Question

Let \( M = \begin{bmatrix} 4 & -3 \\ 1 & 0 \end{bmatrix} \).
Consider the following statements:
P: \(M^8 + M^{12}\) is diagonalizable.
Q: \(M^7 + M^9\) is diagonalizable.
Which of the following statements is correct?

• A. P is TRUE and Q is FALSE
• B. P is FALSE and Q is TRUE
• C. Both P and Q are FALSE
  (D) Both P and Q are TRUE
• D.

Hint:

An \(n \times n\) matrix with \(n\) distinct eigenvalues is always diagonalizable. If a matrix \(A\) is diagonalizable, so is \(p(A)\) for any polynomial \(p\). This provides a very quick way to solve problems like this without needing to compute the polynomial matrices themselves.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This question is about the diagonalizability of a matrix and polynomials of that matrix. A square matrix is diagonalizable if it has a full set of linearly independent eigenvectors. A key theorem states that if a matrix \(A\) is diagonalizable, then any polynomial in \(A\), say \(p(A)\), is also diagonalizable.

Step 2: Key Formula or Approach:
1. Determine if the matrix \(M\) is diagonalizable. An \(n \times n\) matrix is diagonalizable if it has \(n\) distinct eigenvalues.
2. If \(M\) is diagonalizable, then there exists an invertible matrix \(P\) and a diagonal matrix \(D\) such that \(M = PDP^{-1}\).
3. For any polynomial \(p(x)\), we have \(p(M) = p(PDP^{-1}) = P p(D) P^{-1}\). Since \(p(D)\) is also a diagonal matrix, \(p(M)\) is diagonalizable.

Step 3: Detailed Calculation:
First, let's check if \(M\) is diagonalizable by finding its eigenvalues. The characteristic equation is \(\det(M - \lambda I) = 0\).
\[ \det \begin{pmatrix} 4-\lambda & -3 \\ 1 & -\lambda \end{pmatrix} = (4-\lambda)(-\lambda) - (1)(-3) = 0 \]
\[ -4\lambda + \lambda^2 + 3 = 0 \]
\[ \lambda^2 - 4\lambda + 3 = 0 \]
\[ (\lambda - 1)(\lambda - 3) = 0 \]
The eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = 3\).
Since \(M\) is a \(2 \times 2\) matrix with two distinct eigenvalues, it is diagonalizable.

Now, consider the statements:
Statement P: The matrix \(M^8 + M^{12}\) can be written as \(p_1(M)\) where \(p_1(x) = x^8 + x^{12}\). Since \(M\) is diagonalizable and \(p_1(x)\) is a polynomial, the matrix \(p_1(M)\) is also diagonalizable. Thus, P is TRUE.
Statement Q: The matrix \(M^7 + M^9\) can be written as \(p_2(M)\) where \(p_2(x) = x^7 + x^9\). Since \(M\) is diagonalizable and \(p_2(x)\) is a polynomial, the matrix \(p_2(M)\) is also diagonalizable. Thus, Q is TRUE.
Both statements are correct based on the general theorem. The eigenvalues of \(p(M)\) are \(p(\lambda_i)\). In both cases, the resulting eigenvalues are distinct, but even if they were not, the matrix \(p(M)\) would still be diagonalizable (it would be a scalar multiple of the identity matrix, which is diagonal).

Step 4: Final Answer:
Both P and Q are TRUE.

Step 5: Why This is Correct:
The core principle is that any polynomial of a diagonalizable matrix is itself diagonalizable. We first established that \(M\) is diagonalizable because its eigenvalues are distinct. This immediately implies that both \(M^8 + M^{12}\) and \(M^7 + M^9\), being polynomials in \(M\), are also diagonalizable.