# Show that given n distinct eigenvalues, eigenvectors are independent

1. Apr 3, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
Let $T$ be a linear operator on a vector space $V$, and let $\lambda_1,\lambda_2, \dots, \lambda_n$ be distinct eigenvalues of $T$. If $v_1, v_2, \dots , v_n$ are eigenvectors of $T$ such that $\lambda_i$ corresponds to $v_i \ (1 \le i \le k)$, then $\{ v_1, v_2, \dots , v_n \}$ is linearly independent.

2. Relevant equations

3. The attempt at a solution
The proof is by induction.
First, suppose that $n=1$. Then $v_1 \ne 0$ since it is an eigenvector and hence $\{ v_1 \}$ is linearly independent.
Now assume that the theorem holds of $n=k$, and that we have $k+1$ eigenvectors corresponding to distinct eigenvalues. We wish to show that these $k+1$ eigenvectors are linearly independent.
Suppose that $a_1, a_2 , \dots, a_n$ are scalars such that $a_1 v_1 + a_2 v_2 + \cdots + a_{k+1} v_{k+1} = 0$ (1). Apply $T$ to both sides to get $a_1 \lambda_1 v_1 + a_2 \lambda_2 v_2 + \cdots + a_{k+1} \lambda_{k+1} v_{k+1} = 0$ (2).

Now, my plan to continue the proof is to multiply (1) by $\lambda_{k+1}$ and then subtract (2) from it to get a linear relation only in terms of the vectors $v_1, v_2 , \dots , v_k$. However, I am concerned about assuming what $\lambda_{k+1}$ is. Do I need to split the proof into two cases, one where $\lambda_{k+1} = 0$ and one where $\lambda_{k+1} \ne 0$?

2. Apr 3, 2017

### Staff: Mentor

I think so. If you formulate the induction hypothesis as for all $n \leq k$, then in the case $\lambda_{k+1}=0$ you can apply the argument on $\lambda_{k}\neq 0$ and then consider (1) once again before the multiplication by $\lambda_{k}$.

You could as well start with $\lambda_1=0$ and anchor the induction here to both cases.