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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

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Well, you know, when $A$ has linearly independent columns, $A^+ = (A^\ast A)^{-1}A^\ast$. So ...

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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 $$ \mathbf{A} = \mathbf{V} \, \Sigma^{+} \, \color{blue}{\mathbf{U}_{\mathcal{R}}} = \left( a^{2} + b^{2} \right)^{\frac{1}{2}} \mathbf{A}^{*} $$

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