Eigenvectors Corresponding to Distinct Eigenvalues of Linear Operator are Linearly Independent

== Theorem ==
Let $K$ be a field. 
Let $X$ be a vector space over $K$. 
Let $T : X \to X$ be a linear operator.
Let $\lambda_1, \lambda_2, \ldots, \lambda_n \in K$ be distinct eigenvalues of $T$.
Let $x_1, x_2, \ldots, x_n$ be eigenvectors corresponding to $\lambda_1, \lambda_2, \ldots, \lambda_n$. 
Then $\set {x_1, \ldots, x_n}$ is a linearly independent set.
 
== Proof ==
Proof by induction:
Let $\map P n$ be the proposition:
:for any $n$ eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n \in K$ and eigenvectors $x_1, x_2, \ldots, x_n$ corresponding to $\lambda_1, \lambda_2, \ldots, \lambda_n$, we have that: 
::$\set {x_1, \ldots, x_n}$ is a linearly independent set.
=== Basis for the Induction ===
Take $n = 1$.
From Singleton is Linearly Independent, $\set {x_1}$ is linearly independent set.
=== Induction Hypothesis ===
This is our induction hypothesis:
:for any $n$ eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n \in K$ and eigenvectors $x_1, x_2, \ldots, x_n$ corresponding to $\lambda_1, \lambda_2, \ldots, \lambda_n$, we have that: 
::$\set {x_1, \ldots, x_n}$ is a linearly independent set.
We need to show: 
:for any $n + 1$ eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_{n + 1} \in K$ and eigenvectors $x_1, x_2, \ldots, x_{n + 1}$ corresponding to $\lambda_1, \lambda_2, \ldots, \lambda_{n + 1}$, we have that: 
::$\set {x_1, \ldots, x_{n + 1} }$ is a linearly independent set.
=== Induction Step ===
Suppose that $\lambda_{n + 1}$ is an eigenvalue distinct from $\lambda_1, \lambda_2, \ldots, \lambda$ and an eigenvector $x_{n + 1}$ corresponding to $\lambda_{n + 1}$.
Suppose that: 
:$\ds \sum_{k \mathop = 1}^{n + 1} \alpha_k x_k = \mathbf 0_X$
for $\alpha_1, \ldots, \alpha_{n + 1}$.
Applying $T$ to the above equation we have: 
:$\ds \sum_{k \mathop = 1}^{n + 1} \lambda_k \alpha_k x_k = \mathbf 0_X$
from the linearity of $T$. 
Instead multiplying the previous equation by $\lambda_{n + 1}$, we have: 
:$\ds \sum_{k \mathop = 1}^{n + 1} \lambda_{n + 1} \alpha_k x_k = \mathbf 0_X$
So, we have:
:$\ds \sum_{k \mathop = 1}^{n + 1} \lambda_{n + 1} \alpha_k x_k = \sum_{k \mathop = 1}^{n + 1} \lambda_k \alpha_k x_k$
so that:
:$\ds \lambda_{n + 1} \alpha_{n + 1} x_{n + 1} + \lambda_{n + 1} \sum_{k \mathop = 1}^n \alpha_k x_k = \lambda_{n + 1} \alpha_{n + 1} x_{n + 1} + \sum_{k \mathop = 1}^n \lambda_k \alpha_k x_k$
So, we have: 
:$\ds \sum_{k \mathop = 1}^n \alpha_k \paren {\lambda_{n + 1} - \lambda_k} x_k = \mathbf 0_X$
From the induction hypothesis, we have that $x_1, \ldots, x_n$ are linearly independent, so we have: 
:$\alpha_k \paren {\lambda_{n + 1} - \lambda_k} = 0$
for each $1 \le k \le n$. 
Since $\lambda_{n + 1} - \lambda_k \ne 0$, we must have $\alpha_k = 0$ for each $1 \le k \le n$. 
So we have: 
:$\alpha_{n + 1} x_{n + 1} = \mathbf 0_X$
Since $x_{n + 1} \ne 0$ by definition, we have $\alpha_{n + 1} = 0$.
So if: 
:$\ds \sum_{k \mathop = 1}^{n + 1} \alpha_k x_k = \mathbf 0_X$
then $\alpha_1 = \alpha_2 = \cdots = \alpha_{n + 1} = 0$.
So $x_1, \ldots, x_{n + 1}$ are linearly independent.
== Sources ==
* $14.1$: The Resolvent and Spectrum
Category:Eigenvalues
Category:Linear Operators