# Two Matrices are Nonsingular if and only if the Product is Nonsingular

## Problem 562

An $n\times n$ matrix $A$ is called **nonsingular** if the only vector $\mathbf{x}\in \R^n$ satisfying the equation $A\mathbf{x}=\mathbf{0}$ is $\mathbf{x}=\mathbf{0}$.

Using the definition of a nonsingular matrix, prove the following statements.

**(a)** If $A$ and $B$ are $n\times n$ nonsingular matrix, then the product $AB$ is also nonsingular.

**(b)** Let $A$ and $B$ be $n\times n$ matrices and suppose that the product $AB$ is nonsingular. Then:

- The matrix $B$ is nonsingular.
- The matrix $A$ is nonsingular. (You may use the fact that a nonsingular matrix is invertible.)

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## Proof.

### (a) If $A$ and $B$ are $n\times n$ nonsingular matrix, then the product $AB$ is also nonsingular.

Let $\mathbf{v}\in \R^n$ and suppose that $(AB)\mathbf{x}=\mathbf{0}$.

Our goal is to prove that $\mathbf{x}=\mathbf{0}$.

Let $\mathbf{y}:=B\mathbf{x} \in \R^n$. Then we have

\begin{align*}

A\mathbf{y}=A(B\mathbf{x})=(AB)\mathbf{x}=\mathbf{0}.

\end{align*}

Since $A$ is nonsingular, this implies that the vector $\mathbf{y}=\mathbf{0}$.

Hence we have $\mathbf{y}=B\mathbf{x}=\mathbf{0}$.

Since $B$ is nonsingular, this further implies that $\mathbf{x}=\mathbf{0}$.

It follows that if $(AB)\mathbf{x}=\mathbf{0}$, then we must have $\mathbf{x}=\mathbf{0}$.

By definition, this means that the matrix $AB$ is nonsingular.

### (b)-1. If $AB$ is nonsingular, then $B$ is nonsingular.

Suppose that $B\mathbf{x}=\mathbf{0}$. We prove that $\mathbf{x}=\mathbf{0}$.

Since $B\mathbf{x}=\mathbf{0}$, it yields that

\begin{align*}

(AB)\mathbf{x}=A(B\mathbf{x})=A\mathbf{0}=\mathbf{0}.

\end{align*}

As the matrix $AB$ is nonsingular, it follows from $(AB)\mathbf{x}=\mathbf{0}$ that $\mathbf{x}=\mathbf{0}$.

This proves that the matrix $B$ is nonsingular.

### (b)-2. If $AB$ is nonsingular, then $A$ is nonsingular.

By part (1), we know that $B$ is nonsingular, hence it is invertible.

The inverse matrix $B^{-1}$ and the matrix $AB$ are both nonsingular.

Hence it follows from part (a) that the product of $AB$ and $B^{-1}$ is also nonsingular.

Thus,

\[A=(AB)B^{-1}\]
is a nonsingular matrix.

## Nonsingular if and only if Invertible

For the proof of the fact we used in the proof of (b)-2 that a matrix is nonsingular if and only if it is invertible, see the post↴

A Matrix is Invertible If and Only If It is Nonsingular

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