If a Matrix is the Product of Two Matrices, is it Invertible?

Problem 393

(a) Let $A$ be a $6\times 6$ matrix and suppose that $A$ can be written as
\[A=BC,\]
where $B$ is a $6\times 5$ matrix and $C$ is a $5\times 6$ matrix.

Prove that the matrix $A$ cannot be invertible.

(b) Let $A$ be a $2\times 2$ matrix and suppose that $A$ can be written as
\[A=BC,\]
where $B$ is a $ 2\times 3$ matrix and $C$ is a $3\times 2$ matrix.

(a) Prove that the matrix $A$ cannot be invertible.

Since $C$ is a $5 \times 6$ matrix, the equation
\[C\mathbf{x}=\mathbf{0}\]
has a nonzero solution $\mathbf{x}_1$.
(There are more variables than equations in the system $C\mathbf{x}=\mathbf{0}$.)

It follows that we have
\begin{align*}
A\mathbf{x}_1=BC\mathbf{x}_1=B\mathbf{0}=\mathbf{0}.
\end{align*}
Since the vector $\mathbf{x}_1$ is nonzero, the matrix $A$ is a singular matrix, hence $A$ is not invertible.

(b) Can the matrix $A$ be invertible?

The answer is yes. For example consider the following $2\times 3$ matrix $B$ and $3 \times 2$ matrix $C$:
\[B=\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0
\end{bmatrix}, \qquad C=\begin{bmatrix}
1 & 0 \\
0 & 1 \\
0 &0
\end{bmatrix}.\]
Then we have
\begin{align*}
A=BC=\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
0 & 1 \\
0 &0
\end{bmatrix}
=
\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix},
\end{align*}
and the matrix $A$ is invertible.

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Proof.
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Let us think backwards. Suppose that […]

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[…]

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\begin{align*}
S=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\},
\end{align*}
where
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1 \\
3 \\
1
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Definition.
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Let $\mathbf{x}, \mathbf{y}, \mathbf{z}$ are linearly independent $3$-dimensional vectors. Suppose that we have
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1 \\
0 \\
1
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1 \\
0
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1 \\
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0
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h \\
1 \\
-h
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2h \\
3h+1
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(a) Let $A=(a_{ij})$ be an $n\times n$ matrix. Suppose that the entries of the matrix $A$ satisfy the following relation.
\[|a_{ii}|>|a_{i1}|+\cdots +|a_{i\,i-1}|+|a_{i \, i+1}|+\cdots +|a_{in}|\]
for all $1 \leq i \leq n$.
Show that the matrix $A$ is nonsingular.
(b) Let […]