Discussion
Let
\[ P = \begin{bmatrix} 4 & -2 & 2 \\ 6 & -3 & 4 \\ 3 & -2 & 3 \end{bmatrix}, \quad Q = \begin{bmatrix} 3 & -2 & 2 \\ 4 & -4 & 6 \\ 2 & -3 & 5 \end{bmatrix}. \]
The eigenvalues of both \(P\) and \(Q\) are 1, 1, and 2. Which one of the following statements is TRUE?
Answer: B
The true statement is \(P\) is diagonalizable but \(Q\) is NOT diagonalizable.
A matrix is diagonalizable when each eigenvalue has as many independent eigenvectors as its multiplicity. The repeated eigenvalue is 1, so look at the rank of \(M - I\); the eigenspace dimension is \(3 - \text{rank}\).
\[ P - I = \begin{bmatrix} 3 & -2 & 2 \\ 6 & -4 & 4 \\ 3 & -2 & 2 \end{bmatrix} \]
Every row is a multiple of \((3, -2, 2)\), so the rank is 1 and the eigenspace for 1 has dimension 2. With one more eigenvector for 2, \(P\) has three independent eigenvectors and is diagonalizable.
\[ Q - I = \begin{bmatrix} 2 & -2 & 2 \\ 4 & -5 & 6 \\ 2 & -3 & 4 \end{bmatrix} \]
Subtracting twice row 1 from row 2 and row 1 from row 3 leaves \((0, -1, 2)\) twice, so the rank is 2 and the eigenspace for 1 has dimension only 1. \(Q\) is short of an eigenvector and is not diagonalizable, which rules out the other three combinations.
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