# Similar matrices

1. Dec 11, 2012

### pyroknife

If A and B are similar matrices, show that A has an inverse IFF B is invertible.

A=P^-1 * B * P
Where P is an invertible matrix.

(A)^-1 = (P^-1 * B * P)^-1
A^-1 = P^-1 * B^-1 * P

Does this show what the question wanted?

2. Dec 11, 2012

### Dick

Not until you explain in words how it does.

3. Dec 11, 2012

### pyroknife

This shows that B^-1 is similar to A^-1?

I forgot to put this in the statement, the problem specifically asked to "use the "definition" of similar matrices to show this."

My original proof was that if A and b are similar then they must have the same determinant. If B is not invertible then it's determinant is 0 and A will have a determinant of 0 and is therefore not invertible. If B is intervible then it's determinant is not 0 and....A is invertible. But the teacher said I needed to use the definition.

4. Dec 11, 2012

### Dick

Ok, then use the definition. If A is invertible then A^(-1) exists. Use what you've got to give a formula for B^(-1). The show it works. And vice versa.

5. Dec 11, 2012

### pyroknife

Alright so, after this:
(A)^-1 = (P^-1 * B * P)^-1
A^-1 = P^-1 * B^-1 * P
P^A^-1*P^-1=B^-1

Ugh, I guess I don't know what I need to do or maybe that I already have all the stuff, but not have worded it.

6. Dec 11, 2012

### Dick

Ok, so you are claiming B^(-1)=PA^(-1)P^(-1). Check it. B=PAP^(-1) right, from your definition? Multiply your expression for B^(-1) by the expression for B using those forms. What do you get? How does that prove part of what you want?

Last edited: Dec 11, 2012
7. Dec 11, 2012

### pyroknife

Wait from the definition, doesn't B=P^-1 * A *P?

8. Dec 11, 2012

### Dick

No. P is a fixed matrix once you've used the definition. You said A=P^-1 * B * P. So B=PAP^(-1). You are misunderstanding what the definition of similarity means.

9. Dec 11, 2012

### pyroknife

(A)^-1 = (P^-1 * B * P)^-1
A^-1 = P^-1 * B^-1 * P
P^A^-1*P^-1=B^-1

B*P^A^-1*P^-1=B*B^-1=I
PAP^(-1)*P^A^-1*P^-1=I
PA*I*A^-1*P^-1=I
PP^-1=I
I=I

Ok, so you are claiming B^(-1)=PA^(-1)P^(-1). Check it. B=PAP^(-1) right, from your definition? Multiply your expression for B^(-1) by the expression for B using those forms. What do you get? How does that prove part of what you want?

10. Dec 11, 2012

### Dick

Any words to got along with all of those typos to indicate whether you get it or not? This sequence is rightish. PAP^(-1)*P^A^-1*P^-1=PA*I*A^-1*P^-1=PP^-1=I. So?

11. Dec 11, 2012

### pyroknife

If P*P^-1=I Then that means P is invertible. I think I'm just stating the obvious. This showed that multipling B^-1 by B gives the identity matrix which means B is invertible.

12. Dec 11, 2012

### Dick

You've shown something you think is B^(-1) times B is I. That proves that the thing that you think is B^(-1) is, in fact, B^(-1). If CB=I then C is B^(-1). I think this is how the exercise wants you to show it.

13. Dec 12, 2012

### HallsofIvy

Staff Emeritus
By the way, in a more abstract sense, a matrix always represents a linear transformation in some basis. A "similar" matrix represents the same linear transformation in a different basis. Of course, whether or not a linear transformation is invertible is independent of the specific basis in which it is represented.