Determinants

Algebra Level pending

Let A A , B M n × n B \in M \tiny n\times n ( R ) (\mathbb{R}) such that A B B A = A AB-BA=A .

What is the value of det ( A ) \det(A) ?

Clarifications:

M n × n M \tiny n\times n ( R ) (\mathbb{R}) denotes a matrix with n n columns and n n rows, where every element is a real number.


The answer is 0.

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2 solutions

Suppose d e t ( A ) 0 det(A) \neq 0 . Then A A would be invertible and A B A 1 B = I t r ( A B A 1 B ) = t r I = n ABA^{-1}-B=I \Rightarrow tr(ABA^{-1}-B) =tr I = n

But we can also know that t r ( A B A 1 B ) = t r ( A B A 1 ) t r B = t r ( A 1 A B ) = t r B = 0 tr(ABA^{-1}-B) = tr(ABA^{-1})-tr B = tr(A^{-1}AB)=tr B = 0 , which is a contradiction.

Then, d e t ( A ) = 0 det(A)=0

Kushal Bose
Jan 23, 2017

A B B A = A = > A B = B A + A = > A B = A ( B + I ) = > d e t ( A B ) = d e t ( A ( B + I ) ) = > d e t ( A ) . d e t ( B ) = d e t ( A ) . d e t ( B + I ) = > d e t ( A ) ( d e t ( B ) d e t ( B + I ) ) = 0 AB-BA=A \\ => AB=BA+A \\ => AB=A(B+I) \\ => det(AB)=det(A(B+I)) \\ =>det(A).det(B)=det(A).det(B+I) \\ =>det(A)(det(B)-det(B+I))=0 .

It is clear that d e t ( B ) d e t ( B + I ) det(B) \neq det(B+I) .So d e t ( A ) = 0 det(A)=0

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