# Does this work. Matrices

1. Jul 8, 2008

### Saladsamurai

1. The problem statement, all variables and given/known data

(a) Show that for any invertible matrix A,

$(A^{-1})^TA^T=I$ and $A^T(A^{-1})^T=I$

(b) Deduce that AT is invertible and that its inverse is the transpose of $A^{-1}$

(c) Deduce also that if A is symmetric then A-1 is also symmetric.

2. Relevant equations (AB)T=BTAT

3. The attempt at a solution

(a) If A is invertible,

$AA^{-1}=A^{-1}A=I$
$\Rightarrow (AA^{-1})^T=I^T$
$\Rightarrow (AA^{-1})^T=I$
$\Rightarrow (A^{-1})^TA^T=I$

Now for part (b) and (c)

2. Jul 8, 2008

### Dick

Yes, that works GREAT! Doesn't that make b) pretty easy?

3. Jul 8, 2008

### Saladsamurai

Oh yeah. I guess it kind of takes care of it right? Since this last line $(A^{-1})^TA^T=I$ is the definition of an Inverse? That, is: if AB=BA=I then B=A^{-1}

So I also have to show, somehow, that $A^T(A^{-1})^T=I$ as well?

4. Jul 8, 2008

### Dick

The definition does say 'if AB=BA=I'. So you'd better show both. There's nothing hard about it.

5. Jul 9, 2008

### Saladsamurai

If A is invertible, $AA^{-1}=A^{-1}A=I$

$\Rightarrow A^{-1}A=I$
$\Rightarrow (A^{-1}A)^T=I^T$
$\Rightarrow (A^{-1}A)^T=I$
$\Rightarrow A^T(A^{-1})^T=I$

Alright-then

Now how about part (c). . . Deduce also that if A is symmetric then A-1 is also symmetric.

If A is symmetric, A=AT and if A-1 is symmetric, A-1=(A-1)T

Let me think for a minute here...

6. Jul 9, 2008

### dirk_mec1

Remember:

$$(\mathbf{A}^\mathrm{T})^{-1} = (\mathbf{A}^{-1})^\mathrm{T}$$

7. Jul 9, 2008

### Saladsamurai

Is this a property of matrices, or just of exponents in general? Is this just saying that exponents can 'commute'? Sorry if that seems like a stupid question

8. Jul 9, 2008

### dirk_mec1

It's a property of the transpose of matrices I think you need it here.

9. Jul 9, 2008

### Dick

No, you can't 'commute' everything that's written as a superscript. But you have already shown that $$A^T(A^{-1})^T=I$$. That means that the second matrix is the inverse of the first. The inverse of the first is $$(A^T)^{-1}$$. It's pretty easy to show for powers as well.

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