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Expert-verified Found in: Page 246 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Consider an ${\mathbit{m}}{\mathbf{×}}{\mathbit{n}}$ matrix A with ${\mathbit{K}}{\mathbit{e}}{\mathbit{r}}{\mathbf{}}{\mathbf{\left(}}{\mathbit{A}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{\left\{}}\stackrel{\mathbf{\to }}{\mathbf{0}}{\mathbf{\right\}}}$. Show that there exists an ${\mathbit{n}}{\mathbf{×}}{\mathbit{m}}$ matrix B such that ${\mathbit{B}}{\mathbit{A}}{\mathbf{=}}{{\mathbit{l}}}_{{\mathbf{n}}}$.

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

See the step by step solution

## Step 1: Determine the matrix R.

Consider a $m×n$ matrix A.

If ${\mathbit{K}}{\mathbit{e}}{\mathbit{r}}{\mathbf{}}{\mathbf{\left(}}{\mathbit{A}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{\left\{}}\stackrel{\mathbf{\to }}{\mathbf{0}}{\mathbf{\right\}}}$ for a role="math" localid="1659500428667" ${m}{×}{n}$ matrix A then the matrix A is invertible.

If the matrix A is invertible then the matrix role="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 $\mathrm{Ke}r\left(A\right)=\left\{\stackrel{\to }{0}\right\}$, 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 $BA={l}_{n}$. ### Want to see more solutions like these? 