Question

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?

• A. The matrix \(PQ - QP\) is invertible.
• B. The matrix \(PQ - QP\) is nilpotent.
• C. The matrix \(I + PQ - QP\) is invertible, where \(I\) is the \(3 \times 3\) identity matrix.
• D. If \(PQ - QP\) is a nonzero matrix, then \(PQ - QP\) is diagonalizable.

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.

Answer 1

The Correct Option is B, C

Step 1: Understanding the Question.
We are told \(P\) and \(Q\) can be brought to upper triangular form by the same similarity transform \(S\), that is, \(T_1 = S^{-1}PS\) and \(T_2 = S^{-1}QS\) are both upper triangular \(3 \times 3\) matrices. We need to work out what this forces about the commutator \(PQ - QP\).

Step 2: Transform the commutator.
Since \(T_1 = S^{-1}PS\) and \(T_2 = S^{-1}QS\), we get \(P = ST_1S^{-1}\) and \(Q = ST_2S^{-1}\). So
\[ S^{-1}(PQ - QP)S = T_1T_2 - T_2T_1 \]
This means \(PQ - QP\) is similar to the commutator \(T_1T_2 - T_2T_1\) of two upper triangular matrices, so both matrices share the same eigenvalues and the same nilpotency behaviour.

Step 3: Show the commutator of two triangular matrices is strictly upper triangular.
The product of two upper triangular matrices is upper triangular, and the diagonal entry of the product only involves the diagonal entries of the factors: \((T_1T_2)_{ii} = (T_1)_{ii}(T_2)_{ii}\), and likewise \((T_2T_1)_{ii} = (T_2)_{ii}(T_1)_{ii}\). Since ordinary numbers commute, these two diagonal entries are equal, so every diagonal entry of \(T_1T_2 - T_2T_1\) is zero. A \(3\times3\) upper triangular matrix with all zero diagonal entries is strictly upper triangular, and every strictly upper triangular \(3\times3\) matrix \(N\) satisfies \(N^3 = 0\).

Step 4: Conclude PQ - QP is nilpotent.
Since \(T_1T_2 - T_2T_1\) is nilpotent and \(PQ-QP\) is similar to it, \(PQ-QP\) is nilpotent too (similarity preserves \(M^k = 0\), because if \(S^{-1}MS = N\) and \(N^3=0\), then \(M^3 = SN^3S^{-1} = 0\)). So option (B) is TRUE.

Step 5: Use nilpotency to test options (A) and (C).
A nilpotent matrix has all eigenvalues equal to \(0\), so its determinant, the product of its eigenvalues, is \(0\). This means \(PQ-QP\) can never be invertible, so option (A) is FALSE. On the other hand, the eigenvalues of \(I + (PQ-QP)\) are \(1 + 0 = 1\) for each of the three eigenvalues of \(PQ-QP\), so
\[ \det(I + PQ - QP) = 1 \times 1 \times 1 = 1 \neq 0 \]
so \(I + PQ - QP\) is always invertible. Option (C) is TRUE.

Step 6: Test option (D).
Suppose \(PQ-QP\) is a nonzero matrix and, for contradiction, that it is diagonalizable. A diagonalizable matrix is similar to a diagonal matrix built from its own eigenvalues. Since all the eigenvalues of \(PQ-QP\) are \(0\) (Step 5), that diagonal matrix would be the zero matrix, so \(PQ-QP\) itself would have to be zero, contradicting the assumption that it is nonzero. So a nonzero nilpotent matrix is never diagonalizable, and option (D) is FALSE.

Final Answer:
The true statements are (B) and (C).
\[ \boxed{PQ-QP \text{ is nilpotent, and } I+PQ-QP \text{ is invertible}} \]