Proof of Theorem EDELI

If $p=1$, then the set $S=\set{\vect{x}_1}$ is linearly independent since eigenvectors are nonzero (Definition EEM), so assume for the remainder that $p\geq 2$.

We will prove this result by contradiction (Proof Technique CD). Suppose to the contrary that $S$ is a linearly dependent set. Define $S_i=\set{\vectorlist{x}{i}}$ and let $k$ be an integer such that $S_{k-1}=\set{\vectorlist{x}{k-1}}$ is linearly independent and $S_k=\set{\vectorlist{x}{k}}$ is linearly dependent. We have to ask if there is even such an integer $k$? First, since eigenvectors are nonzero, the set $\set{\vect{x}_1}$ is linearly independent. Since we are assuming that $S=S_p$ is linearly dependent, there must be an integer $k$, $2\leq k\leq p$, where the sets $S_i$ transition from linear independence to linear dependence (and stay that way). In other words, $\vect{x}_k$ is the vector with the smallest index that is a linear combination of just vectors with smaller indices.

Since $\set{\vectorlist{x}{k}}$ is a linearly dependent set there must be scalars, $\scalarlist{a}{k}$, not all zero (Definition LI), so that \begin{align*} \zerovector=\lincombo{a}{x}{k} \end{align*}

Then, \begin{align*} \zerovector &=\left(A-\lambda_kI_n\right)\zerovector &&\knowl{./knowls/theorem.ZVSM.knowl}{\text{Theorem ZVSM}}\\ &=\left(A-\lambda_kI_n\right)\left(\lincombo{a}{x}{k}\right) &&\knowl{./knowls/definition.RLD.knowl}{\text{Definition RLD}}\\ &=\left(A-\lambda_kI_n\right)a_1\vect{x}_1+ \cdots+ \left(A-\lambda_kI_n\right)a_k\vect{x}_k &&\knowl{./knowls/theorem.MMDAA.knowl}{\text{Theorem MMDAA}}\\ &=a_1\left(A-\lambda_kI_n\right)\vect{x}_1+ \cdots+ a_k\left(A-\lambda_kI_n\right)\vect{x}_k &&\knowl{./knowls/theorem.MMSMM.knowl}{\text{Theorem MMSMM}}\\ &=a_1\left(A\vect{x}_1-\lambda_kI_n\vect{x}_1\right)+ \cdots+ a_k\left(A\vect{x}_k-\lambda_kI_n\vect{x}_k\right) &&\knowl{./knowls/theorem.MMDAA.knowl}{\text{Theorem MMDAA}}\\ &=a_1\left(A\vect{x}_1-\lambda_k\vect{x}_1\right)+ \cdots+ a_k\left(A\vect{x}_k-\lambda_k\vect{x}_k\right) &&\knowl{./knowls/theorem.MMIM.knowl}{\text{Theorem MMIM}}\\ &=a_1\left(\lambda_1\vect{x}_1-\lambda_k\vect{x}_1\right)+ \cdots+ a_k\left(\lambda_k\vect{x}_k-\lambda_k\vect{x}_k\right) &&\knowl{./knowls/definition.EEM.knowl}{\text{Definition EEM}}\\ &=a_1\left(\lambda_1-\lambda_k\right)\vect{x}_1+ \cdots+ a_k\left(\lambda_k-\lambda_k\right)\vect{x}_k &&\knowl{./knowls/theorem.MMDAA.knowl}{\text{Theorem MMDAA}}\\ &=a_1\left(\lambda_1-\lambda_k\right)\vect{x}_1+ \cdots+ a_{k-1}\left(\lambda_{k-1}-\lambda_k\right)\vect{x}_{k-1}+ a_k\left(0\right)\vect{x}_k &&\knowl{./knowls/property.AICN.knowl}{\text{Property AICN}}\\ &=a_1\left(\lambda_1-\lambda_k\right)\vect{x}_1+ \cdots+ a_{k-1}\left(\lambda_{k-1}-\lambda_k\right)\vect{x}_{k-1}+ \zerovector &&\knowl{./knowls/theorem.ZSSM.knowl}{\text{Theorem ZSSM}}\\ &=a_1\left(\lambda_1-\lambda_k\right)\vect{x}_1+ \cdots+ a_{k-1}\left(\lambda_{k-1}-\lambda_k\right)\vect{x}_{k-1} &&\knowl{./knowls/property.Z.knowl}{\text{Property Z}} \end{align*}

This equation is a relation of linear dependence on the linearly independent set $\set{\vectorlist{x}{k-1}}$, so the scalars must all be zero. That is, $a_i\left(\lambda_i-\lambda_k\right)=0$ for $1\leq i\leq k-1$. However, we have the hypothesis that the eigenvalues are distinct, so $\lambda_i\neq\lambda_k$ for $1\leq i\leq k-1$. Thus $a_i=0$ for $1\leq i\leq k-1$.

This reduces the original relation of linear dependence on $\set{\vectorlist{x}{k}}$ to the simpler equation $a_k\vect{x}_k=\zerovector$. By Theorem SMEZV we conclude that $a_k=0$ or $\vect{x}_k=\zerovector$. Eigenvectors are never the zero vector (Definition EEM), so $a_k=0$. So all of the scalars $a_i$, $1\leq i\leq k$ are zero, contradicting their introduction as the scalars creating a nontrivial relation of linear dependence on the set $\set{\vectorlist{x}{k}}$. With a contradiction in hand, we conclude that $S$ must be linearly independent.