# Properties of Nonsingular and Singular Matrices

## Problem 25

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

Otherwise $A$ is called

*.*

**singular****(a)** Show that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular.

**(b)** Show that if $A$ is nonsingular, then the column vectors of $A$ are linearly independent.

**(c)** Show that an $n \times n$ matrix $A$ is nonsingular if and only if the equation $A\mathbf{x}=\mathbf{b}$ has a unique solution for any vector $\mathbf{b}\in \R^n$.

**Restriction**

Do not use the fact that a matrix is nonsingular if and only if the matrix is invertible.

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Contents

- Problem 25
- Proof.
- (a) Show that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular.
- (b) Show that if $A$ is nonsingular, then the column vectors of $A$ are linearly independent.
- (c) Show that $A$ is nonsingular if and only if $A\mathbf{x}=\mathbf{b}$ has a unique solution for any $\mathbf{b}\in \R^n$.

- Related Question.

## Proof.

### (a) Show that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular.

Consider the equation $AB\mathbf{x}=\mathbf{0}$

Let $\mathbf{y}=B\mathbf{x}$. Then we have $A\mathbf{y}=\mathbf{0}$.

Then $\mathbf{y}$ must be the zero vector since $A$ is nonsingular.

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

Since $B$ is nonsingular, we must have $\mathbf{x}=\mathbf{0}$.

Hence the only solution to $AB\mathbf{x}=\mathbf{0}$ is the zero vector, and we conclude that $AB$ is nonsingular.

### (b) Show that if $A$ is nonsingular, then the column vectors of $A$ are linearly independent.

Suppose that $A$ is an $n\times n$ nonsingular matrix. Let $A_i$ be the $i$-th column vector of $A$ for $i=1,2,\dots, n$.

Suppose that we have a linear combination

\[x_1 A_1+x_2 A_2+\cdots +x_n A_n=\mathbf{0} \tag{*}.\]
This equation can be written as

\[ A\mathbf{x}=\mathbf{0},\]
where $A=[A_1A_2\dots A_n]$ and $\mathbf{x}=\begin{bmatrix}

x_1 \\

x_2 \\

\vdots\\

x_n

\end{bmatrix}$.

Since $A$ is nonsingular, we must have $\mathbf{x}=\mathbf{0}$, hence $x_1=x_2=\cdots =x_n=0$.

Thus only solution to (*) is zero coefficients. Thus the vectors $A_i$ are linearly independent.

### (c) Show that $A$ is nonsingular if and only if $A\mathbf{x}=\mathbf{b}$ has a unique solution for any $\mathbf{b}\in \R^n$.

#### $(\Leftarrow)$

If $A\mathbf{x}=\mathbf{b}$ has a unique solution for any $\mathbf{b}$, then choose $\mathbf{b}=\mathbf{0}$.

Then $A\mathbf{x}=\mathbf{0}$ has a unique solution.

Note that this equation has the trivial solution $\mathbf{x}=\mathbf{0}$.

Thus $\mathbf{x}=\mathbf{0}$ is the unique solution of $A\mathbf{x}=\mathbf{0}$, hence $A$ is nonsingular.

#### $(\Rightarrow)$

Assume that $A$ is nonsingular. Let $\mathbf{b}$ be a vector in $\R^n$ and consider the equation $A\mathbf{x}=\mathbf{b}$.

We prove that

- the existence of a solution, and
- the uniqueness of the solution.

#### 1. The existence of a solution

We show that $A\mathbf{x}=\mathbf{b}$ has at least one solution .

Let us write $A=[A_1A_2\dots A_n]$, where $A_i$ is the $i$-th column vector of the matrix $A$.

Then the set $\{A_1, A_2, \dots, A_n, \mathbf{b} \}$ contains $n+1$ vectors in $R^n$. Thus the set must be linearly dependent.

So there exist $c_i$, $i=1,2, \dots, n+1$, such that not all of $c_i$ are zero and we have

\[c_1A_1+c_2 A_2+\cdots c_n A_n+c_{n+1} \mathbf{b}=\mathbf{0} \tag{**}.\]
We claim that $c_{n+1} \neq 0$. If $c_{n+1}=0$, then we have

\[c_1A_1+c_2 A_2+\cdots c_n A_n=\mathbf{0}\]
and by part (b), the vectors $A_i$ are linearly independent, hence $c_i=0$ for all $i$.

This contradicts the choice of $c_i$. Thus we must have $c_{n+1}\neq 0$.

Since $c_{n+1}\neq 0$, we divide (**) by $c_{n+1}$ and solve for $\mathbf{b}$, and obtain

\[ \mathbf{b}=-\frac{c_1}{c_{n+1}} A_1 – \cdots -\frac{c_n}{c_{n+1}}A_n.\]
Therefore the equation $A\mathbf{x}=\mathbf{b}$ has solution

\[x_i=-\frac{c_i}{c_{n+1}}.\]

#### 2. The uniqueness of the solution

Now we prove that $A\mathbf{x}=\mathbf{b}$ has a unique solution.

Suppose that $\mathbf{x}_1$ and $\mathbf{x}_2$ are solutions.

Thus we have

\[A \mathbf{x}_1=\mathbf{b}=A\mathbf{x}_2.\]
Then we have $A(\mathbf{x}_1-\mathbf{x}_2)=\mathbf{0}$.

Since $A$ is nonsingular, we must have $\mathbf{x}_1-\mathbf{x}_2=\mathbf{0}$, hence $\mathbf{x}_1=\mathbf{x}_2$.

This proves the uniqueness of the solution.

## Related Question.

The converse statment of (a) is also true:

See the post ↴

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

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## 1 Response

[…] 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 […]