# Linear independence of eigenvectors; a sufficient condition

## Branko Ćurgus

Theorem. If $\vec{v_1}, \ldots, \vec{v_r}$ are eigenvectors that correspond to distinct eigenvalues $\lambda_1,\ldots,\lambda_r$ of an $n\times n$ matrix $A$, then the set $\bigl\{ \vec{v_1}, \ldots, \vec{v_r} \bigr\}$ is linearly independent.
Proof. In this proof we will assume that $r=3$. Assume:
1. $\lambda_1 \neq \lambda_2, \lambda_1 \neq \lambda_3$ and $\lambda_2 \neq \lambda_3$.
2. $\vec{v_1}, \vec{v_2}, \vec{v_3}$ are nonzero vectors.
3. $A\vec{v_1} = \lambda_1 \vec{v_1}, A\vec{v_2} = \lambda_2 \vec{v_2}, A\vec{v_3} = \lambda_3 \vec{v_3}$.
To prove that $\vec{v_1}, \vec{v_2}, \vec{v_3}$ are linearly independent we have to prove the following implication $\alpha_1 \vec{v_1} + \alpha_2 \vec{v_2}+ \alpha_3 \vec{v_3} = \vec{0} \quad \Rightarrow \quad \alpha_1 = \alpha_2 = \alpha_3 = 0.$ To prove this implication we assume $$\label{eq1} \alpha_1 \vec{v_1} + \alpha_2 \vec{v_2}+ \alpha_3 \vec{v_3} = \vec{0}.$$ The first step is to apply $A$ to both sides of \eqref{eq1} to get \begin{equation*} A\bigl(\alpha_1 \vec{v_1} + \alpha_2 \vec{v_2}+ \alpha_3 \vec{v_3}\bigr) = A\vec{0}. \end{equation*} Next, we remember that this is a linear algebra class, so we use the linearity property of $A$ to get \begin{equation*} \alpha_1 A\vec{v_1} + \alpha_2 A\vec{v_2}+ \alpha_3 A\vec{v_3} = \vec{0}. \end{equation*} By the assumption 3 the last equality becomes $$\label{eq2} \alpha_1 \lambda_1 \vec{v_1} + \alpha_2 \lambda_2 \vec{v_2}+ \alpha_3 \lambda_3 \vec{v_3} = \vec{0}.$$ Next we multiply both sides of \eqref{eq1} by $\lambda_3$ to get $$\label{eq3} \alpha_1 \lambda_3 \vec{v_1} + \alpha_2 \lambda_3 \vec{v_2}+ \alpha_3 \lambda_3 \vec{v_3} = \vec{0}.$$ Now we subtract \eqref{eq3} from \eqref{eq2} to get $$\label{eq4} \alpha_1 (\lambda_1-\lambda_3) \vec{v_1} + \alpha_2 (\lambda_2-\lambda_3) \vec{v_2} = \vec{0}.$$ The equality \eqref{eq4} is similar to \eqref{eq1}, just with fewer eigenvectors. So, we repeat the steps from before; apply $A$ to \eqref{eq4} to get $$\label{eq5} \alpha_1 (\lambda_1-\lambda_3) \lambda_1 \vec{v_1} + \alpha_2 (\lambda_2-\lambda_3)\lambda_2 \vec{v_2} = \vec{0},$$ and multiply \eqref{eq4} by $\lambda_2$ to get $$\label{eq6} \alpha_1 (\lambda_1-\lambda_3) \lambda_2 \vec{v_1} + \alpha_2 (\lambda_2-\lambda_3)\lambda_2 \vec{v_2} = \vec{0}.$$ Subtracting \eqref{eq6} from \eqref{eq5} yields $$\label{eq7} \alpha_1 (\lambda_1-\lambda_3) (\lambda_1-\lambda_2) \vec{v_1} = \vec{0}.$$ Since $\lambda_1 - \lambda_2 \neq 0, \lambda_1 - \lambda_3\neq 0$ and $\vec{v_1} \neq \vec{0}$, \eqref{eq7} implies $$\label{eq8} \alpha_1 = 0.$$ With \eqref{eq8}, equation \eqref{eq1} becomes $$\label{eq9} \alpha_2 \vec{v_2}+ \alpha_3 \vec{v_3} = \vec{0}.$$ Starting from \eqref{eq9} and repeating similar steps as before we can prove $\alpha_2 = 0, \qquad \alpha_3 = 0.$ This completes the proof.