If A,B are matrices, B^2=B then rk(AB-BA)<=rk(AB+BA)



Given two matrices \( A \) and \( B \) where \( B^2 = B \) (indicating that \( B \) is idempotent), we need to prove that:

\[
\text{rank}(AB - BA) \leq \text{rank}(AB + BA)
\]

### **Proof Outline:**

1. **Idempotent Matrix Property**:
   Since \( B^2 = B \), \( B \) is idempotent. This implies that \( B \) acts as a projection operator, meaning that \( B \) projects vectors onto a subspace.

2. **Key Observation**:
   Since \( B \) is idempotent, it satisfies:
   \[
   (I - B)^2 = I - 2B + B^2 = I - B
   \]
   This shows that \( I - B \) is also idempotent, and hence another projection matrix.

3. **Consider Decomposition**:
   Decompose the space into two subspaces:
   - The image (range) of \( B \), denoted by \( \text{Im}(B) \).
   - The kernel (null space) of \( B \), denoted by \( \text{Ker}(B) \).

   Let \( V \) be the vector space that \( A \) and \( B \) act on. Then, \( V \) can be decomposed as:
   \[
   V = \text{Im}(B) \oplus \text{Ker}(B)
   \]

4. **Block Matrix Representation**:
   In a basis adapted to this decomposition, \( B \) can be written as a block matrix:
   \[
   B = \begin{pmatrix} I & 0 \\ 0 & 0 \end{pmatrix}
   \]
   Similarly, \( A \) can be written in block form corresponding to this decomposition:
   \[
   A = \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix}
   \]

5. **Matrix Products**:
   Now, consider the products \( AB \) and \( BA \):
   \[
   AB = \begin{pmatrix} A_{11} & 0 \\ A_{21} & 0 \end{pmatrix}, \quad BA = \begin{pmatrix} A_{11} & A_{12} \\ 0 & 0 \end{pmatrix}
   \]
   Therefore,
   \[
   AB - BA = \begin{pmatrix} 0 & -A_{12} \\ A_{21} & 0 \end{pmatrix}, \quad AB + BA = \begin{pmatrix} 2A_{11} & A_{12} \\ A_{21} & 0 \end{pmatrix}
   \]

6. **Rank Inequality**:
   To prove \( \text{rank}(AB - BA) \leq \text{rank}(AB + BA) \), notice that \( AB - BA \) is an off-diagonal matrix, while \( AB + BA \) contains both the diagonal and off-diagonal terms.

   The rank of a matrix is the number of linearly independent rows (or columns). Since the off-diagonal terms of \( AB - BA \) are contained within \( AB + BA \), and \( AB + BA \) additionally has diagonal components, the rank of \( AB - BA \) must be less than or equal to the rank of \( AB + BA \).

Thus, we conclude that:

\[
\text{rank}(AB - BA) \leq \text{rank}(AB + BA)
\]

### Do you want details or have any questions?

### Further Questions:
1. What are the implications of a matrix being idempotent in other contexts?
2. How does the decomposition of vector space help in simplifying matrix operations?
3. Can this result be generalized to non-idempotent matrices?
4. What are some real-world applications of this kind of matrix inequality?
5. How does the concept of matrix rank influence the solution of systems of linear equations?

**Tip:** Understanding the block matrix decomposition is a powerful technique for simplifying many matrix algebra problems.

Explore a proof involving matrix algebra and idempotent matrices where rk(AB - BA) <= rk(AB + BA). Understand the concept of matrix rank and its implications in linear algebra. Suitable for advanced undergraduate students and those interested in matrix theory.

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