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

Joined: Sat Nov 14, 2015 6:32 am
Posts: 127
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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PostPosted: 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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