Does this work. Matrices

[Saladsamurai]

Homework Statement

(a) Show that for any invertible matrix A,

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

(b) Deduce that A^T 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.

Homework Equations

(AB)^T=B^TA^T

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)

 

[Dick]

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

 

[Saladsamurai]

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

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?

 

[Dick]

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?

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

 

[Saladsamurai]

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

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

Let me think for a minute here...

 

[dirk_mec1]

Remember:

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

 

[Saladsamurai]

Remember:

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

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

​

 

[dirk_mec1]

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

​

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

 

[Dick]

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

​

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.