Question

Consider an $\mathit{m}\mathbf{\times }\mathit{n}$ matrix A with$\mathit{K}\mathit{e}\mathit{r}\mathbf{}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\overrightarrow{\mathbf{0}}\mathbf{\right\}}$. Show that there exists an$\mathit{n}\mathbf{\times }\mathit{m}$ matrix B such that$\mathit{B}\mathit{A}\mathbf{=}{\mathit{l}}_{\mathbf{n}}$.

Short answer

The matrix is ${\left(A^{T}A\right)}^{-1}-A^T$.

Step-by-step solution

Determine the matrix R.

Consider a$m\times n$ matrix A.

If $\mathit{K}\mathit{e}\mathit{r}\mathbf{}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\overrightarrow{\mathbf{0}}\mathbf{\right\}}$ for arole="math" localid="1659500428667" $m\times n$ matrix A then the matrix A is invertible.

If the matrix A is invertible then the matrixrole="math" localid="1659500442729" $A^T$ is invertible.

If the matrices A and B is invertible then the matrix AB is invertible.

As the value of $\mathrm{Ke}r\left(A\right)=\left\{\overrightarrow{0}\right\}$, by the definition the matrix A is invertible.

By the definition, the matrices$A^T$ and$A^{T}A$ are invertible.

Assume the matrix $BA={\left(A^{T}A\right)}^{-1}$, simplify the matrix BA as follows.

$\begin{array}{l}BA=A^{-1}\left\{l_n\right\}A\\ =A^{-1}{\left(A^T\right)}^{-1}{A}^{T}A\\ =A^{-1}\left\{{\left(A^T\right)}^{-1}{A}^T\right\}A\\ BA=A^{-1}\left(l_n\right)A\end{array}$

Further, simplify the equation as follows.

$\begin{array}{l}BA=A^{-1}\left\{l_n\right\}A\\ =A^{-1}A\\ BA=l_n\end{array}$

Hence, for the matrix$B={\left(A^{T}A\right)}^{-1}-A^T$ the equation $BA=l_n$.