$T$ has an eigenvalue of $3$ or $-3$.

Question

Given a Linear operator $T$ satisfying $Tv = 3w$ and $Tw = 3v$ for some non-zero vectors $v$ and $w$ i am required to show that $T$ has an eigen value of either $3$ or $-3$.

Is it sufficient then to say that because $T(u+v) = Tu+Tv = 3w+3v = 3(v+w)$. $3$ is an eigenvalue of $T$.

Answer 1

No, it is not sufficient, because $u+v$ might be equal to $0$.

You should always suspect that there's something wrong with a proof when it proves more than what is being asked. You are asked to prove that $3$ or $-3$ is an eigenvalue and your proof, if correct, would prove that $3$ is an eigenvalue.

Answer 2

As noted by Jose Carlos, your argument is not correct. But you can argue as follws: $T^2v=T(Tv)=T(3w)=3Tw=3(3v)=9v$, hence $9$ is an eigenvalue of $T^2$.

The "spectral mapping theorem" shows now that $T$ has an eigenvalue $ \lambda$ such that $\lambda^2=9$, thus $ \lambda= \pm 3$.

Answer 3

Another argument:

We have $T(w+v) = 3(w+v)$ and $T(w-v)=-3(w-v)$.

But it is impossible that $w+v=0=w-v$, since $w \ne 0 \ne v$.

Answer 4

We have $T^2v = 9v$ and so $(T+3I)(T-3I)v=0$.

If $(T-3I)v=0$, then $3$ is an eigenvalue of $T$ with eigenvector $v$.

If $(T-3I)v\ne0$, then $-3$ is an eigenvalue of $T$ with eigenvector $(T-3I)v$.