Question

Let \[ A = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}. \] Consider the linear map \(T_A : M_4(\mathbb{R}) \to M_4(\mathbb{R})\) defined by \[ T_A(X) = AX - XA. \] Then the dimension of the range of \(T_A\) is _________.

Hint:

When \(A\) has repeated eigenvalues, analyzing \(T_A(X) = AX - XA\) is simplified by expressing \(X\) in block form corresponding to eigenvalue groups.

Answer 1

Correct Answer: 8

Step 1: Understand the structure of \(A.\)
Matrix \(A\) has eigenvalues \(1, 1, -1, -1.\) It can be written in block diagonal form: \[ A = \begin{pmatrix} I_2 & 0 \\ 0 & -I_2 \end{pmatrix}. \]
Step 2: Express \(X\) in block form.
Let \[ X = \begin{pmatrix} B & C \\ D & E \end{pmatrix}, \] where \(B, C, D, E\) are \(2\times2\) real matrices.
Step 3: Compute \(T_A(X).\)
\[ T_A(X) = AX - XA = \begin{pmatrix} I_2 & 0 \\ 0 & -I_2 \end{pmatrix} \begin{pmatrix} B & C \\ D & E \end{pmatrix} - \begin{pmatrix} B & C \\ D & E \end{pmatrix} \begin{pmatrix} I_2 & 0 \\ 0 & -I_2 \end{pmatrix}. \] This gives: \[ T_A(X) = \begin{pmatrix} 0 & 2C \\ -2D & 0 \end{pmatrix}. \]
Step 4: Determine the range.
The image consists of all matrices of the above form, where \(C, D\) are arbitrary \(2 \times 2\) matrices. Thus: \[ \dim(\text{Range } T_A) = \dim(C) + \dim(D) = 4 + 4 = 8. \]
Step 5: Conclusion.
Hence, the dimension of the range of \(T_A\) is \(\boxed{8}.\)