Finding all left inverses of a matrix
I have to find all left inverses of a matrix
$$A = \begin{bmatrix} 2&-1 \\ 5 & 3\\ -2& 1 \end{bmatrix}$$
I created a matrix to the left of $A$,
$$\begin{bmatrix} a &b &c \\ d &e &f \end{bmatrix} \begin{bmatrix} 2&-1 \\ 5 & 3\\ -2& 1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}$$
and I got the following system of equations:
\begin{array} {lcl} 2a+5b-2c & = & 1 \\-a+3b+c & = & 0 \\ 2d+5e-2f & = & 0 \\ -d+3e+f & = & 1 \end{array}
After this step, I am unsure how to continue or form those equations into a solvable matrix, and create a left inverse matrix from the answers of these equations.
$\endgroup$ 03 Answers
$\begingroup$Find the left inverse for the rank $\rho = 2$ matrix $$ \mathbf{A} = \left[ \begin{array}{rr} 2 & -1 \\ 5 & 3 \\\ -2 & 1 \end{array} \right] $$ The approach is straightforward. Compute the singular value decomposition $$ \mathbf{A} = \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \tag{1} $$ and use this to construct the Moore-Penrose pseudoinverse matrix $$ \mathbf{A}^{+} = \mathbf{V} \, \Sigma^{+} \, \mathbf{C}^{*} \tag{2} $$
How do we know the pseudoinverse matrix is a left inverse? See generalized inverse of a matrix and convergence for singular matrix, What forms does the Moore-Penrose inverse take under systems with full rank, full column rank, and full row rank?
The singular value decomposition is completed using the recipe for the row space in this post: SVD and the columns — I did this wrong but it seems that it still works, why?
Compute the SVD
1. Construct product matrix$$ \begin{align} \mathbf{W} &= \mathbf{A}^{*} \mathbf{A}\\ &= % A* \left[ \begin{array}{rcr} 2 & 5 & -1 \\ -1 & 3 & 2 \\ \end{array} \right] % A \left[ \begin{array}{rr} 2 & -1 \\ 5 & 3 \\ -2 & 1 \\ \end{array} \right] \\ % W &= \left[ \begin{array}{cc} 33 & 11 \\ 11 & 11 \\ \end{array} \right] \end{align} $$
2. Solve for eigenvalues
The characteristic polynomial is given by $$ p(\lambda) = \lambda^{2} - \lambda \text{ tr }\mathbf{W} + \det \mathbf{W} $$ The trace and determinant are $$ \text{tr }\mathbf{W} = 44, \qquad \det \mathbf{W} = (33-11)11=242 $$ Therefore $$ p(\lambda) = \lambda^{2} - 44 \lambda + 242 $$ The roots are the eigenvalues: $$ \lambda \left( \mathbf{W} \right) = \left\{ 11 \left(2 + \sqrt{2}\right), 11 \left(2-\sqrt{2}\right) \right\} $$ The matrix of singular values is $$ \mathbf{S} = \sqrt{11} \left( \begin{array}{cc} \sqrt{2 + \sqrt{2}} & 0 \\ 0 & \sqrt{2 - \sqrt{2}} \\ \end{array} \right) $$ The sabot matrix is $$ \Sigma = \left[ \begin{array}{c} \mathbf{S} \\ \mathbf{0} \\ \end{array} \right] = \sqrt{11} \left[ \begin{array}{cc} \sqrt{2 + \sqrt{2}} & 0 \\ 0 & \sqrt{2 - \sqrt{2}} \\ 0 & 0 \\ \end{array} \right] $$
3. Solve for eigenvectors
First eigenvector
Solving $$ \begin{align} \left( \mathbf{W} - \lambda_{1} \mathbf{I}_{2} \right) w_{1} &= \mathbf{0} \\ % \left[ \begin{array}{cc} 33-11 \left(2+\sqrt{2}\right) & 11 \\ 11 & 11-11 \left(2+\sqrt{2}\right) \\ \end{array} \right] % \left[ \begin{array}{cc} w_{x} \\ w_{y} \\ \end{array} \right] % &= % \left[ \begin{array}{cc} 0 \\ 0 \\ \end{array} \right] % \end{align} $$ produces $$ w_{1} = \left[ \begin{array}{cc} 1+\sqrt{2} \\ 1 \\ \end{array} \right] $$
