Question

Let \( A \) be an \( n \times n \) invertible matrix and \( C \) be an \( n \times n \) nilpotent matrix. If \[ X = \begin{pmatrix} X_{11} & X_{12} \\ X_{21} & X_{22} \end{pmatrix} \] is a \( 2n \times 2n \) matrix (each \( X_{ij} \) is \( n \times n \)) that commutes with the \( 2n \times 2n \) matrix \[ B = \begin{pmatrix} A & 0 \\ 0 & C \end{pmatrix}, \] then

• A. \( X_{11} \) and \( X_{22} \) are necessarily zero matrices.
• B. \( X_{12} \) and \( X_{21} \) are necessarily zero matrices.
• C. \( X_{11} \) and \( X_{21} \) are necessarily zero matrices.
• D. \( X_{12} \) and \( X_{22} \) are necessarily zero matrices.

Hint:

When a matrix commutes with a block-diagonal matrix, its off-diagonal blocks vanish if one diagonal block is invertible and the other is nilpotent.

Answer 1

The Correct Option is B

Step 1: Commutation condition.
Given that \( XB = BX \), we write explicitly: \[ \begin{pmatrix} X_{11} & X_{12} \\ X_{21} & X_{22} \end{pmatrix} \begin{pmatrix} A & 0 \\ 0 & C \end{pmatrix} = \begin{pmatrix} A & 0 \\ 0 & C \end{pmatrix} \begin{pmatrix} X_{11} & X_{12} \\ X_{21} & X_{22} \end{pmatrix}. \]
Step 2: Expand both sides.
Left-hand side: \[ \begin{pmatrix} X_{11}A & X_{12}C \\ X_{21}A & X_{22}C \end{pmatrix}, \qquad \text{Right-hand side: } \begin{pmatrix} AX_{11} & AX_{12} \\ CX_{21} & CX_{22} \end{pmatrix}. \] Equating corresponding blocks gives: \[ X_{11}A = AX_{11}, \quad X_{22}C = CX_{22}, \quad X_{12}C = AX_{12}, \quad X_{21}A = CX_{21}. \]
Step 3: Analyze the off-diagonal blocks.
Since \(A\) is invertible and \(C\) is nilpotent, consider \( X_{12}C = AX_{12} \). Multiply on the right by \( C^k = 0 \) (for some \( k \)): \[ X_{12}C^k = A^k X_{12} = 0 \quad \Rightarrow \quad X_{12} = 0. \] Similarly, \( X_{21}A = CX_{21} \) implies: \[ C^k X_{21} = 0 = A^k X_{21} \quad \Rightarrow \quad X_{21} = 0. \]
Step 4: Conclusion.
Thus, \( X_{12} \) and \( X_{21} \) are zero matrices, while \( X_{11} \) and \( X_{22} \) commute with \( A \) and \( C \), respectively. Therefore, (B) is correct.