# 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:

1. The matrix $B$ is nonsingular.
2. The matrix $A$ is nonsingular. (You may use the fact that a nonsingular matrix is invertible.) Add to solve later

## 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 Add to solve later

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