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The question is asking to find the rank of a \( 5 \times 5 \) diagonalizable matrix \( A \) given its eigenvalues: \( 0, 0, 0, 5, 2 \). Here's how to solve it:

### Step 1: Understand the Eigenvalues and Rank
The eigenvalues of a matrix determine its rank because:
- The rank of a matrix is the number of **non-zero rows in its row-reduced echelon form**.
- The number of non-zero eigenvalues of a diagonalizable matrix corresponds to the rank of the matrix.

### Step 2: Count Non-Zero Eigenvalues
From the given eigenvalues \( 0, 0, 0, 5, 2 \):
- There are **two non-zero eigenvalues** (\( 5 \) and \( 2 \)).

### Step 3: Rank of \( A \)
Since there are 2 non-zero eigenvalues, the rank of \( A \) is **2**.

### Final Answer:
\[
\text{rank}(A) = 2
\]

Would you like me to explain this further or discuss related concepts? Here are some follow-up questions you might find useful:

1. Why does the number of non-zero eigenvalues correspond to the rank of the matrix?
2. How does diagonalizability of a matrix affect the interpretation of eigenvalues?
3. What happens to the rank of \( A \) if one more eigenvalue is \( 0 \)?
4. Can a non-diagonalizable matrix have the same rank in this case? Why or why not?
5. What is the connection between the nullity and rank of a matrix in this context?

**Tip**: Always check the eigenvalues and multiplicities when dealing with rank and nullity problems in diagonalizable matrices!

A diagonalizable 5 × 5 matrix A has eigenvalues 0, 0, 0, 5, 2 (i.e., 0 is a triple eigenvalue and 5, 2 are simple eigenvalues). Find the rank of A.

Learn how to determine the rank of a 5 × 5 diagonalizable matrix with eigenvalues 0, 0, 0, 5, 2. This problem explores key linear algebra concepts like diagonalization, eigenvalues, and the rank-nullity theorem. Suitable for university-level students.

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