10 True of False Problems about Nonsingular / Invertible Matrices

False. For example, let $A=I$ and $B=-I$. Then both matrices are nonsingular but $A+B=O$ is singular.

False. As a counterexample, consider
\[A=\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix}.\]
Then $A$ has no zero rows or columns, yet is does not have the inverse matrix as the determinant of $A$ is zero.

True. Let
\[B=\begin{bmatrix}
B_1 & B_2 &\dots & B_n \\
\end{bmatrix},\]
where $B_i$ is the $i$-th column vector of $B$ for $i=1, \dots, n$.
Suppose that we have a linear combination
\[c_1B_1+c_2B_2+\cdots+c_n B_n=\mathbf{0}\]
for some scalars $c_1, c_2, \dots, c_n$.
Then we can write it as
\[\begin{bmatrix}
B_1 & B_2 &\dots & B_n \\
\end{bmatrix}\begin{bmatrix}
c_1 \\
c_2 \\
\vdots \\
c_n
\end{bmatrix}=\mathbf{0}.\]

Since $B\mathbf{x}=\mathbf{0}$ has only the trivial solution, we must have
\[\begin{bmatrix}
c_1 \\
c_2 \\
\vdots \\
c_n
\end{bmatrix}=\mathbf{0}.\]
Hence $c_1=c_2=\cdots=c_n=0$, and the column vectors $B_1, B_2, \dots, B_n$ are linearly independent.

False. As a counterexample, consider the $3\times 2$ matrix
\[A=\begin{bmatrix}
1 & 0 \\
1 & 0 \\
0 &1
\end{bmatrix}.\]
Then the equation $A\mathbf{x}=\mathbf{0}$ has only the trivial solution $\mathbf{x}=\mathbf{0}$, and yet the first and the second row of $A$ are linearly dependent.

True. The reduced row echelon form of an invertible matrix is the identity matrix, which is invertible.

False. Suppose $A$ is nonsingular such that $A^2=O$.
Since $A$ is nonsingular, it is invertible.
Hence we have
\begin{align*}
A=A^{-1}A^2=A^{-1}O=O,
\end{align*}
and the matrix $A$ must be the zero matrix.

False. For example, consider
\[A=\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix} \text{ and } B=\begin{bmatrix}
1 & 0\\
0& 2
\end{bmatrix}.\]
Then these matrices are invertible as their determinants are $\det(A)=1\neq 0$ and $\det(B)=2\neq 0$.
However they do not commute since
\begin{align*}
AB=\begin{bmatrix}
1 & 2\\
0& 2
\end{bmatrix} \text{ and } BA=\begin{bmatrix}
1 & 1\\
0& 2
\end{bmatrix}.
\end{align*}

True. Note that we have
\[A=A^{-1}A^2=A^{-1}I=A^{-1}.\]
Similarly, we have $B^{-1}=B$.

It follows that
\begin{align*}
(AB)^{-1}=B^{-1}A^{-1}=BA.
\end{align*}

True. For each $i=1, 2, \dots, n$, let $\mathbf{e}_i\in \R^n$ be the $n$-dimensional vector whose $i$-th entry is $1$ and $0$ elsewhere.

Then $A\mathbf{e}_i$ is the $i$-th column vector of the matrix $A$.

Since by assumption $A\mathbf{e}_i=\mathbf{0}$, we see that the $i$-th column of $A$ is the zero vector.
As this is true for any $i=1, \dots, n$, we conclude that $A$ is the zero matrix.

True. Consider a linear combination
\[c_1(A\mathbf{v}_1)+c_2(A\mathbf{v}_2)=\mathbf{0},\]
where $c_1, c_2$ are scalars.

It yields that
\[A(c_1\mathbf{v}_1+c_2\mathbf{v}_2)=\mathbf{0}.\]

Since $A$ is nonsingular, we have
\[c_1\mathbf{v}_1+c_2\mathbf{v}_2=\mathbf{0}.\]

Because the vectors $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent, we have
\[c_1=c_2=0.\]
Thus, the vectors $A\mathbf{v}_1, A\mathbf{v}_2$ are linearly independent vectors.