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

Problems and solutions in Linear Algebra

Problem 561

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

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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
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,
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.

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