+ = (A^\ast A)^{-1}A^\ast$. So ...
A: Example $\mathbf{Q} \mathbf{R}$
$$
\begin{align}
  \mathbf{A} &= \mathbf{Q} \, \mathbf{R} \\
% A
\left[
\begin{array}{ccc}
 a & 0 & 0 \\
 0 & a  & 0 \\
 0 & 0 & a  \\
 b & 0 & 0 \\
 0 & b & 0 \\
 0 & 0 & b \\
\end{array}
\right]
%
&=
% Q
\left( a^{2}+b^{2} \right)^{-\frac{1}{2}}
\left[
\begin{array}{ccc}
 a & 0 & 0 \\
 0 & a & 0 \\
 0 & 0 & a  \\
 b & 0 & 0 \\
 0 & b & 0 \\
 0 & 0 & b \\
\end{array}
\right]
% R
\left( a^{2}+b^{2} \right)^{\frac{1}{2}}
\left[
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 1 \\
\end{array}
\right]
%
\end{align}
$$
Example SVD
The singular value decomposition will use both $\color{blue}{range}$ and $\color{red}{null}$ spaces.
$$
\begin{align}
  \mathbf{A} &= 
\left[ \begin{array}{c|c}
\color{blue}{\mathbf{U}_{\mathcal{R}}} & 
\color{red} {\mathbf{U}_{\mathcal{N}}}
\end{array} \right]
\, \Sigma \, \color{blue}{\mathbf{V}^{*}} \\
%
&=
% U
\left( a^{2}+b^{2} \right)^{-\frac{1}{2}}
\left[
\begin{array}{ccc|ccc}
 \color{blue}{0} & \color{blue}{0} & \color{blue}{a} & \color{red}{0} & \color{red}{0} & \color{red}{-a} \\
 \color{blue}{0} & \color{blue}{a} & \color{blue}{0} & \color{red}{0} & \color{red}{-a} & \color{red}{0} \\
 \color{blue}{a} & \color{blue}{0} & \color{blue}{0} & \color{red}{-a} & \color{red}{0} & \color{red}{0} \\
 \color{blue}{0} & \color{blue}{0} & \color{blue}{b} & \color{red}{0} & \color{red}{0} & \color{red}{b} \\
 \color{blue}{0} & \color{blue}{b} & \color{blue}{0} & \color{red}{0} & \color{red}{b} & \color{red}{0} \\
 \color{blue}{b} & \color{blue}{0} & \color{blue}{0} & \color{red}{b} & \color{red}{0} & \color{red}{0} \\
\end{array}
\right]
% S
\left( a^{2}+b^{2} \right)^{\frac{1}{2}}
\left[
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 1 \\\hline
 0 & 0 & 0 \\
 0 & 0 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right]
% V
\left[
\begin{array}{ccc}
 \color{blue}{0} & \color{blue}{0} & \color{blue}{1} \\
 \color{blue}{0} & \color{blue}{1} & \color{blue}{0} \\
 \color{blue}{1} & \color{blue}{0} & \color{blue}{0} \\
\end{array}
\right]
%
\end{align}
$$
The pseudoinverse matrix is
$$
\begin{align}
  \mathbf{A}^{+} &= 
 \color{blue}{\mathbf{V}} \,
\Sigma^{+} 
\left[ \begin{array}{c}
\color{blue}{\mathbf{U}_{\mathcal{R}}^{*}} \\
\color{red} {\mathbf{U}_{\mathcal{N}}^{*}}
\end{array} \right]
%
=
\left( a^{2}+b^{2} \right)^{\frac{1}{2}}
\left[
\begin{array}{cccccc}
 a & 0 & 0 & b & 0 & 0 \\
 0 & a & 0 & 0 & b & 0 \\
 0 & 0 & a & 0 & 0 & b \\
\end{array}
\right]
%
\end{align}
$$

Derivation $\mathbf{Q}\, \mathbf{R}$
The $\mathbf{Q}\, \mathbf{R}$ is computationally cheaper than the SVD. It provides an orthonormal basis for the column space. The column vectors of the target matrix are already orthogonal; they just need normalization:
$$
 \mathbf{Q} = \left( a^{2} + b^{2} \right)^{-\frac{1}{2}} \mathbf{A}.
$$ 
We can bypass the usual Gram-Schmidt process.
The $\mathbf{R}$ matrix is upper-triangular. The the supradiagonal terms describe the projections of the column vectors of $\mathbf{Q}$ on the column vectors of $\mathbf{A}$. For example
$$
 \mathbf{R}_{1,2} = \mathbf{Q}^{*}_{1} \mathbf{A}_{2}
$$
Because the columns or $\mathbf{A}$ were already orthogonal, the supradiagonal terms are $0$. The diagonal terms hold the lengths of the column vectors of $\mathbf{A}$
$$
  \mathbf{Q}_{k,k} = \lVert \mathbf{A}_{k} \rVert, \quad k = 1, n.
$$
The target matrix $\mathbf{A}$ has column vectors of uniform length $\sqrt{a^{2} + b^{2}}$.
Derivation SVD
The SVD starts by resolving the eigensystem of the product matrix:
$$
\mathbf{A}^{*} \mathbf{A} = 
\left( a^{2} + b^{2} \right)
\left[
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 1 \\
\end{array}
\right]
$$
The eigenvalue spectrum is
$$
  \lambda_{k} = a^{2} + b^{2}, \quad k = 1, n
$$
There are no zero eigenvalues. There is no need to order the spectrum. We can harvest the singular values directly
$$
 \sigma_{k} = \sqrt{\lambda_{k}\left( \mathbf{A}^{*} \mathbf{A} \right)} 
= \sqrt{ a^{2} + b^{2} }, \quad k=1,n
$$
The algebraic multiplicity $(n)$ matches the geometric multiplicity for the eigenvalue. For the eigenvectors we choose the simplest set, the unit vectors in the identity matrix. Using the relationship 
$$
  \mathbf{A} \, \mathbf{V} = \mathbf{U} \, \Sigma
$$
we find that 
$$
\color{blue}{\mathbf{U}_{\mathcal{R}}} =  \left( a^{2} + b^{2} \right)^{-\frac{1}{2}} \mathbf{A}
$$
To summarize the SVD
$$
 \mathbf{V} = \mathbf{I}_{n}, \quad \Sigma = \left( a^{2} + b^{2} \right)^{\frac{1}{2}}
\left[ \begin{array}{c}
  \mathbf{I}_{n} \\
  \mathbf{0}
\end{array} \right], \quad 
\color{blue}{\mathbf{U}_{\mathcal{R}}} =  \left( a^{2} + b^{2} \right)^{-\frac{1}{2}} \mathbf{A}
$$
The psuedoinverse 
$$
 \mathbf{A} = \mathbf{V} \, \Sigma^{+} \, \color{blue}{\mathbf{U}_{\mathcal{R}}} = \left( a^{2} + b^{2} \right)^{\frac{1}{2}} \mathbf{A}^{*}
$$
