given that v is a eigen vector of T, prove that is equivalent to this two statements separately:

Question

1. if and only if, span{v} = S, then $T(S) \subset S$
2. add to that, that v is a eigen vector of $T \in L(V,V)$ (collection of all linear transformation from V to V) if and only iff is a eigen vector of (T-$\alpha$*I) for all $\alpha \in K$

my attemp:

1) From left to right: if v is a eigen vector of T, then it's trivial that T(v)=$\alpha_v$v,in which $\alpha_v$ is the eigenvalue associated with v. Then, since every element on span{v} is lineally independent, then $(\alpha_v)v \in S$ for any $\alpha_v$ in K

From right to left: the right side is telling us there exist linear transformation T such that $T(span{v}) \in span{v}$ then that means that T(span{v}) its a linear combination of span{v}. Then v must be a eighvector of v

2) With respect to this one, there's is something I don't understand yet: if v es a eigenvector of a linear transformation that belongs to the collection of al LT that goes from V to V, then why (T-$\alpha$*I) works for every $\alpha \in K$ and not to the respective eigenvalues of v? what's that im not getting?

thanks, I know this question is between a proof-verification and a real question, but since the excersise is put together, i wanted to ask both. I really want to know what i am getting wrong or if i am actually doin it.

Answer 1

Let $b\in K$ be the eigen value of $v$. if $b=0$ then $T(span(v))=\{0\}$, the null space, and is contained in every subspace, so we need $b \neq 0$.

And $(T - \alpha^*I)v=(b-\alpha^*)v$ so we better not have $\alpha=b^*$ because $v$ will not be an egien vector given, again, that the "eigen value" would be $0$.

Answer 2

Your proof for 1) is not rigorous. We must also assume that $v\ne 0$ in the definition of $S$ or we get $S=\{0\}$ which has no eigenvectors of $T$ because it has no non-zero vectors.

For 1):

We have that $S=\text{span(v)}=\{kv: k\in K\}$ where $K$ is the scalar field for $V$.

Suppose $T(S)\subset S$.

Then $T(v)=kv$ for some $k\in K$ which means $k$ and $v$ are an eigenvalue, eigenvector pair.

Suppose $v$ is an eigenvector for $T$.

Then for any $u\in S$, $u=kv$ for some $k\in K$ and $T(u)=T(kv)=kT(v)=k\lambda v\in S$ where $\lambda$ is the eigenvalue associated with $v$.

For 2):

Suppose $v$ is an eigenvector of $T$ and let $\alpha$ be an element of $K$. Then $$(T-\alpha I)v=T(v)-\alpha v=\lambda v -\alpha v=(\lambda-\alpha)v$$ so $v$ is an eigenvector for $T-\alpha I$

Suppose $v$ is an eigenvector of $T-\alpha I$.

Then $$(T-\alpha I)v=\lambda v\implies Tv=\alpha v+\lambda v=(\alpha+\lambda)v$$ so $v$ is an eigenvector for $T$.