Why are Eigenvectors from Distinct Eigenvalues Linearly Independent?
In this post we will give a proof of the following theorem.
| Theorem |
|---|
| If $\vec v_1, \vec v_2, \ldots , \vec v_r$ are eigenvectors of matrix $A$ that correspond to distinct eigenvalues $\lambda_1, \lambda_2, \ldots , \lambda_r$, then the set of eigenvectors $\{\vec v_1, \vec v_2, \ldots , \vec v_r\}$ is linearly independent. |
Proof
We will prove the above theorem by contradiction. Suppose that the set of vectors $\vec v_1, \vec v_2, \ldots \vec v_p$ are linearly dependent and are associated with distinct eigenvalues, where $p\le r$ is the smallest integer so that the first $p-1$ vectors are independent. Because the vectors are linearly dependent we can write that
$$c_1\vec v_1 + c_2\vec v_2 + \ldots + c_{p-1}\vec v_{p-1} = \vec v_p \quad (1)$$
Multiplying the above equation by $A$ and using that $A\vec v_i = \lambda_i \vec v_i$ we obtain:
\begin{align}c_1A\vec v_1 + c_2A\vec v_2 + \ldots + c_{p-1}A\vec v_{p-1} &= A\vec v_p \\ c_1\lambda_1\vec v_1 + c_2\lambda_2\vec v_2 + \ldots + c_{p-1}\lambda_{p-1}\vec v_{p-1} &= \lambda_p \vec v_p \quad\quad (2)\end{align}
Multiplying (1) by $\lambda_p$ and subtracting (2) yields:
\begin{align} c_1(\lambda_1-\lambda_p)\vec v_1 + c_2(\lambda_2-\lambda_p)\vec v_2 + \ldots + c_{p-1}(\lambda_{p-1}-\lambda_p)\vec v_{p-1} &= (\lambda_p -\lambda_p)\vec v_p = \vec 0\end{align}
All of the eigenvalues are distinct, so the $\lambda_i – \lambda_p$ terms are all non-zero. Therefore $c_i=0$ for $i=1,2, \ldots , p-1$. But that means that in Equation (1) that $\vec v_p = \vec 0$, which isn’t possible because $\vec v_p$ is an eigenvector. So the set of vectors $\vec v_1, \vec v_2, \ldots \vec v_p$ must be linearly independent. ■