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 Post subject: The determinant is zeroPosted: Tue Oct 25, 2016 6:49 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 156
Location: Melbourne, Australia
Given two real square matrices $A, B$ such that $AB - BA =A$ prove that $\det A =0$.

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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 Post subject: Re: The determinant is zeroPosted: Fri Jul 28, 2017 11:43 pm

Joined: Tue Nov 24, 2015 7:47 pm
Posts: 13
I think i got a solution :

First suppose $A$ is invertible so $\exists A^{-1}$
I will prove by induction that $A^nB-BA^n=nA^n$ $\forall n \in \mathbb{N}$

The $n=1$ follows trivially from the hypotheses .

Suppose it holds for all $k \leq n$ i will prove it for $k=n+1$

$A^nB-BA^n=nA^n$

So multiplying by both sides with $A^{-1}$

$A^{n-1}B-A^{-1}BA^n=nA^{n-1}$ (1)

$A^nBA^{-1}-BA^{n-1}=nA^{n-1}$ (2)

I add (1)+(2) and i get
$(A^{n-1}B-BA^{n-1})+(A^nBA^{-1}-A^{-1}BA^n)=2nA^{n-1} \Leftrightarrow A^nBA^{-1}-A^{-1}BA^n=(n+1)A^{n-1}$

Now i multiply with $A$ from right and left respectively and thus i get
$A^{n+1}B-BA^{n+1}=(n+1)A^{n+1}$ (3) which is exactly what we needed to prove thus
$A^nB-BA^n=nA^n \quad , \quad \forall n \in \mathbb{N}$

Hence , by taking trace on (3) we get $tr(A^n)=0 \quad , \quad \forall n \in \mathbb{N}$
So $A$ is nilponent thus $\exists m \in \mathbb{N}$ such that $A^m=0$ so $det(A^m)=0 \Rightarrow det(A)=0$ which obviously contradicts with our original hypotheses that $A$ is invertible. Hence $det(A)=0$ .

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