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Determinant of nilpotent matrices

Let P, Q be nilpotent matrices such that PQ + P+Q=0 . Evaluate the determinant

    \[\Delta  = \det \left( \mathbb{I}  + 2 P + 3 Q \right)\]

Solution

Lemma: If A and B are nilpotent matrices that commute and a,b are scalars, then aA + bB is nilpotent.

Proof: Since A and B commute, they are simultaneously triangularizable. Let S be an invertible matrix such that A = STS^{-1} and B = SUS^{-1}, where T and U are upper triangular. Note that since A and B are nilpotent, T and U must have zeros down the main diagonal. Hence aT + bU is upper triangular with zeros along the main diagonal which means that it’s nilpotent. Finally aA + bB = S(aT + bU)S^{-1} and so aA + bB is nilpotent.

We have \left(P+\mathbb{I}\right)\left(Q+\mathbb{I}\right) = \mathbb{I} \implies \left(Q+\mathbb{I} \right) \left(P+\mathbb{I}\right) = \mathbb{I}. Then equating the two left hand sides and simplifying gives us PQ = QP. Thus by the lemma we know that 2P + 3Q is nilpotent, i.e., it’s eigenvalues are all zero. It follows that the eigenvalues of \mathbb{I} + 2P + 3Q are all one and so \det \left(\mathbb{I}+2P+3Q \right) = 1.

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