Discussion
Consider a matrix P whose only eigenvectors are the multiples of \(\begin{bmatrix} 1 \\ 4 \end{bmatrix}\).
Consider the following statements.
(I) P does not have an inverse
(II) P has a repeated eigenvalue
(III) P cannot be diagonalized
Which one of the following options is correct?
Answer: D
Only II and III are necessarily true.
The eigenvectors live in two dimensions, so \(P\) is a 2 by 2 matrix. If it had two different eigenvalues, each would bring its own eigenvector, and the two would be independent. Since every eigenvector is a multiple of \((1, 4)\), there is only one eigenvalue, repeated (II).
A 2 by 2 matrix is diagonalizable only when it has two independent eigenvectors. Here there is only one direction, so \(P\) cannot be diagonalized (III).
Statement I need not hold. For example \(P = \begin{bmatrix} 5 & -1 \\ 16 & -3 \end{bmatrix}\) has the single eigenvalue 1, its eigenvectors are the multiples of \((1, 4)\), and its determinant is \(-15 + 16 = 1\), so it is invertible. That rules out the options that include I and the option with II alone.
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