Matrix Proof

  • Thread starter pyroknife
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  • #1
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Show that if A is invertible, then At is invertible
and (At )-1
= (A-1)t .


If A is invertible, then det(A)≠0
det(A)=det(A^t)
thus det(A^t)≠0



I'm not really sure how to prove the second part. It's an identity that I remembered, but don't know how to prove.

I'll take a crack at it though:

Multiply A^t to both sides
gives I=A^t (A^-1)^t
For the RHS, can i multiply A^t to A^-1 or does the ^t prevent me from doing that?
 

Answers and Replies

  • #2
jbunniii
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If A is invertible, then there exists a matrix B such that AB = I and BA = I. What is the natural thing to do here, to both sides of each equation?
 
  • #3
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If A is invertible, then there exists a matrix B such that AB = I and BA = I. What is the natural thing to do here, to both sides of each equation?
I'm not sure what you are asking.

(At )-1 = (A-1)t .

If AB=I, then obviously, A is the inverse of B or vise versa.

So In this case I let A=(A^t)^-1
Thus A^=((A^t)^-1)^-1 ? <<<<multiply this to both sides? If so, the left said reduces to the I matrix, but the right side still looks complicated.
 
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  • #4
jbunniii
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I'm not sure what you are asking.

(At )-1 = (A-1)t .

If AB=I, then obviously, A is the inverse of B or vise versa.

So In this case I let A=(A^t)^-1
Thus A^=((A^t)^-1)^-1 ? <<<<multiply this to both sides? If so, the left said reduces to the I matrix, but the right side still looks complicated.
If A is invertible, then there exists a matrix B such that AB = I and BA = I. Try transposing both sides of each of these equations.
 
  • #5
Dick
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This can be a little hard to see. You already have I=A^t (A^-1)^t. This tells you that A^t has an inverse. What is it? Now take a deep breath. The inverse of A^t is called (A^t)^(-1).

Now that I look back I'm not sure you even got that far. Take junniii's advice. Find the transpose of A^(-1)A=I.
 
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  • #6
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This can be a little hard to see. You already have I=A^t (A^-1)^t. This tells you that A^t has an inverse. What is it? Now take a deep breath. The inverse of A^t is called (A^t)^(-1).
OH!

So in this form

I=A^t [(A^-1)^t]

To get to the identity matrix A^T has to be multiplied by its inverse (A^T)^-1. The quantity [(A^-1)^t] must equal (A^T)^-1.


So simple, yet hard.
 
  • #7
Dick
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OH!

So in this form

I=A^t [(A^-1)^t]

To get to the identity matrix A^T has to be multiplied by its inverse (A^T)^-1. The quantity [(A^-1)^t] must equal (A^T)^-1.


So simple, yet hard.
Yes, really simple. But it can be completely opaque if you don't remember what inverse means and that inverses are unique.
 

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