Suppose A^k=0 for some integer k is greater than or equal to 1

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SUMMARY

In the discussion, it is established that if \( A^k = 0 \) for some integer \( k \geq 1 \), then matrix \( A \) is not invertible. The proof utilizes a contradiction approach, assuming \( A \) is invertible with an inverse \( B \). By manipulating the equation \( (A^k)B = 0 \) and substituting \( B \) with \( A^{-1} \), it leads to a contradiction, confirming that \( A \) cannot be invertible.

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Chris Rorres
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I was wondering if anyone could give me some hints on this

Suppose A^k=0 for some integer k is greater than or equal to 1. Prove that A is not invertible.
 
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Try a proof by contradiction. Assume it is invertible with inverse B. Then AB=1. (A^k)B=0B=0
But B=A^-1. So you can simplify (A^k)B=(A^k)(A^-1)=...
 


Nice!
 

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