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Zero determinant of a matrix

Suppose that for the complex square matrices A, B it holds

(1)   \begin{equation*} AB - BA =A \end{equation*}

Prove that \det A =0.

Solution

If \det A\ne0, then A is invertible. So we get

    \[I=A^{-1}A=A^{-1}(AB-BA)=B-A^{-1}BA\]

But this equality is impossible: taking trace on both sides, we get

n=\text{Tr}(I)=\text{Tr}(B-A^{-1}BA)=\text{Tr}(B)-\text{Tr}(A^{-1}BA)=\text{Tr}(B)-\text{Tr}(B)=0

We get a contradiction, which shows that A is singular, i.e. \det A=0.

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