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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.

Dylan C. Beck
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Clinton
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5 Answers5

19

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

Jonathan
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    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 – Clinton May 12 '13 at 02:26
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    @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 $BA^T=A^TB=I$. – Alex Becker May 12 '13 at 02:27
  • @AlexBecker ah. thanks! – Clinton May 12 '13 at 02:28
7

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!

Chinnapparaj R
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Ashish Bani
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5

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 :
$$B^T=(A^{-1})^T=(A^T)^{-1}$$

user103828
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amir hossein afshar
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1

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: $$\begin{align}(\text{adj} (A))^T&=\text{adj} (A^T);\\ (cA)^T&=c(A)^T;\\ |A|&=|A^T|.\end{align}$$

farruhota
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$A^T(A^T)^{-1} = I$ ; $ eq^n 1$

taking transpose on both side of $ eq^n 1 $

$[(A^T)^{-1}]^T A = I $; $eq^n 1.1$

[Since $(AB)^T= B^TA^T $ & $I^T=I$ , $(A^T)^T=A$]

Now to hold above $eq^n 1.1$ it must satisfy the relation below

$[(A^T)^{-1}]^T = A^{-1} $ ; $eq^n 2 $

taking transpose on $eq^n 2 , $ we get $ (A^T)^{-1} = (A^{-1})^T $

Gopal1110
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