If a matrix has an inverse, then the transpose also has an inverse

Question

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?

Answer 1

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 $(A^{-1})^T$).

Answer 2

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.

Answer 3

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}