Question

Let \[ P=\begin{bmatrix}4 & -2 & 2\\ 6 & -3 & 4\\ 3 & -2 & 3\end{bmatrix}, \qquad Q=\begin{bmatrix}3 & -2 & 2\\ 4 & -4 & 6\\ 2 & -3 & 5\end{bmatrix}. \] The eigenvalues of both $P$ and $Q$ are $1,1,2$. Which one of the following statements is TRUE?

• A. Both $P$ and $Q$ are diagonalizable
• B. $P$ is diagonalizable but $Q$ is NOT diagonalizable
• C. $P$ is NOT diagonalizable but $Q$ is diagonalizable
• D. Both $P$ and $Q$ are NOT diagonalizable

Hint:

When eigenvalues are known, skip full diagonalization: compare $\dim\ker(A-\lambda I)$ to the algebraic multiplicity. For size $3$ with eigenvalues $1,1,2$, you only need $\rank(A-I)$.

Answer 1

The Correct Option is B

Step 1: Criterion.
A \(3 \times 3\) matrix with eigenvalues \(1, 1, 2\) is diagonalizable \( \Leftrightarrow \) the geometric multiplicity of \( \lambda = 1 \) equals its algebraic multiplicity 2, i.e. \[ \dim \ker(A - I) = 2. \] 
Step 2: Check \(P\).
Compute \( \text{rank}(P - I) \): \[ P - I = \begin{bmatrix} 3 & -2 & 2 \\ 6 & -4 & 4 \\ 3 & -2 & 2 \end{bmatrix}. \] Row-reduction shows \( \text{rank}(P - I) = 1 \Rightarrow \dim \ker(P - I) = 3 - 1 = 2 \). Hence the eigenspace for \( \lambda = 1 \) has dimension 2 (and the eigenspace for \( \lambda = 2 \) has dimension 1), so \(P\) is diagonalizable. 
Step 3: Check \(Q\).
\[ Q - I = \begin{bmatrix} 2 & -2 & 2 \\ 4 & -5 & 6 \\ 2 & -3 & 4 \end{bmatrix}. \] Row-reduction gives \( \text{rank}(Q - I) = 2 \Rightarrow \dim \ker(Q - I) = 3 - 2 = 1 < 2 \). Thus the geometric multiplicity of \( \lambda = 1 \) is only 1; \(Q\) is not diagonalizable. 
Final Answer:\(P\) diagonalizable, \(Q\) not diagonalizable