Eigenvectors Corresponding to Distinct Eigenvalues of Linear Operator are Linearly Independent

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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.

$\Box$

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.

$\Box$

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.

$\blacksquare$


Sources

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