Second eigenvector
Solving $$ \begin{align} \left( \mathbf{W} - \lambda_{2} \mathbf{I}_{2} \right) w_{2} &= \mathbf{0} \\ % \left[ \begin{array}{cc} 33-11 \left(2-\sqrt{2}\right) & 11 \\ 11 & 11-11 \left(2-\sqrt{2}\right) \\ \end{array} \right] % \left[ \begin{array}{cc} w_{x} \\ w_{y} \\ \end{array} \right] % &= % \left[ \begin{array}{cc} 0 \\ 0 \\ \end{array} \right] % \end{align} $$ produces $$ w_{2} = \left[ \begin{array}{cc} 1 - \sqrt{2} \\ 1 \\ \end{array} \right] $$
4. Assemble the domain matrix for the row space$$ \begin{align} \mathbf{V} &= \left[ \begin{array}{cc} \frac{w_{1}}{\lVert w_{1} \rVert} & \frac{w_{2}}{\lVert w_{2} \rVert} \\ \end{array} \right] \\ &= \left[ \begin{array}{cc} % c1 \left( 2 \left(2 + \sqrt{2} \right) \right)^{-\frac{1}{2}} \left[ \begin{array}{c} 1 + \sqrt{2} \\ 1 \end{array} \right] & % c2 \left( 2 \left(2 - \sqrt{2} \right) \right)^{-\frac{1}{2}} \left[ \begin{array}{c} 1 - \sqrt{2} \\ 1 \end{array} \right] \end{array} \right] \end{align} $$
5. Compute the domain matrix for the column space
Compute the column vectors for $k=1,2$: $$ \mathbf{U}_{k} = \sigma^{-1}_{k} \mathbf{A} \mathbf{V}_{k} $$
$$ \begin{align} \mathbf{U} &= \left[ \begin{array}{ccc} % c1 \left( \sqrt{22} \left(\sqrt{2}+2\right) \right)^{-1} \left[ \begin{array}{r} 1 + 2\sqrt{2} \\ 8 + 5\sqrt{2} \\ -1 - 2\sqrt{2} \\ \end{array} \right] & % c2 \left( \sqrt{22} \left(\sqrt{2} - 2\right) \right)^{-1} \left[ \begin{array}{r} 1 - 2\sqrt{2} \\ 8 - 5\sqrt{2} \\ -1 + 2\sqrt{2} \\ \end{array} \right] & % c3 \left( 2 \right)^{-\frac{1}{2}} \color{gray}{\left[ \begin{array}{r} 1 \\ 0 \\ -1 \\ \end{array} \right]} % \end{array} \right] \end{align} $$ The third column vector, the gray null space vector, was added by inspection.
Construct the pseudoinverse
Following the prescription in $(2)$ $$ \mathbf{A}^{+} = \mathbf{V} \, \Sigma^{+} \, \mathbf{U}^{*} = \left[ \begin{array}{rcr} 3 & 2 & -3 \\ -5 & 4 & 5 \\ \end{array} \right] $$
Verify pseudoinverse
Left inverse
$$ \mathbf{A}^{+}\mathbf{A} = \left[ \begin{array}{rcr} 3 & 2 & -3 \\ -5 & 4 & 5 \\ \end{array} \right] \left[ \begin{array}{rr} 2 & -1 \\ 5 & 3 \\ -2 & 1 \\ \end{array} \right] = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right] = \mathbf{I}_{2} $$ Therefore, $\mathbf{A}^{+}$ is a left inverse.
$$ \mathbf{A} \mathbf{A}^{+} = \left[ \begin{array}{rr} 2 & -1 \\ 5 & 3 \\ -2 & 1 \\ \end{array} \right] \left[ \begin{array}{rcr} 3 & 2 & -3 \\ -5 & 4 & 5 \\ \end{array} \right] = \frac{1}{2} \left[ \begin{array}{rcr} 1 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 1 \\ \end{array} \right] \ne \mathbf{I}_{3} $$ Therefore, $\mathbf{A}^{+}$ is not a right inverse.
$\endgroup$ $\begingroup$Hint:
solve the first two equations using $c$ as a parameter and find $b=\frac1{11}$ , $a=c+\frac3{11}$
do the same in last two equation and you have the result.
$\endgroup$ $\begingroup$Okay. I got the following matrix as my final answer as the general form of all left inverses of A:
\begin{bmatrix} a & \frac1{11} &a-\frac3{11} \\ d & \frac2{11} &d+\frac5{11} \end{bmatrix}
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