# A Matrix is Invertible If and Only If It is Nonsingular

## Problem 26

In this problem, we will show that the concept of non-singularity of a matrix is equivalent to the concept of invertibility.

That is, we will prove that:

**(a)**Show that if $A$ is invertible, then $A$ is nonsingular.

**(b)**Let $A, B, C$ be $n\times n$ matrices such that $AB=C$.

Prove that if either $A$ or $B$ is singular, then so is $C$.

**(c)**Show that if $A$ is nonsingular, then $A$ is invertible.

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## Definition (Nonsingular Matrix)

An $n \times n$ matrix $A$ is called **nonsingular** if the equation $A\mathbf{x}=\mathbf{0}$ has only the zero solution $\mathbf{x}=\mathbf{0}$.

For basic properties of a nonsingular matrix, see the problem Properties of nonsingular and singular matrices.

The result of this problem will be used in the proof below.

## Proof.

### (a) If $A$ is invertible, then $A$ is nonsingular

Suppose that $A$ is invertible. This means that we have the inverse matrix $A^{-1}$ of $A$.

Consider the equation $A\mathbf{x}=\mathbf{0}$. We show that this equation has only zero solution.

Multiplying it by $A^{-1}$ on the left, we obtain \begin{align*}

A^{-1}A\mathbf{x}&=A^{-1}\mathbf{0}\\

\Rightarrow \mathbf{x}=\mathbf{0}

\end{align*}

Hence $A$ is nonsingular.

### (b) If either $A$ or $B$ is singular, then so is $C$

Suppose first that the matrix $B$ is singular.

Then there exists nonzero vector $\mathbf{b}\neq\mathbf{0}$ such that $B \mathbf{b}=\mathbf{0}$. Then we have $C\mathbf{b}=AB\mathbf{b}=A\mathbf{0}=\mathbf{0}$.

Since $\mathbf{b}$ is a nonzero vector, the matrix $C$ is singular.

Next, we assume that $B$ is nonsingular and $A$ is singular.

Since $A$ is singular, there exists nonzero vector $\mathbf{y_0}$ such that $A \mathbf{y}_0=\mathbf{0}$.

Then consider the equation $B\mathbf{x}=\mathbf{y}_0$ has a unique solution $\mathbf{x}_0$ by part (c) of Properties of nonsingular and singular matrices.

The vector $\mathbf{x}_0$ is nonzero because $\mathbf{y}_0$ is nonzero.

Then we have

\begin{align*}

C\mathbf{x}_0 =AB \mathbf{x}_0= A \mathbf{y}_0=\mathbf{0}.

\end{align*}

Therefore $C \mathbf{x}=\mathbf{0}$ has nonzero solution $\mathbf{x}_0$, hence $C$ is singular.

### (c) If $A$ is nonsingular, then $A$ is invertible

Suppose that $A$ is nonsingular. Let $\mathbf{e}_i$ be the $n$-dimensional vector whose entries are all $0$ but $1$ in the $i$th place.

Then the equations $A\mathbf{x}=\mathbf{e}_i$ has a unique solution $\mathbf{x}_i$ for $i=1, \dots, n$.

Create a matrix $B$ whose $i$-th column vector is $\mathbf{x}_i$, namely $B=[\mathbf{x}_1 \mathbf{x}_2 \dots \mathbf{x}_n]$.

Then we have $AB=[\mathbf{e}_1\mathbf{e}_2\dots \mathbf{e}_n]=I_n$.

Hence $B$ is the right inverse of $A$.

Note that the identity matrix $I_n$ is nonsingular. Thus by part (c) of Properties of nonsingular and singular matrices, the matrix $B$ must be nonsingular as well.

We repeat the above argument using $B$ instead of $A$. Then there exist a matrix $C$ such that $BC=I_n$. We claim that $C=A$.

To see this, multiply $AB=I_n$ by $C$ on the right, we get $ABC=C$.

Since $BC=I_n$, we get $A=C$.

In summary, we obtained $AB=I_n$ and $BA=I_n$. Thus $A$ is invertible with the inverse $B$.

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