> [!NOTE] Theorem > Let $A$ be a [[Real Square Matrices|real square matrix]] of order $n.$ Let $\underline{v}_{1},\dots,\underline{v}_{r}\in \mathbb{R}^{n}$ be [[Eigenpair of Real Square Matrix|eigenvectors]] corresponding to distinct eigenvalues $\lambda_{1},\dots\lambda_{r}$ of $A.$ Then $\underline{v}_{1},\underline{v}_{2},\dots,\underline{v}_{r}$ are [[Linearly Independent Subset of Real n-Space|linear independent]]. **Proof**: Suppose for contradiction that $\{ \underline{v}_{1},\underline{v}_{2},\dots,\underline{v}_{r} \}$ is linearly dependent. Let $k$ be the smallest positive integer such that $\underline{v}_{k}\in \langle \underline{v}_{1},\dots,\underline{v}_{k-1} \rangle.\tag{1}$Then there exists $\mu_{1},\mu_{2},\dots,\mu_{k-1}$ such that $\underline{v}_{k}=\mu_{1}\underline{v}_{1}+\dots+\mu_{k-1}\underline{v}_{k-1}.\tag{2}$Left-multiplying by $A$ gives $ \begin{aligned} A \underline{v}_k & =A\left(\mu_1 \underline{v}_1+\cdots+\mu_{k-1} \underline{v}_{k-1}\right) \\ & =\mu_1 A \underline{v}_1+\cdots+\mu_{k-1} A \underline{v}_{k-1} \end{aligned} $ so that, since each $\underline{v}_i$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda_i$, we have $\lambda_k \underline{v}_k=\mu_1 \lambda_1 \underline{v}_1+\cdots+\mu_{k-1} \lambda_{k-1} \underline{v}_{k-1}\tag{3}$ Multiplying both sides of $(2)$ by $\lambda_k$ and then subtracting $(3)$ gives $ \underline{0}=\mu_1\left(\lambda_k-\lambda_1\right) \underline{v}_1+\cdots+\mu_{k-1}\left(\lambda_k-\lambda_{k-1}\right) \underline{v}_{k-1} $ Now $\underline{v}_1, \ldots, \underline{v}_{k-1}$ are linearly independent since $k$ was chosen to be the smallest positive integer satisfying $(1)$, so $\mu_1=\cdots=\mu_{k-1}=0$. Substituting these values for $\mu_i$ in $(2)$ gives $\underline{v}_k=\underline{0}$. But this is a contradiction: $\underline{v}_k$ is an eigenvector, so $\underline{v}_i \neq \underline{0}$.