Proof. In this proof we will assume that $r=3$. Assume:
-
$\lambda_1 \neq \lambda_2, \lambda_1 \neq \lambda_3$ and $\lambda_2 \neq \lambda_3$.
-
$\vec{v_1}, \vec{v_2}, \vec{v_3}$ are nonzero vectors.
-
$A\vec{v_1} = \lambda_1 \vec{v_1}, A\vec{v_2} = \lambda_2 \vec{v_2}, A\vec{v_3} = \lambda_3 \vec{v_3}$.
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
\begin{equation} \label{eq1}
\alpha_1 \vec{v_1} + \alpha_2 \vec{v_2}+ \alpha_3 \vec{v_3} = \vec{0}.
\end{equation}
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
\begin{equation} \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}.
\end{equation}
Next we multiply both sides of \eqref{eq1} by $\lambda_3$ to get
\begin{equation} \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}.
\end{equation}
Now we subtract \eqref{eq3} from \eqref{eq2} to get
\begin{equation} \label{eq4}
\alpha_1 (\lambda_1-\lambda_3) \vec{v_1} + \alpha_2 (\lambda_2-\lambda_3) \vec{v_2} = \vec{0}.
\end{equation}
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
\begin{equation} \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},
\end{equation}
and multiply \eqref{eq4} by $\lambda_2$ to get
\begin{equation} \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}.
\end{equation}
Subtracting \eqref{eq6} from \eqref{eq5} yields
\begin{equation} \label{eq7}
\alpha_1 (\lambda_1-\lambda_3) (\lambda_1-\lambda_2) \vec{v_1} = \vec{0}.
\end{equation}
Since $\lambda_1 - \lambda_2 \neq 0, \lambda_1 - \lambda_3\neq 0$ and $\vec{v_1} \neq \vec{0}$, \eqref{eq7} implies
\begin{equation}\label{eq8}
\alpha_1 = 0.
\end{equation}
With \eqref{eq8}, equation \eqref{eq1} becomes
\begin{equation} \label{eq9}
\alpha_2 \vec{v_2}+ \alpha_3 \vec{v_3} = \vec{0}.
\end{equation}
Starting from \eqref{eq9} and repeating similar steps as before we can prove
\[
\alpha_2 = 0, \qquad \alpha_3 = 0.
\]
This completes the proof.