Let `psi_A` be defined as trace of a matrix `A` which is sum of diagonal elements of a square matrix. Which of the following is true?

Options

A. (a) `psi_(A+B)=psi_(A-B)`
B. (b) `psi_(A+B)=psi_(AB)`
C. (c) `psi_(AB)=psi_(BA)`
D. (d) `psi_(A-B)=psi_(BA)`

Answer

To solve the problem, we need to analyze the trace of matrices and how it behaves under addition and multiplication. Let's denote the trace of a matrix \( A \) as \( \text{tr}(A) \).

### Step-by-Step Solution:

1. **Define the Matrices**:
Let matrix \( A \) have diagonal elements \( X, Y, Z \) and matrix \( B \) have diagonal elements \( A, B, C \).
\[ A = \begin{pmatrix} X & 0 & 0 \\ 0 & Y & 0 \\ 0 & 0 & Z \end{pmatrix}, \quad B = \begin{pmatrix} A & 0 & 0 \\ 0 & B & 0 \\ 0 & 0 & C \end{pmatrix} \]

2. **Calculate \( A + B \)**:
The sum of matrices \( A \) and \( B \) is:
\[ A + B = \begin{pmatrix} X + A & 0 & 0 \\ 0 & Y + B & 0 \\ 0 & 0 & Z + C \end{pmatrix} \]
The trace of \( A + B \) is:
\[ \text{tr}(A + B) = (X + A) + (Y + B) + (Z + C) = X + Y + Z + A + B + C \]

3. **Calculate \( A - B \)**:
The difference of matrices \( A \) and \( B \) is:
\[ A - B = \begin{pmatrix} X - A & 0 & 0 \\ 0 & Y - B & 0 \\ 0 & 0 & Z - C \end{pmatrix} \]
The trace of \( A - B \) is:
\[ \text{tr}(A - B) = (X - A) + (Y - B) + (Z - C) = X + Y + Z - A - B - C \]

4. **Calculate \( AB \)**:
The product of matrices \( A \) and \( B \) is:
\[ AB = \begin{pmatrix} X & 0 & 0 \\ 0 & Y & 0 \\ 0 & 0 & Z \end{pmatrix} \begin{pmatrix} A & 0 & 0 \\ 0 & B & 0 \\ 0 & 0 & C \end{pmatrix} = \begin{pmatrix} XA & 0 & 0 \\ 0 & YB & 0 \\ 0 & 0 & ZC \end{pmatrix} \]
The trace of \( AB \) is:
\[ \text{tr}(AB) = XA + YB + ZC \]

5. **Calculate \( BA \)**:
The product of matrices \( B \) and \( A \) is:
\[ BA = \begin{pmatrix} A & 0 & 0 \\ 0 & B & 0 \\ 0 & 0 & C \end{pmatrix} \begin{pmatrix} X & 0 & 0 \\ 0 & Y & 0 \\ 0 & 0 & Z \end{pmatrix} = \begin{pmatrix} AX & 0 & 0 \\ 0 & BY & 0 \\ 0 & 0 & CZ \end{pmatrix} \]
The trace of \( BA \) is:
\[ \text{tr}(BA) = AX + BY + CZ \]

6. **Compare Traces**:
Now we can summarize:
- \( \text{tr}(A + B) = X + Y + Z + A + B + C \)
- \( \text{tr}(A - B) = X + Y + Z - A - B - C \)
- \( \text{tr}(AB) = XA + YB + ZC \)
- \( \text{tr}(BA) = AX + BY + CZ \)

From the calculations, we can see that:
\[ \text{tr}(AB) = \text{tr}(BA) \]

### Conclusion:
The correct answer is that the trace of the product of two matrices is invariant under the order of multiplication:
\[ \text{tr}(AB) = \text{tr}(BA) \]