Let $A$ be an $n\times n$ singular matrix.
Then prove that there exists a nonzero $n\times n$ matrix $B$ such that
\[AB=O,\]
where $O$ is the $n\times n$ zero matrix.

Recall that an $n \times n$ matrix $A$ is called singular if the equation
\[A\mathbf{x}=\mathbf{0}\]
has a nonzero solution $\mathbf{x}\in \R^n$.

Proof.

Since $A$ is singular, there exists a nonzero vector $\mathbf{b} \in \R^n$ such that
\[A\mathbf{b}=\mathbf{0}.\]

Then we define the $n\times n$ matrix $B$ whose first column is the vector $\mathbf{b}$ and the other entries are zero. That is
\[B=\begin{bmatrix}
\mathbf{b} & \mathbf{0} & \cdots & \mathbf{0}
\end{bmatrix}.\]
Since the vector $\mathbf{b}$ is nonzero, the matrix $B$ is nonzero.

With this choice of $B$, we have
\begin{align*}
AB&=A\begin{bmatrix}
\mathbf{b} & \mathbf{0} & \cdots & \mathbf{0}
\end{bmatrix} \\
&=\begin{bmatrix}
A\mathbf{b} & A\mathbf{0} & \cdots & A\mathbf{0}
\end{bmatrix}\\
&=\begin{bmatrix}
\mathbf{0} & \mathbf{0} & \cdots & \mathbf{0}
\end{bmatrix}
=O.
\end{align*}
Hence we have proved that $AB=O$ with the nonzero matrix $B$.

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