Short Answer

Consider an $m \times n$ matrix A with $K e r (A) = {\vec{0}}$ . Show that there exists an $n \times m$ matrix B such that $B A = l_{n}$ .

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

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Step by Step Solution

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TABLE OF CONTENTS

Step 1: Determine the matrix R.

Consider a $m \times n$ matrix A.

If $K e r (A) = {\vec{0}}$ for a role="math" localid="1659500428667" $m \times n$ 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 $Ke r (A) = {\vec{0}}$ , by the definition the matrix A is invertible.

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

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

$\begin{array}{l} B A = A^{- 1} \{l_{n}\} A \\ = A^{- 1} {(A^{T})}^{- 1} A^{T} A \\ = A^{- 1} \{{(A^{T})}^{- 1} A^{T}\} A \\ B A = A^{- 1} (l_{n}) A \end{array}$

Further, simplify the equation as follows.

$\begin{array}{l} B A = A^{- 1} \{l_{n}\} A \\ = A^{- 1} A \\ B A = l_{n} \end{array}$

Hence, for the matrix $B = {(A^{T} A)}^{- 1} - A^{T}$ the equation $B A = l_{n}$ .

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Linear Algebra With Applications

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Short Answer

Consider an $m \times n$ matrix A with $K e r (A) = {\vec{0}}$ . Show that there exists an $n \times m$ matrix B such that $B A = l_{n}$ .

Step by Step Solution

Step 1: Determine the matrix R.

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Linear Algebra With Applications

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Short Answer

Consider an m×n matrix A with Ker (A)={0→}. Show that there exists an n×m matrix B such that BA=ln.

Step by Step Solution

Step 1: Determine the matrix R.

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Consider an $m \times n$ matrix A with $K e r (A) = {\vec{0}}$ . Show that there exists an $n \times m$ matrix B such that $B A = l_{n}$ .