Sachchidanand Prasad

30-06-2024

Problem: Let $T: V \to W$ be a linear transformation between two finite dimensional vector spaces $V$ and $W$. Check whether the following statements are true or false. If it is true prove it and if false, then provide a counterexample.
  • If $Tv = \lambda v$ for some $\lambda \in \mathbb{R} $, then $v$ is an eigenvector with an eigenvalue $\lambda $.
  • If $v_1$ and $v_2$ are linearly independent eigenvectors corresponding to the eigenvalues $\lambda _1$ and $\lambda _2$, respectively. Then $\lambda _1 \neq \lambda _2$.
  • An eigenspace of $T$ is a null space of a certain matrix.
linear-algebra eigenvalues eigenvectors
Solution:
  • The statement is False. To see this, note that if $v = 0$, then $T v = 0 = \lambda v$ is true for any $\lambda $. We recall that $\lambda $ is an eigenvalue if there exists a nonzero vector $v \in V$ such that $Tv = \lambda v$.
  • The statement is False. In order to see a counter example, take $V = W = \mathbb{R} ^2$ and let $T$ is the identity map. Then we have \begin{align*} T \left( e_1 = \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \right) &= \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \text{ and } T \left( e_2 = \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \right) = \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}. \end{align*} It is clear that $e_1$ and $e_2$ are linearly independent eigenvectors with the same eigenvalue $1$.
  • The statement is True. Recall that if $E_\lambda $ denotes the eigenspace corresponding to the eigenvalue $\lambda $, then \[ E_\lambda = \mathrm{Null} (T - \lambda I). \] Thus, the eigenspace is the null space of the matrix $T - \lambda I$.