Question 36

The eigenvectors of matrix $P$ are all scalar multiples of $\begin{bmatrix} 1 \\ 4 \end{bmatrix}$. This means $P$ is a $2 \times 2$ matrix with only 1 linearly independent eigenvector.

Evaluate the statements:

• Statement I ($P$ does not have an inverse): A matrix can have a defective eigenspace while remaining invertible (e.g., $P = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}$ has eigenvalues $\lambda = 2, 2$ with determinant $4 \neq 0$). Thus, I is not necessarily true.
• Statement II ($P$ has a repeated eigenvalue): If $P$ had two distinct eigenvalues, the eigenvectors corresponding to distinct eigenvalues would be linearly independent, yielding two linearly independent eigenvectors. Since all eigenvectors are multiples of a single vector, $P$ cannot have distinct eigenvalues. Hence, $P$ must have a repeated eigenvalue. II is necessarily true.
• Statement III ($P$ cannot be diagonalized): A $2 \times 2$ matrix is diagonalizable if and only if it has 2 linearly independent eigenvectors. Since $P$ has only 1 linearly independent eigenvector, it is defective and cannot be diagonalized. III is necessarily true.

Therefore, only II and III are necessarily true, matching option D.