# Thread: different eigenvalues

1. ## different eigenvalues

Suppose you find n different eigenvalues for an nxn matrix. Are you then guaranteed, that the eigenvectors corresponding to those eigenvalues are linearly independent?
If so, from what does that follow?

2. ## Re: different eigenvalues

Yes. Suppose you have n different eigenvalues $c_1, \ldots, c_n$ of an endomorfism $f: V\to V$ corresponding to the eigenvectors $v_1, \ldots, v_n$ then the set $\{v_1, \ldots, v_n\}$ is linearly independent.

Suppose the eigenvectors are linearly dependent, then we know there's a first vector $v_k$ in the row which can be written as a linear combination of the previous vectors $v_1, \ldots, v_{k-1}$. Thus $\forall a_i \in K$:
$v_k=a_1v_1+\ldots+a_{k-1}v_{k-1}$
and $v_1, \ldots, v_{k-1}$ are linearly independent.
We have:
$f(v_k)=f(a_1v_1+\ldots+a_{k-1}v_{k-1})=a_1c_1v_1+\ldots+a_{k-1}c_{k-1}v_{k-1}$ and
$f(v_k)=c_kv_k=c_ka_1v_1+\ldots+c_ka_{k-1}v_{k-1}$

Thus:
$a_1c_1v_1+\ldots+a_{k-1}c_{k-1}v_{k-1}=c_ka_1v_1+\ldots+c_ka_{k-1}v_{k-1}$
$\Rightarrow (c_1-c_k)a_1v_1+\ldots+(c_{k-1}-c_k)a_{k-1}v_{k-1}=0$

But because $v_1,\ldots,v_{k-1}$ are linearly independent alle the coefficients are equal to zero, so:
$(c_1-c_k)a_1=\ldots=(c_{k-1}-c_k)a_{k-1}=0$
But because all the eigenvalues are different $a_1=\ldots=a_{k-1}=0$ and so $v_k=0$, this is a contradiction because $v_k$ wa supposed to be an eigenvector so our statement is wrong and the eigenvectors $v_1, \ldots,v_k$ are indeed linearly independent.