The discussion revolves around the properties of similar matrices, specifically focusing on the relationship between the invertibility of two similar matrices, A and B. The original poster seeks to demonstrate that A is invertible if and only if B is invertible, using the definition of similar matrices.
Participants are actively engaging with the problem, with some providing guidance on how to approach the proof using the definition of similarity. There is a mix of interpretations and attempts to clarify the relationships between the matrices, but no explicit consensus has been reached on the final proof.
Some participants mention the need to adhere to specific definitions and the importance of demonstrating properties through formal reasoning rather than relying on intuitive arguments. There are also references to the implications of determinants in relation to invertibility.
pyroknife said: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?
pyroknife said: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 said: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 said: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 said: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 said:Wait from the definition, doesn't B=P^-1 * A *P?
Dick said: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 said:(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?
Dick said: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 said: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.