Show Invertibility of Similar Matrices: A & B

[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?

 

[Dick]

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?

Not until you explain in words how it does.

 

[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.

 

[Dick]

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.

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.

 

[pyroknife]

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.

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.

 

[Dick]

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.

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?

 

[pyroknife]

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?

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

 

[Dick]

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

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.

 

[pyroknife]

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.

(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?

 

[Dick]

(A)^-1 = (P^-1 * B * P)^-1
A^-1 = P^-1 * B^-1 * P
P^A^-1*P^-1=B^-1B*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=IOk, 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?

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?

 

[pyroknife]

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?

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.

 

[Dick]

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.

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.

 

[HallsofIvy]

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.