How Does the Determinant Change When Applying a Similarity Transformation?

[mlarson9000]

Homework Statement

If A and C are nxn matricies, with C invertible, prove that det(A)=det((C^-1)AC).

Homework Equations

The Attempt at a Solution

I think the way to go is to show that if A=(C^-1)AC, then det(A)=det((C^-1)AC), but I'm not sure how to show A=(C^-1)AC. I know CC^-1=(C^-1)C=I, but I just can't see how to put this all together.

 

[g_edgar]

Do you know some formula for determinant of a product?

 

[mlarson9000]

Do you know some formula for determinant of a product?

det((C^-1)AC)=det(C^-1)*det(A)*det(C)=(1/det(C))*det(C)*det(A)=(det(C)/det(C))*det(A)=det(A).

Is that it?

 

[HallsofIvy]

Well, his question was "Do you know some formula for determinant of a product?".

I assume your answer is "Yes"!

In that case, so is mine.

 

[epenguin]

det((C^-1)AC)=det(C^-1)*det(A)*det(C)=(1/det(C))*det(C)*det(A)=(det(C)/det(C))*det(A)=det(A).

Is that it?

I don't think so - you seem to be confusing the inverse C^-1 of a matrix C with the reciprocal 1/det(C) of the number det(C) which does not come into this at all.

g_edgar asked do you know any formula for a determinant of a product of matrices, e.g. of the determinant of MN where M, N are both n X n matrices.

You can treat the right hand side using a couple of such formulae.

Your proposed matrix equation A = C^-1AC (equivalent to CA = AC ) is not true in general.

 

[Office_Shredder]

I don't think so - you seem to be confusing the inverse C^-1 of a matrix C with the reciprocal 1/det(C) of the number det(C) which does not come into this at all.

det(Identity matrix) = 1 = det(CC^-1 = det(C)*det(C^-1) and hence det(C^-1) = 1/det(C)