MHB Determinants and Rank: Solving for det(A) Given Rank(A)=4

[Yankel]

Hello

I have a question, I think I solved it, and I would like to confirm...

Let A be a 4X4 matrix with and let rank(A)=4.
It is known that det(A^2) = det(-A)
Is det(A)=-1 ?

I think the answer is no.

det(A)*det(A)=det(-A)
det(A)*det(A)=(-1)^4 * det(A) = det(A)
det(A)*det(A) = det(A)

this is true only when det(A)=1 or det(A)=0. The later can't be since rank(A)=4, so det(A)=1. Am I correct ?

(Thinking)

 

[CaptainBlack]

Hello

I have a question, I think I solved it, and I would like to confirm...

Let A be a 4X4 matrix with and let rank(A)=4.
It is known that det(A^2) = det(-A)
Is det(A)=-1 ?

I think the answer is no.

det(A)*det(A)=det(-A)
det(A)*det(A)=(-1)^4 * det(A) = det(A)
det(A)*det(A) = det(A)

this is true only when det(A)=1 or det(A)=0. The later can't be since rank(A)=4, so det(A)=1. Am I correct ?

(Thinking)

For a matrix \(\bf{A}\) of dimension \(2n\times 2n,\ n \in \mathbb{N}_+\): \(\det(-{\bf{A}})=\det({\bf{A}})\).

So you have the equality \([\det({\bf{A}})]^2=\det({\bf{A}})\), and also \(\det({\bf{A}}) \ne 0\)

CB