Prove A is invertible iff A^T is invertible and, in that case, (A^{-1})^T = (A^T)^{-1}

[DJKHALID$]

Linear algebra is killing me...please help!!!!

4. Recall from lectures that if $A$ and $B$ are matrices with $AB$ defined, then $(AB)^T = B^T A^T$. Use this property to prove that $A$ is invertible if and only if $A^T$ is invertible and, in that case, $(A^{-1})^T = (A^T)^{-1}$.

 

[stapel]

4. Recall from lectures that if $A$ and $B$ are matrices with $AB$ defined, then $(AB)^T = B^T A^T$. Use this property to prove that $A$ is invertible if and only if $A^T$ is invertible and, in that case, $(A^{-1})^T = (A^T)^{-1}$

Please reply with a clear listing of your thoughts and efforts so far, so that we may begin working with you. (Read Before Posting)

Thank you!

 

[Steven G]

I would first try using the hint with B = A^-1
So I = I^T = (A*A^-1)^T.... now you continue.

Use the fact that if C*D = I, then D = C^-1

 

[blamocur]

4. Recall from lectures that if $A$ and $B$ are matrices with $AB$ defined, then $(AB)^T = B^T A^T$. Use this property to prove that $A$ is invertible if and only if $A^T$ is invertible and, in that case, $(A^{-1})^T = (A^T)^{-1}$.

I don't see linear algebra killing you: you haven't shown any effort yet.

 

[Steven G]

I don't see linear algebra killing you: you haven't shown any effort yet.

Maybe it already killed him? I hope not.

 

[khansaheb]

@Steven G said:

I would first try using the hint with B = A^-1

@Steven G, where do you see that hint?

 

[Steven G]

@Steven G said:


@Steven G, where do you see that hint?

The hint is in code. It says Recall from lectures ...

 

[Steven G]

@Steven G said:


@Steven G, where do you see that hint?

In addition, I just noticed that the problem also said...use this property to prove.....

 

[khansaheb]

..use this property to prove.....

Which property?

 

[Dr.Peterson]

Which property?

The property stated in the previous sentence.

This is called an antecedent ...

 

[khansaheb]

The property stated in the previous sentence.

This is called an antecedent ...

Does that imply

hint with B = A^-1

The property mentioned in the previous sentence (in the op) was: (AB)^T = B^TA^T

 

[Steven G]

...and that is the property that you use to prove the theorem.
Can we please see your proof?
I started my proof, but you don't see how to finish it? I'll message the whole proof to you if you like.

 

[khansaheb]

...and that is the property that you use to prove the theorem.

But not the hint:

using the hint with B = A^-1

Or do you?

 

[Steven G]

How can you say that the author suggesting that you use a particular theorem to prove a fact is not a hint?

 

[Steven G]

I would first try using the hint with B = A^-1

The property mentioned in the previous sentence (in the op) was: (AB)^T = B^TA^T

There is a B in the hint. I suggested that the OP replaces B with A^-1.
Seriously, what is the problem with my calling something a hint??

 

[Steven G]

I = I^T = (A*A^-1)^T = (A^-1)^T*(A^T)
So (A^-1)^T = (A^T)^-1