By definition, $A^{\trans}$ is a nonsingular matrix if the only solution to
\[A^{\trans}\mathbf{x}=\mathbf{0}\]
is the zero vector $\mathbf{x}=\mathbf{0}$ in $\R^n$.

Proof.

Suppose that
\[A^{\trans}\mathbf{v}=\mathbf{0} \tag{*}\]
for an $n$-dimensional vector $\mathbf{v}$.
We prove that $\mathbf{v}=\mathbf{0}$.

Since $A$ is nonsingular, there exists a vector $\mathbf{u}\in \R^n$ such that
\[A\mathbf{u}=\mathbf{v}.\]
(For those who know the inverse matrix, the vector $\mathbf{u}$ is given by $\mathbf{u}=A^{-1}\mathbf{v}$. The fact can be proved without using the inverse matrix, though.)

Hence we obtain from (*)
\[A^{\trans}A\mathbf{u}=\mathbf{0}.\]
It follows that
\[\mathbf{u}^{\trans}A^{\trans}A\mathbf{u}=\mathbf{u}^{\trans}\mathbf{0}=0.\]
By the property of the transpose, we have
\[\mathbf{u}^{\trans}A^{\trans}=(A\mathbf{u})^{\trans}.\]
Thus we have
\[0=(A\mathbf{u})^{\trans}A\mathbf{u}=\|A\mathbf{u}\|^2.\]
This yields the length $\|A\mathbf{u}\|=0$, and hence $A\mathbf{u}=\mathbf{0}$.
Since $\mathbf{v}=A\mathbf{u}$, we conclude that $\mathbf{v}=\mathbf{0}$.

Therefore if $A^{\trans}\mathbf{v}=\mathbf{0}$, then $\mathbf{v}=\mathbf{0}$.
This shows that the transpose $A^{\trans}$ is nonsingular.

Find the Inverse Matrix of a Matrix With Fractions
Find the inverse matrix of the matrix
\[A=\begin{bmatrix}
\frac{2}{7} & \frac{3}{7} & \frac{6}{7} \\[6 pt]
\frac{6}{7} &\frac{2}{7} &-\frac{3}{7} \\[6pt]
-\frac{3}{7} & \frac{6}{7} & -\frac{2}{7}
\end{bmatrix}.\]
Hint.
You may use the augmented matrix […]

Construction of a Symmetric Matrix whose Inverse Matrix is Itself
Let $\mathbf{v}$ be a nonzero vector in $\R^n$.
Then the dot product $\mathbf{v}\cdot \mathbf{v}=\mathbf{v}^{\trans}\mathbf{v}\neq 0$.
Set $a:=\frac{2}{\mathbf{v}^{\trans}\mathbf{v}}$ and define the $n\times n$ matrix $A$ by
\[A=I-a\mathbf{v}\mathbf{v}^{\trans},\]
where […]

Subspaces of Symmetric, Skew-Symmetric Matrices
Let $V$ be the vector space over $\R$ consisting of all $n\times n$ real matrices for some fixed integer $n$. Prove or disprove that the following subsets of $V$ are subspaces of $V$.
(a) The set $S$ consisting of all $n\times n$ symmetric matrices.
(b) The set $T$ consisting of […]

Diagonalizable by an Orthogonal Matrix Implies a Symmetric Matrix
Let $A$ be an $n\times n$ matrix with real number entries.
Show that if $A$ is diagonalizable by an orthogonal matrix, then $A$ is a symmetric matrix.
Proof.
Suppose that the matrix $A$ is diagonalizable by an orthogonal matrix $Q$.
The orthogonality of the […]

Find the Distance Between Two Vectors if the Lengths and the Dot Product are Given
Let $\mathbf{a}$ and $\mathbf{b}$ be vectors in $\R^n$ such that their length are
\[||\mathbf{a}||=||\mathbf{b}||=1\]
and the inner product
\[\mathbf{a}\cdot \mathbf{b}=\mathbf{a}^{\trans}\mathbf{b}=-\frac{1}{2}.\]
Then determine the length $||\mathbf{a}-\mathbf{b}||$.
(Note […]

Eigenvalues of a Matrix and its Transpose are the Same
Let $A$ be a square matrix.
Prove that the eigenvalues of the transpose $A^{\trans}$ are the same as the eigenvalues of $A$.
Proof.
Recall that the eigenvalues of a matrix are roots of its characteristic polynomial.
Hence if the matrices $A$ and $A^{\trans}$ […]

Rank and Nullity of a Matrix, Nullity of Transpose
Let $A$ be an $m\times n$ matrix. The nullspace of $A$ is denoted by $\calN(A)$.
The dimension of the nullspace of $A$ is called the nullity of $A$.
Prove the followings.
(a) $\calN(A)=\calN(A^{\trans}A)$.
(b) $\rk(A)=\rk(A^{\trans}A)$.
Hint.
For part (b), […]

Let $\mathbf{v}$ be a nonzero vector in $\R^n$. Then the dot product $\mathbf{v}\cdot \mathbf{v}=\mathbf{v}^{\trans}\mathbf{v}\neq 0$. Set $a:=\frac{2}{\mathbf{v}^{\trans}\mathbf{v}}$ and define the...