Eigenvectors of Distinct Eigenvalues are Independent

Theorem

• Suppose $T(v_i)-{\lambda }_i{v}_i$ for $i=1,\dots ,n$ are distinct eigenvalues and eigenvectors of T
• The set of vectors $\{v_1,\dots ,v_n\}$ is Linearly Independent

Proof

• Suppose for sake of contradiction that $\{v_1,\dots ,v_n\}$ is dependent
• Then, there is a shortest possible dependence $0=a_1{v}_1+..+a_k{v}_k$ for $k\le n$
• We apply $T$ to both sides and get
• $0=T(0)=a_{1}T(v_1)+\cdots +a_{k}T(v_k)$
• $=a_1{\lambda }_1{v}_1+\cdots +a_k{\lambda }_k{v}_k$
• We know that we have distinct eigenvalues.
• Therefore, there are atleast two values of ${\lambda }_i$ meaning $\exists {\lambda }_k\ne 0$
• Then, we divide out by that ${\lambda }_k$ to create a shorter dependence
• $0=\frac{a_1{\lambda }_1}{{\lambda }_k}{v}_1+\cdots +\frac{a_k{\lambda }_k}{{\lambda }_k}{v}_k$
• Then, we subtract our two equations $a_1{v}_1+..+a_k{v}_k-\frac{a_1{\lambda }_1}{{\lambda }_k}{v}_1+\cdots +\frac{a_k{\lambda }_k}{{\lambda }_k}{v}_k$
• $=a_1(1-\frac{{\lambda }_1}{{\lambda }_k})v_1+...+a_{k-1}(1-\frac{{\lambda }_{k-1}}{{\lambda }_k})v_k$
• We know that this equation is linearly dependent as some $a_i$ for $1\le i\le k-1$ is non-zero. and $1-\frac{{\lambda }_i}{{\lambda }_k}=0$
• Then, ${\lambda }_i={\lambda }_k$. Then, there is no shortest dependence
• $\square$