Multiplying the submatrix of a right inverse

Question

I've generated a 3 by 5 matrix using Matlab:

$$ A=\begin{pmatrix} 21 & 3 & 4 & 4 & 17 \\ 23 & 7 & 25 & 11 & 1 \\ 4 & 14 & 24 & 23 & 22 \\ \end{pmatrix} $$

A right inverse to this matrix is

$$ A^{-1}=\begin{pmatrix} 0.0315 & 0.0191 & -0.0252\\ 0 & 0 & 0\\ -0.0300 & 0.0236 & 0.0221\\ 0 & 0 & 0\\ 0.0270 & -0.0292 & 0.0259\\ \end{pmatrix} $$

This inverse was found using this method: column 1 of the inverse is equal to the solution of $$ Ax=\begin{pmatrix}1\\0\\0\\\end{pmatrix}$$ Column 2 of the inverse is the solution to $$ Ax=\begin{pmatrix}0\\1\\0\\\end{pmatrix}$$ And so on.

Now, let B be a matrix made of the non-empty rows of $A^{-1}$. In this case,

$B=\begin{pmatrix} 0.0315 & 0.0191 & -0.0252\\ -0.0300 & 0.0236 & 0.0221\\ 0.0270 & -0.0292 & 0.0259\\ \end{pmatrix}$

Multiplying $BA$ gives the following

$ BA=\begin{pmatrix} 1.0000 & -0.1241 & 0.0000 & -0.2427 & 0.0000\\ 0.0000 & 0.3846 & 1.0000 & 0.6480 & 0.0000\\ 0.0000 & 0.2393 & 0.0000 & 0.3827 & 1.0000 \\ \end{pmatrix} $

As you can see, a submatrix of $BA$ is the identity matrix. What is going on? I've tried with multiple matrix and this result always shows up.

Answer 1

Think of $A$ as a linear map $T_A \colon \mathbb{R}^5 \rightarrow \mathbb{R}^3$. The matrix $A$ has rank $3$ so $T_A$ is onto. The inverse you have generated has two zero rows (the second and forth). This corresponds to the fact that if you restrict $T_A$ to $V = \operatorname{span} \{ e_1, e_3, e_5 \}$ then $T_A|_{V} \colon V \rightarrow \mathbb{R}^3$ is an isomorphism and the matrix $B$ (which is obtained from your "$A^{-1}$" by deleting the second and fourth row) represents the inverse map $\left( T_A|_{V} \right)^{-1}$ with respect to the basis $\mathcal{B}_2 = (e_1,e_2,e_3)$ of $\mathbb{R}^3$ and the basis $\mathcal{B}_1 = (e_1, e_3, e_5)$ of $V$. Let us call the full basis of $\mathbb{R}^3$ by $\mathcal{C} = (e_1,e_2,e_3,e_4,e_5)$.

We have:

$$ BA = \left[ \left( T_A|_{V} \right)^{-1} \right]^{\mathcal{B}_2}_{\mathcal{B}_1} [T_A]^{\mathcal{C}}_{\mathcal{B}_2} = \left[ \left( T_A|_{V} \right)^{-1} \circ T_A\right]^{\mathcal{C}}_{\mathcal{B}_1}. $$

But $\left( \left( T_A|_{V} \right)^{-1} \circ T_A \right)(v) = v$ for all $v \in V$ and in particular for $v = e_1,e_3,e_5$. Hence, the first, third and fifth columns of $BA$ will be $e_1,e_2,e_3$ which corresponds to your observation that a submatrix of $BA$ is the $3 \times 3$ identity matrix.