If a matrix has an inverse, then the transpose also has an inverse I just have simple one question.
"If a matrix $A$ has an inverse, then the transpose $A^T$also has an inverse."
I think the statement is 'true'.
Because
$AA^{-1} = I \\
(AA^{-1})^{T} = I^T \\
(AA^{-1})^T = I \\
(A^{-1})^TA^T = I \\$
since $(A^{-1})^TA^T=\operatorname{Id}$, $A^T$ has an inverse (which is $(A^{-1})^T$).
Is it right?
 A: Yes, it is right. The statement is true (not just 'true' as you wrote). At the end, instead of writing “since $A^{-1}$ exists, $(A^{-1})^T$also exists”, I would have written that, since $(A^{-1})^TA^T=\operatorname{Id}$, $A^T$ has an inverse (which is $(A^{-1})^T$).
A: Yes and no. A proof requires a reference to the fact that if $AC = CA = I$, then $A$ is nonsingular, and $A^{-1} = C$. 
By definition, you have $AA^{-1} = A^{-1}A = I$. The transposition rule, i.e. $(AB)^T = B^T A^T$, now implies that $$I = I^T = (AA^{-1})^T = (A^{-1})^T A^T = (A^{-1}A)^T = A^T (A^{-1})^T.$$
It now follows that $A^T$ is nonsingular with inverse $(A^T)^{-1} = (A^{-1})^T$.
While the connection might be obvious to you, be wary of written statements with  no explicit logical connectors, such as implication/biimplication symbols.
A: This result essentially follows by the uniqueness of the inverse of a matrix.
If $U\cdot X= I$ then $U=X^{-1}$. In fact, 
\begin{align}
\left.
\begin{array}{c}
U\cdot X=I\\
X^{-1}\cdot X=I\\
\end{array}
\right\}
\implies &
U\cdot X- X^{-1}X=0
\\
\implies &
(U-X^{-1})\cdot X=0
\\
\implies &
(U-X^{-1})\cdot X\cdot X^{-1}=0\cdot X^{-1}
\\
\implies &
(U-X^{-1})\cdot I=0
\\
\implies &
U-X^{-1}=0
\\
\implies &
U=X^{-1}
\end{align}
A: You could also prove using determinants, since $ det A = det A^T $ 
A: Suppose $A \in M_n(R)$. If $R/J(R)$ is commutative and $A$ has an inverse, then so does its transpose. Otherwise the transpose of $A$ may not have an inverse. See Journal of Algebra 322 (2009) 1627–1636.
