Understanding proof that ##(A^T)^{-1} = (A^{-1})^T##

[Incand]

Theorem
If A is non singular then $(A^T)^{-1} = (A^{-1})^T$

Proof
The first part of proof shows that the inverse unambiguously decided. Then
$A^T(A^T)^{-1} = I$
and
$I = A^{-1}A = (A^{-1}A)^T = A^T(A^{-1})^T$
Where the second step is possible because $I = I^T$. From the equations above
$A^T(A^T)^{-1} = A^T(A^{-1})^T \Longrightarrow (A^T)^{-1} = (A^{-1})^T$
The text claims the last step is possible thanks to the inverse being unabigously decided. Why does this allow us to use cancellation?

 

[micromass]

Multiply both sides of $A^T (A^T)^{-1} = A^T (A^{-1})^T$ by $(A^T)^{-1}$ on the left.

 

[Incand]

Multiply both sides of $A^T (A^T)^{-1} = A^T (A^{-1})^T$ by $(A^T)^{-1}$ on the left.

But why do we need to show that the inverse is well defined or is that unnecessary?
And how do i know that $(A^T)^{-1}$ exists? If $A$ is square it's obvious but for the general case I'm not sure I understand why it follows.

 

[Delta2]

How is "non singular" defined in your book, because the way i know it, is that the definition for non singular matrix is invertible and square matrix.

 

[Incand]

How is "non singular" defined in your book, because the way i know it, is that the definition for non singular matrix is invertible and square matrix.

I'm sure you're right. I was just confused yesterday i guess. Actually this is more of repetition question of linear algebra for a numerical analysis courseso the book doesn't really define it. I just seem to have forgotten a bit of linear algebra it seems.

But as I understand it now the part in the proof that shows that an inverse is well defined isn't really necessary?
$A^T(A^T)^{-1}=I$
$A^T(A^{-1})^T=I$
If the inverse is well defined then that implies that $(A^T)^{-1} =(A^{-1})^T$ but left multiplying with $(A^T)^{-1}$ would show the same thing since if $A$ is invertible than so is $A^T$ (equal determinant).

 

[Delta2]

But as I understand it now the part in the proof that shows that an inverse is well defined isn't really necessary?

It is necessary in order for the statement of the theorem to have proper and complete meaning. If the inverse is not unique (i suppose that's what you mean when you say the inverse is well defined) then which of the two or more inverse matrices you choose when you state $(A^T)^{-1}$?

If the inverse is well defined then that implies that $(A^T)^{-1} =(A^{-1})^T$ ...

What proof the book gives in the first part, as i said above all i know is that when we say that the inverse is well defined it means that it is unique, i don't understand how you "dare" to say that inverse is well defined implies that $(A^T)^{-1} =(A^{-1})^T$ where do you get that implication from?

 

[Incand]

What proof the book gives in the first part, as i said above all i know is that when we say that the inverse is well defined it means that it is unique, i don't understand how you "dare" to say that inverse is well defined implies that $(A^T)^{-1} =(A^{-1})^T$ where do you get that implication from?

From that $A^T(A^T)^{-1} = I$ tells us that $(A^T)^{-1}$ is an inverse to $A^T$ (or by definition).
And the other equation the same thing, $(A^{-1})^T$ have to be an inverse to $A^T$ since $A^T(A^{-1})^T = I$. Since both of them are inverses too $A^T$ and the inverse was unique/well defined they have to be equal. right?

 

[Delta2]

Not exactly, matrix multiplication is not commutative so you have to prove that $A^T(A^{-1})^T=(A^{-1})^TA^T=I$ in order to be able to say that $(A^{-1})^T$ is an inverse to $A^T$.

 

[Incand]

Not exactly, matrix multiplication is not commutative so you have to prove that $A^T(A^{-1})^T=(A^{-1})^TA^T=I$ in order to be able to say that $(A^{-1})^T$ is an inverse to $A^T$.

Do you know if it possible for an invertible matrix to have a left (or right) inverse that's not equal to the inverse? Or did you mean that a complete proof would have to prove that that's the case?

 

[Delta2]

well yes a complete proof would have to prove that :). Its easy to prove anyway so i guess that's why the book omited it.

Anyway i guess we all got it now why we need the unique of the inverse, it is used in the proof afterall.

 

[Incand]

well yes a complete proof would have to prove that :). Its easy to prove anyway so i guess that's why the book omited it.

Anyway i guess we all got it now why we need the unique of the inverse, it is used in the proof afterall.

Yepp! Thanks for explaining!

 

[Fredrik]

Since the problem has been solved, I think it should be OK if I post my solution.

The inverse of a matrix A is a matrix B such that AB=BA=I. (Note that this only makes sense if A and B are both n×n-matrices, with the same n). Inverses are unique, because if B and C are both inverses of A, we have AB=BA=I and AC=CA=I, and therefore
$$B=IB=(CA)B=C(AB)=CI=C.$$ Now, if you find a matrix B such that AB=I, is that enough to ensure both that A is invertible and that B is the inverse of A? The answer is yes, because
$$AB=I\ \Rightarrow\ \det A\cdot\det B=1\ \Rightarrow\ \det A\neq 0,$$ so if AB=I, we know that A is invertible, that it has only one inverse, and that
$$A^{-1}=A^{-1}I=A^{-1}(AB)=(A^{-1}A)B=IB=B.$$
These results tell us that to prove that $A^T$ is invertible and that its inverse is $(A^{-1})^T$, it's sufficient to prove that $A^T(A^{-1})^T=I$.
$$A^T(A^{-1})^T =(A^{-1}A)^T=I^T=I.$$ If you view the first two results above as already proved theorems, then this little calculation is all you have to do to.