# A Singular Matrix and Matrix Equations $A\mathbf{x}=\mathbf{e}_i$ With Unit Vectors ## Problem 561

Let $A$ be a singular $n\times n$ matrix.
Let
$\mathbf{e}_1=\begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \mathbf{e}_2=\begin{bmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{bmatrix}, \dots, \mathbf{e}_n=\begin{bmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{bmatrix}$ be unit vectors in $\R^n$.

Prove that at least one of the following matrix equations
$A\mathbf{x}=\mathbf{e}_i$ for $i=1,2,\dots, n$, must have no solution $\mathbf{x}\in \R^n$. Add to solve later

## Proof.

Assume on the contrary that each matrix equation $A\mathbf{x}=\mathbf{e}_i$ has a solution.
Let $\mathbf{b}_i\in \R^n$ be a solution of $A\mathbf{x}=\mathbf{e}_i$ for each $i=1, \dots, n$.
That is, we have
$A\mathbf{b}_i=\mathbf{e}_i.$ Let $B=[\mathbf{b}_1, \mathbf{b}_2, \dots, \mathbf{b}_n]$ be the $n\times n$ matrix whose $i$-th column vector is $\mathbf{b}_i$.

Then we have
\begin{align*}
AB&=A[\mathbf{b}_1, \mathbf{b}_2, \dots, \mathbf{b}_n]\\[6pt] &=[A\mathbf{b}_1, A\mathbf{b}_2, \dots, A\mathbf{b}_n]\\[6pt] &=[\mathbf{e}_1, \mathbf{e}_2, \dots, \mathbf{e}_n]=I,
\end{align*}
where $I$ is the $n\times n$ identity matrix.

Since $I$ is the nonsingular matrix, the matrix $A$ must also be nonsingular.
However this contradicts the assumption that $A$ is singular.
It follows that at least one of the matrix equations $A\mathbf{x}=\mathbf{e}_i$ has no solution. Add to solve later

### More from my site

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

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

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...

Close