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 $\det \left(\mathbb{I}+2P+3Q \right) = 1$ .