Let \(P\) and \(Q\) be \(3 \times 3\) nonzero real matrices. Assume that there exists a \(3 \times 3\) real nonsingular matrix \(S\) such that \(S^{-1}PS\) and \(S^{-1}QS\) are both upper triangular. Which of the following statements is/are TRUE?
Show Hint
Simultaneous triangularization forces the commutator \(PQ-QP\) to be strictly upper triangular in that basis, so it is nilpotent with all eigenvalues \(0\); use this to check invertibility, shifting by \(I\), and diagonalizability.
The matrix \(I + PQ - QP\) is invertible, where \(I\) is the \(3 \times 3\) identity matrix.
If \(PQ - QP\) is a nonzero matrix, then \(PQ - QP\) is diagonalizable.
Show Solution
The Correct Option isB, C
Solution and Explanation
Step 1: Understanding the Concept.
Simultaneous triangularization by the same $S$ means $P$ and $Q$ can each be split, in the basis given by $S$, into a diagonal part and a strictly upper triangular part: $T_1 = D_1 + N_1$ and $T_2 = D_2 + N_2$, where $D_1, D_2$ are diagonal (carrying the eigenvalues of $P$ and $Q$) and $N_1, N_2$ are strictly upper triangular.
Step 2: Expand the commutator using this split.
\[ T_1T_2 - T_2T_1 = (D_1+N_1)(D_2+N_2) - (D_2+N_2)(D_1+N_1) \]
Expanding and grouping, this equals $[D_1,D_2] + [D_1,N_2] + [N_1,D_2] + [N_1,N_2]$, where $[A,B]=AB-BA$. Two diagonal matrices always commute, so $[D_1,D_2] = 0$.
Step 3: Track the remaining terms.
A diagonal matrix times a strictly upper triangular matrix, in either order, is still strictly upper triangular, and a commutator of two strictly upper triangular matrices is also strictly upper triangular. So every one of the three remaining terms is strictly upper triangular, and their sum $T_1T_2-T_2T_1$ is strictly upper triangular as well.
Step 4: Nilpotency and its consequences.
Any $3\times3$ strictly upper triangular matrix $N$ satisfies $N^3=0$, so it is nilpotent, with every eigenvalue equal to $0$. Since $PQ-QP$ is similar to $T_1T_2-T_2T_1$ (via the same $S$), it is nilpotent too, and all three of its eigenvalues are $0$. This directly makes option (B) true.
Step 5: Check the determinant-based options.
Because the eigenvalues of $PQ-QP$ are all $0$, $\det(PQ-QP) = 0 \times 0 \times 0 = 0$, so $PQ-QP$ is singular, not invertible, making option (A) false. Shifting by the identity shifts every eigenvalue up by $1$, so the eigenvalues of $I+PQ-QP$ are $1,1,1$, giving $\det(I+PQ-QP)=1 \neq 0$. So $I+PQ-QP$ is always invertible, making option (C) true.
Step 6: Check diagonalizability in option (D).
A nonzero matrix all of whose eigenvalues are $0$ can be diagonalizable only if it is similar to the zero matrix, which forces the matrix itself to be zero. That contradicts the "nonzero" assumption in option (D), so a nonzero nilpotent matrix is never diagonalizable, and option (D) is false.
Step 7: Final Answer.
The statements that are always true are (B), $PQ-QP$ is nilpotent, and (C), $I+PQ-QP$ is invertible.
\[ \boxed{\text{(B) and (C)}} \]
