$T:\Bbb{R}^2\rightarrow\Bbb{R}^2$ has 2 distinct eigenvalues. Showing that $v$ or $T(v)− \lambda_1v$ are eigenvectors of $T$

I will assume that $v\ne0$; otherwise, the statement is false.

Let $v_1$ be an eigenvector corresponding to the eigenvalue $\lambda_1$ and let $v_2$ be an eigenvector corresponding to the eigenvalue $\lambda_2$. If $v\in\Bbb R^2$, then $v$ can be written as $\alpha_1v_1+\alpha_2v_2$. There are two possibilites now:

  • $\alpha_2=0$: then $v=\alpha_1v_1$ and therefoe it is an eigenvector corresponding to the eigenvalue $\lambda_1$.
  • $\alpha_2\ne0$: then $T(v)-\lambda_1v=\alpha_2(\lambda_2-\lambda_1)v_2$ and therefore $T(v)-\lambda_1v$ is an eigenvector corresponding to the eigenvalue $\lambda_2$.

In the second case, consider $$ T \bigl( T(v)-\lambda_1 v \bigr) = T^2(v) - \lambda_1 T(v) . $$ By Cayley–Hamilton, $T^2(v)$ can be replaced with $(\lambda_1+\lambda_2)T(v) - \lambda_1 \lambda_2 v$.

Can you continue from there?

Tags:

Linear Algebra

Eigenvalues Eigenvectors

Linear Transformations

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