8
$\begingroup$

Prove $(A^T)^{-1}$ = $(A^{-1})^T$ for any invertible matrix "A".

I actually don't know where to start - I do not think I can just apply index laws.

Any help is cool! Thanks.

$\endgroup$
13
$\begingroup$

Try taking the transpose of the equation $$AA^{-1}=I.$$

$\endgroup$
  • 2
    $\begingroup$ using that, I got $(A^{-1})^T A^T = I$ and $A^T (A^{-1})^T = I$, then $(A^{-1})^T A^T = A^T (A^{-1})^T$, but what next? Sorry if I've been asking too much - been stoning at this for a while $\endgroup$ – Clinton May 12 '13 at 2:26
  • 2
    $\begingroup$ @Clinton Which means that $(A^T)^{-1}=(A^{-1})^T$, by the definition of inverses. $(A^T)^{-1}$ is defined to be the matrix $B$ such that $BA^T=A^TB=I$. $\endgroup$ – Alex Becker May 12 '13 at 2:27
  • $\begingroup$ @AlexBecker ah. thanks! $\endgroup$ – Clinton May 12 '13 at 2:28
5
$\begingroup$

first we define a "B" matrix like below : $$B = (A^{-1})$$ so : $$B^T=(A^{-1})^T *$$ and we can write : $$AB = I$$ then $$(AB)^T=I^T \textrm{ and } B^T A^T =I$$ From this equation we can say that : $$B^T=(A^T)^{-1}$$ and finally from * we can write :
$$B^T=(A^{-1})^T=(A^T)^{-1}$$

$\endgroup$
  • $\begingroup$ Here is some MathJax tutorial $\endgroup$ – ASB Feb 22 '15 at 16:01
2
$\begingroup$

As we know $AA^{-1}=I$. Now taking transpose both sides, we get $$(AA^{-1})^T = I^T$$ which implies $$[ (A^{-1})^T ](A^T) = I$$ Now multiply both sides with $[(A^T)^{-1}]$ at right side, $$[(A^{-1})^T]{(A^T)(A^T)^{-1}} = (A^T)^{-1}$$ Here $(A^T)(A^T)^{-1}$ will form identity $I$, Since we know $AA^{-1} = I$, Therefore $$(A^{-1})^T = (A^T)^{-1}$$ Hence Proved!

$\endgroup$
0
$\begingroup$

Alternatively: $$A^{-1}=\frac{\text{adj} (A)}{|A|}$$ Transpose: $$\begin{align}(A^{-1})^T&=\left(\frac{\text{adj}(A)}{|A|}\right)^T=\\ &=\frac{(\text{adj} (A))^T}{|A|}=\\ &=\frac{\text{adj} (A^T)}{|A|}=\\ &=\frac{\text{adj} (A^T)}{|A^T|}=\\ &=(A^T)^{-1}.\end{align}$$ The following properties of adjoint and transpose were used: $$\begin{align}(\text{adj} (A))^T&=\text{adj} (A^T);\\ (cA)^T&=c(A)^T;\\ |A|&=|A^T|.\end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.