Let $A$ be an $n\times n$ invertible matrix. Then prove the transpose $A^{\trans}$ is also invertible and that the inverse matrix of the transpose $A^{\trans}$ is the transpose of the inverse matrix $A^{-1}$.
Namely, show that
\[(A^{\trans})^{-1}=(A^{-1})^{\trans}.\]

By definition of inverse matrices, if there is an $n\times n$ matrix $B$ such that
\[A^{\trans}B=I \text{ and } B A^{\trans}=I,\]
where $I$ is the $n\times n$ identity matrix, then $A^{\trans}$ is invertible and its inverse is $B$, that is, $B=(A^{\trans})^{-1}$.

We claim that we can take $(A^{-1})^{\trans}$ for this $B$.
In fact, we have
\begin{align*}
A^{\trans}(A^{-1})^{\trans}=(A^{-1}A)^{\trans}=I^{\trans}=I.
\end{align*}

Here in the first equality, we used the fact about transpose matrices that
\[(CD)^{\trans}=D^{\trans}C^{\trans}\]
for any matrices $C, D$ such that the matrix product $CD$ is well-defined.

Similarly, we have
\[(A^{-1})^{\trans}A^{\trans}=(AA^{-1})^{\trans}=I^{\trans}=I.\]

This proves the transpose $A^{\trans}$ is invertible and that $(A^{-1})^{\trans}$ is the inverse matrix of $A^{\trans}$, that is,
\[(A^{\trans})^{-1}=(A^{-1})^{\trans}\]
as required.

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(a) Prove that $A$ is invertible.
(b) Prove that $A^{-1}$ is symmetric.
(c) Prove that $A^{-1}$ is positive-definite.
(MIT, Linear Algebra Exam Problem)
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Hint.
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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
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A square matrix $A$ is called idempotent if $A^2=A$.
Show that a square invertible idempotent matrix is the identity matrix.
Proof.
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Let $A$ be an $n\times n$ nonsingular matrix.
Prove that the transpose matrix $A^{\trans}$ is also nonsingular.
Definition (Nonsingular Matrix).
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[…]

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\[\mathbf{v}^{\trans}\mathbf{u}\neq -1.\]
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Let $A$ be an $n\times n$ matrix such that $A^k=I_n$, where $k\in \N$ and $I_n$ is the $n \times n$ identity matrix.
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