GATE CS 2018 Q26 — Eigenvalues and Eigenvectors, LU Decomposition

Since the given eigenvector $\begin{bmatrix} 1 \\ 4 \end{bmatrix}$ is a $2 \times 1$ vector, $P$ is a $2 \times 2$ matrix.

Let us analyze each statement:

• Statement II is necessarily true: If a $2 \times 2$ matrix $P$ has two distinct eigenvalues, it must have two linearly independent eigenvectors (one corresponding to each eigenvalue). Since the only eigenvectors of $P$ are multiples of $\begin{bmatrix} 1 \\ 4 \end{bmatrix}$, there is only $1$ linearly independent eigenvector. Thus, the eigenvalues cannot be distinct, meaning $P$ must have a repeated eigenvalue.
• Statement III is necessarily true: An $n \times n$ matrix is diagonalizable if and only if it has $n$ linearly independent eigenvectors. Since $P$ is a $2 \times 2$ matrix and has only $1$ linearly independent eigenvector, it cannot be diagonalized.
• Statement I is NOT necessarily true: A matrix with a repeated eigenvalue can still be invertible as long as the eigenvalue is non-zero. For example, consider the matrix: $$P = \begin{bmatrix} 2 & 0 \\ 8 & 2 \end{bmatrix}$$ The characteristic equation is $(\lambda - 2)^2 = 0$, giving a repeated eigenvalue $\lambda = 2$.
  The eigenvectors are found by solving $(P - 2I)x = 0$: $$\begin{bmatrix} 0 & 0 \\ 8 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies x_1 = 0$$ Thus, the only eigenvectors are multiples of $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$. Since $\det(P) = 4 \neq 0$, $P$ is invertible.
  By a simple change of basis, we can construct an invertible matrix whose only eigenvectors are multiples of $\begin{bmatrix} 1 \\ 4 \end{bmatrix}$. Hence, $P$ can be invertible.

Therefore, only II and III are necessarily true.