# The Product of Two Nonsingular Matrices is Nonsingular

## Problem 479

Prove that if $n\times n$ matrices $A$ and $B$ are nonsingular, then the product $AB$ is also a nonsingular matrix.

(The Ohio State University, Linear Algebra Final Exam Problem)

## Definition (Nonsingular Matrix)

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

## Proof.

We give two proofs. The first one uses a property of the determinants of matrices, and the second one uses the definition of nonsingular matrices.

### Proof 1. (Using Determinant)

Recall that a matrix is nonsingular if and only if its determinant is not zero.

Since $A$ and $B$ are nonsingular, we know that
$\det(A)\neq 0 \text{ and } \det(B) \neq 0.$ Then we have using the multiplicative property of the determinant
\begin{align*}
\det(AB)=\det(A)\det(B)\neq 0.
\end{align*}

Since the determinant of the product $AB$ is not zero, we conclude that $AB$ is a nonsingular matrix.

### Proof 2. (Using Definition of Nonsingular Matrices)

Suppose that $A, B$ are nonsingular matrices.
This means that if $A\mathbf{x}=\mathbf{0}$ for some the vector $\mathbf{x}\in \R^n$, then we must have $\mathbf{x}=\mathbf{0}$.
Same for $B$.

Suppose that we have $(AB)\mathbf{x}=\mathbf{0}$ for some vector $\mathbf{x}\in \R^n$.
Let $\mathbf{v}=B\mathbf{x}\in \R^n$. Then we have
$A\mathbf{v}=AB\mathbf{x}=\mathbf{0}.$ So since $A$ is a nonsingular matrix, we have $\mathbf{v}=\mathbf{0}$, namely, $B\mathbf{x}=\mathbf{0}$.

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

In summary, whenever $(AB)\mathbf{x}=\mathbf{0}$, we have $\mathbf{x}=\mathbf{0}$.
Therefore, the matrix $AB$ is nonsingular.

## Final Exam Problems and Solution. (Linear Algebra Math 2568 at the Ohio State University)

This problem is one of the final exam problems of Linear Algebra course at the Ohio State University (Math 2568).

The other problems can be found from the links below.

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