Independent Sets from Eigenspaces

Theorem

• Suppose $V$ is a $n$ dimensional $\mathbb{R}$ vector space with distinct $\mathbb{R}$ eigenvalues ${\lambda }_1,{\lambda }_2$
• If $S_1\subset E_{{\lambda }_1}$, $S_2\subset E_{{\lambda }_2}$ are sets of independent vectors
• Then, $S_1\cup S_2$ is independent

Proof

1. Let $S_1=\{u_1,\dots ,u_k\}$
2. Let $S_2=\{v_{k+1},\dots ,v_l\}$
3. Suppose for the sake of contradiction that $S_1\cup S_2=\{u_1,\dots ,u_k,v_{k+1},\dots ,v_l\}$ is dependent
4. Then, we have $a_1{u}_1+a_k{u}_k+\cdots +a_{k+1}{v}_{k+1}+\cdots +a_l{v}_l=0$ s.t $a_i\ne 0$
5. One of the values ${\lambda }_1\ne {\lambda }_2$ must be non-zero. Assume ${\lambda }_1\ne 0$
6. First, note that $T(a_1{u}_1+\cdots +a_k{u}_k+a_{k+1}{v}_{k+1}+\cdots +a_l{v}_l)=T(0)=0$
7. $⟺a_1{\lambda }_1{u}_1+\cdots +a_k{\lambda }_1{u}_k+a_{k+1}{\lambda }_2{v}_{k+1}+\cdots +a_l{\lambda }_2{v}_l=0$
8. We can mutliply the original dep by ${\lambda }_1$ and get: $a{\lambda }_1{u}_1+\cdots +a_1{\lambda }_1{u}_k+a_{k+1}{\lambda }_1{v}_{k+1}+\cdots +a_l{\lambda }_1{v}_l=0$
9. Then, $a_1{\lambda }_1{u}_1+\cdots +a_k{\lambda }_1{u}_k+a_{k+1}{\lambda }_2{v}_{k+1}+\cdots +a_l{\lambda }_2{v}_l-a{\lambda }_1{u}_1+\cdots +a_1{\lambda }_1{u}_k+a_{k+1}{\lambda }_1{v}_{k+1}+\cdots +a_l{\lambda }_1{v}_l=a_{k+1}({\lambda }_2-{\lambda }_1)v_{k+1}+\cdots +a_l({\lambda }_2-{\lambda }_1)v_l=0$
10. This means that $a_{k+1}({\lambda }_2-{\lambda }_1)v_{k+1}+\cdots +a_l({\lambda }_2-{\lambda }_1)v_l=0$
11. Note that ${\lambda }_2-{\lambda }_1\ne 0$
12. Thus, this gives a dependence for vectors $\{v_{k+1},\dots ,v_l\}$. This contradicts that $S_2$ is a linearly independent set.
13. The argument for ${\lambda }_2\ne 0$ is the same …
14. Thus, by cases: $S_1\cup S_2=\{u_1,\dots ,u_k,v_{k+1},\dots ,v_l\}$ is indep