To determine the answer, let’s analyze the problem with a counterexample:

Key Observations:

The problem likely asks whether all eigenvectors of a 2x2 matrix are multiples of a single vector if the matrix has at least one eigenvector.

An eigenvector (v) satisfies (Av = \lambda v) for some scalar (\lambda). However, matrices can have multiple linearly independent eigenvectors.

Counterexample:

Consider the diagonal matrix (A = \begin{bmatrix}2 & 0 \ 0 & 3\end{bmatrix}):

• (A\begin{bmatrix}1 \ 0\end{bmatrix} = 2\begin{bmatrix}1 \ 0\end{bmatrix}) (eigenvector (\begin{bmatrix}1 \0\end{bmatrix}) with (\lambda=2)).
• (A\begin{bmatrix}0 \1\end{bmatrix} =3\begin{bmatrix}0 \1\end{bmatrix}) (eigenvector (\begin{bmatrix}0 \1\end{bmatrix}) with (\lambda=3)).

These eigenvectors are not multiples of each other.

Conclusion: The statement is false.

Answer: B. False.

(\boxed{B})

作者声明:本文包含人工智能生成内容。