# The Matrix $[A_1, \dots, A_{n-1}, A\mathbf{b}]$ is Always Singular, Where $A=[A_1,\dots, A_{n-1}]$ and $\mathbf{b}\in \R^{n-1}$.

## Problem 560

Let $A$ be an $n\times (n-1)$ matrix and let $\mathbf{b}$ be an $(n-1)$-dimensional vector.

Then the product $A\mathbf{b}$ is an $n$-dimensional vector.

Set the $n\times n$ matrix $B=[A_1, A_2, \dots, A_{n-1}, A\mathbf{b}]$, where $A_i$ is the $i$-th column vector of $A$.

Prove that $B$ is a singular matrix for any choice of $\mathbf{b}$.

## Definition/Hint.

An $n\times n$ matrix $B$ is **nonsingular** if $A\mathbf{x}=\mathbf{0}, \mathbf{x}\in \R^n$ implies that $\mathbf{x}=\mathbf{0}$.

Otherwise, the matrix $B$ is called **singular**.

Namely, the matrix $B$ is singular if there exists a nonzero vector $\mathbf{v}$ such that $B\mathbf{v}=\mathbf{0}$.

You may use the fact that a matrix is nonsingular if and only if its column vectors are lienarly independent.

We give two proofs. The first one uses the definition of a singular matrix. The second one uses the fact mentioned above.

## Proof (Using Defition).

Let

\[\mathbf{b}=\begin{bmatrix}

b_1 \\

b_2 \\

\vdots \\

b_{n-1}

\end{bmatrix}.\]
Then we have

\[A\mathbf{b}=b_1A_1+b_2A_2+\cdots+b_{n-1}A_{n-1} \tag{*}\]
using the column vectors $A_i$ of $A$.

Let us define the $n$-dimensional vector $\mathbf{v}$ to be

\[\mathbf{v}=\begin{bmatrix}

b_1 \\

b_2 \\

\vdots \\

b_{n-1}\\

-1

\end{bmatrix}.\]

Since the last entry of $\mathbf{v}$ is $-1$, the vector $\mathbf{v}$ is nonzero.

We calculate

\begin{align*}

B\mathbf{v}&=[A_1, A_2, \dots, A_{n-1}, A\mathbf{b}] \begin{bmatrix}

b_1 \\

b_2 \\

\vdots \\

b_{n-1}\\

-1

\end{bmatrix}\\[6pt]
&=b_1A_1+b_2A_2+\cdots+b_{n-1}A_{n-1}+(-1)A\mathbf{b}\\[6pt]
&\stackrel{(*)}{=} b_1A_1+b_2A_2+\cdots+b_{n-1}A_{n-1}-(b_1A_1+b_2A_2+\cdots+b_{n-1}A_{n-1})\\

&=\mathbf{0},

\end{align*}

where $\mathbf{0}$ is the $n$-dimensional zero vector.

Since we have $B\mathbf{v}=\mathbf{0}$ for a nonzero vector $\mathbf{v}$, we conclude that the matrix $B$ is singular.

## Proof (Using the Fact).

Note that a square matrix is singular if and only if its columns vectors are linearly dependent.

We prove that the column vectors $A_1, A_2, \cdots, A_{n-1}, A\mathbf{b}$ of the matrix $B$ are linearly dependent.

Let

\[\mathbf{b}=\begin{bmatrix}

b_1 \\

b_2 \\

\vdots \\

b_{n-1}

\end{bmatrix}.\]
Then we have

\[A\mathbf{b}=b_1A_1+b_2A_2+\cdots+b_{n-1}A_{n-1}.\]

Thus, we have the linear combination of column vectors of $B$

\[b_1A_1+b_2A_2+\cdots+b_{n-1}A_{n-1}-A\mathbf{b}=\mathbf{0}.\]
Since the coefficient in front of the vector $A\mathbf{b}$ is $-1$, the left hand side is a nontrivial linear combination of column vectors of $B$.

This implies that the column vectors of $B$ are linearly dependent, hence the matrix $B$ is singular.

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