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

Linear algebra problems and solutions

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 $\mathbf{b}$.

 
LoadingAdd to solve later

Sponsored Links

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.


LoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Linear Algebra
Problems and solutions in Linear Algebra
Prove $\mathbf{x}^{\trans}A\mathbf{x} \geq 0$ and determine those $\mathbf{x}$ such that $\mathbf{x}^{\trans}A\mathbf{x}=0$

For each of the following matrix $A$, prove that $\mathbf{x}^{\trans}A\mathbf{x} \geq 0$ for all vectors $\mathbf{x}$ in $\R^2$. Also, determine...

Close