# Thread: [SOLVED] Proof question for eigenvectors

1. ## [SOLVED] Proof question for eigenvectors

The vectors $v_1, v_2, ..., v_k$ in the vector space $R^n$ are eigenvectors for the matrix $A$ corresponding to the eigenvalues $\lambda_1, \lambda_2, ..., \lambda_k$ (not necessarily distinct).

If these vectors are linearly independent and $\lambda_{k+1}$ is distinct from all of the other $\lambda$'s, prove that $\{ v_1, v_2, ..., v_k, v_{k+1} \}$ is linearly independent for any eigenvector $v_{k+1}$ corresponding to $\lambda_{k+1}$.

(Hint: if not, we have $v_{k+1}=c_1v_1+c_2v_2+...+c_kv_k$ for some constants and $Av_{k+1}$ can easily be written in two ways.)

2. suppose that $\{v_1,\dots,v_k,v_{k+1}\}$ is not linearly independant, hence we can write

$v_{k+1}=c_1 v_1+\dots +c_k v_k$
now consider the following

$Av_{k+1}=\lambda_{k+1}v_{k+1}=\lambda_{k+1}\left(c _1 v_1+\dots +c_k v_k\right)$

and we also have

$Av_{k+1}=A\left(c_1 v_1+\dots +c_k v_k\right)=\lambda_1 c_1 v_1+\dots +\lambda_k c_k v_k$
clearly $Av_{k+1}-Av_{k+1}=0$

hence

$\left(\lambda_{k+1} -\lambda_1\right)c_1 v_1+\dots+\left(\lambda_{k+1} -\lambda_k\right)c_k v_k=0$

and since $\{v_1,\dots,v_k\}$ is linearly independant we know this equality can only hold when the constants are zero, hence

$\lambda_{k+1}=\lambda_{i}$ for each $i=1,2,\dots,k$

so $\lambda_{k+1}$ is not distinct.

So this concludes the proof with a contrapositive argument.