Matrix determinants and inverses

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Homework Help Overview

The discussion revolves around proving a property of determinants involving two n x n matrices, A and S, where S is invertible. The specific statement to be shown is that det(S-1AS) = det(A). Participants explore the implications of the hint provided regarding the relationship between the determinants of S and its inverse.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the determinant product property, det(AB) = det(A)det(B), and the identity matrix's determinant being 1. There are attempts to manipulate the expression det(S-1AS) using the hint about S-1S = In.

Discussion Status

Several participants have provided insights and guidance on how to approach the problem, emphasizing the importance of separating the determinants and considering the properties of the determinant function. There is an ongoing exploration of the implications of the hint and the correctness of various approaches.

Contextual Notes

Some participants express uncertainty about the initial steps and the implications of the hint, indicating a need for clarification on the relationship between the determinants of S and S-1. There is also a question raised about the validity of certain manipulations in the context of matrix multiplication.

bakin
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Homework Statement


If A and S are n x n matrices with S invertible, show that det(S-1AS)=det(A). [HINT: Since S-1S=In, how are det(S-1) and det(S) related?]


Homework Equations





The Attempt at a Solution



Not sure. The only thing I can think of doing is substituting S-1S=In into det(S-1AS), so you have det(InA)=det(A), but I really don't know. Can somebody get me started so we can work through it?
 
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Sure. You should have proved det(AB)=det(A)*det(B). det(I)=1. Start from there.
 
bakin said:

Homework Statement


If A and S are n x n matrices with S invertible, show that det(S-1AS)=det(A). [HINT: Since S-1S=In, how are det(S-1) and det(S) related?]


Homework Equations





The Attempt at a Solution



Not sure. The only thing I can think of doing is substituting S-1S=In into det(S-1AS), so you have det(InA)=det(A), but I really don't know. Can somebody get me started so we can work through it?

You don't seem to have payed any attention to the hint! "S-1S=In, how are det(S-1) and det(S) related?" I presume you know that det(AB)= det(A)det(B).
 
Is it:

det(S-1AS) = det(S-1S)det(A)

=det(In)det(A)

=1det(A)

=det(A)?
 
That's true. But don't make it look like you think the matrices commute. S^(-1)AS is NOT necessarily equal to S^(-1)SA. But det(S^(-1)AS)=det(S^(-1))*det(A)*det(S). Now you can rearrange.
 
So first put them in 3 separate determinants like you did, then move the S-1 next to S, then put it back together? Go backwards from two determinants multiplying each other to a single determinant, which would be equal to det(S-1S) ?
 
bakin said:
So first put them in 3 separate determinants like you did, then move the S-1 next to S, then put it back together? Go backwards from two determinants multiplying each other to a single determinant, which would be equal to det(S-1S) ?

That's it exactly.
 
Thanks for the help :smile:

Just on a curious note, are there examples of when doing what I did in post #5 would be wrong, or is it just an incorrect way of solving?
 

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