Proving Similarity of Matrices: A^2=C to B^2=C

[eyehategod]

I need help with this proof. Can anyone lead me in the right direction?

Let A be an nxn matrix such that A^2=C.
Prove that if B~A, then B^2=C.

 

[Dick]

Let A=diag(1,2), B=diag(2,1). A^2=diag(1,4), B^2=diag(4,1). A~B. A^2 is not equal to B^2. Is there something you aren't telling us about C or do you want to prove A^2~B^2?

 

[eyehategod]

I gave you exactly what the book says

 

[Dick]

I gave you exactly what the book says

If that's exactly what the book says, then you can't prove it. Because it's false.

 

[eyehategod]

what if C were to be 0. Would that proof make sense?

 

[Dick]

what if C were to be 0. Would that proof make sense?

It makes all of the difference in the world. If A^2=0 and A~B then B^2=(PAP^(-1))(PAP^(-1)). What's that?

 

[eyehategod]

that would be:
B^2=P^(-1)A^(2)P
B^2=P^(1)0P=0

 

[Dick]

It would also be very different if C were any other multiple of the identity matrix. Wouldn't it?

 

[eyehategod]

THe book really has instead of A^2=C its A^2=O.But I can't tell if its zero or the letter O. THe O is at a slant if that means anything

 

[Dick]

that would be:
B^2=P^(-1)A^(2)P
B^2=P^(1)0P=0

Yes.

 

[Dick]

THe book really has instead of A^2=C its A^2=O.But I can't tell if its zero of the letter O. THe O is at a slant if that means anything

I don't know. But as I said any multiple of the identity would work as well as 0.