# Proving eigenvector of addition operator

If $v$ is an eigenvector of operators $S$ and $T$, then $v$ is also an eigenvector of $aS + bT$ , $a, b \in F$

I know that if I let $v_1,\space v_2 \in V$. By definition of an eigenvector, $T(v_1) = \lambda_1 v_1$, $S(v_2) = \lambda_2 v_2$,

But should I define $\lambda_1, \lambda_2$ as $a,b$, respectively instead? I'm also a bit confused about adding the vectors, as if it were under the same operator, I could have vectors $v_1 + v_2$ simply having the linear operator $T(v_1 + v_2)$ = $T(v_1) + T(v_2)$, which proves. So I'm assuming a similar process, but I'm not exactly sure.

• "$v_1\in T$" and "$v_2\in S$" don't make sense -- $S$ and $T$ are operators, not sets. – Henning Makholm Feb 5 '15 at 0:27
• Oh, you're right. I was mixing them up to thinking I had a vector space. I suppose that I don't even need to vectors then, just one right? – user180708 Feb 5 '15 at 0:29
• Correct, you need not (and should not) speak about two vectors $v_1$ and $v_2$. The assumption in the problem is that you have a single vector $v$, and that vector happens to be both an eigenvector of $S$ and an eigenvector of $T$ (but possibly with different eigenvalues). – Henning Makholm Feb 5 '15 at 0:30

That $v$ is an eigenvector of $S$ and $T$ means there exist scalars $\alpha$ and $\beta$ such that $S(v)=\alpha\, v$ and $T(v)=\beta\, v$. It follows that $$(a\,S+b\,T)(v)=a\,S(v)+b\,T(v)=a\alpha\,v+b\beta\,v=(a\alpha+b\beta)\,v$$ This proves that $v$ is an eigenvector of $a\,S+b\,T$ with eigenvalue $a\alpha+b\beta$.