How to express det (a(A-B) - bI)

Asked Mar 23 '22 at 08:21

Active Apr 01 '22 at 16:12

Viewed 130 times

I was looking over the solution of a problem and don't understand how they expanded the determinant.

The problem:

Let $A$, $B$ be square $2 \times 2$ real matrices, such that $(A-B)^2 =O$ (where $O$ is the $2 \times 2$ matrix with all elements $0$). Prove that $\det (A^2-B^2)=(\det(A)-\det(B))^2$.

So far I can prove that: $\det(A-B) =0$, $Tr(A-B) =0$ so $Tr(A) =Tr(B) := a.$

We let $b :=\det A-\det B$.

From Cayley-Hamilton we have

$A^2-aA+\det(A)I=O$

$B^2-aB+\det(B)I=O$

It follows that $\det(A^2-B^2)=\det(a(A-B)-bI)$.

I looked over the given solution and to arrive at the needed conclusion they expanded the determinant in the following manner:

$\det(a(A-B)-bI)=a^2\det(A-B)-abTr(A-B)+b^2=b^2$

I suppose they used a form of Cayley-Hamilton again but I can't figure out how.

Note: $I$ represents the $2 \times 2$ identity matrix

edited Apr 01 '22 at 16:12

user26857

53,190

asked Mar 23 '22 at 08:21

Rose2357

https://math.stackexchange.com/questions/673934/expressing-the-determinant-of-a-sum-of-two-matrices Might this help? – Raad Shaikh Mar 23 '22 at 08:51
Since $A-B$ is nilpotent, $\det(a(A-B)-bI)=b^2$. – user1551 Mar 24 '22 at 07:53
For a 2 by 2 matrix, you do not need to invoke the Cayley-Hamilton to get that formula. You just work it out by hand and it will work out fine. – Daniel Akech Thiong Mar 30 '22 at 04:49
Your notation is quite confusing, as $a$ and $b$ look like an entry of $A$ and $B$. I would have used $\alpha$ and $\beta$ instead. – Rodrigo de Azevedo Apr 02 '22 at 08:08

How to express det (a(A-B) - bI)

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