# Small question about Eigenvectors as Basis for $V$

If $\beta = (v_{1}, .. v_{n})$ is a basis for vectorspace $V$, such that $v_{j}$ is an Eigenvector of operator $T \in \text{End}(V)$ for $j = 1,..,n$

Is it correct then to say that all the vectors in $V$ are eigenvectors from $T$ since they are all in the Span of $(v_{1}, .. v_{n})$ and so the entire vector space is $T$-invariant?

If we have another operator $G \in \text{End}(V)$ such that each $v_{j} \in \beta$ is also an Eigenvector of $G$, but perhaps with different Eigenvalues than those of $T$, then we can say that $T$ and $G$ are commutative, because on all $v \in V$, the action of both operators is really just a scaling, so it doesn't matter by which scalar I multiple first?

I am trying to come up with a proof for a problem, but feel like I am missing something.

• I think you went too fast to notice that $T(v_1+v_2)=\lambda_1 v_1+\lambda_2 v_2 \neq \lambda (v_1+v_2)$ for some $\lambda$. May 14, 2018 at 6:48
• What does $\text{End}(V)$ mean? May 14, 2018 at 6:51
• @NicNic8 Endomorphism, $\text{End}(V)$ is space of linear operators from $V$ to itself May 14, 2018 at 6:59
• @Bill O'Haran yes, if $v_{1}$ and $v_{2}$ are Eigenvectors, it doesn't necessarily mean $v_{1} + v_{2}$ is. I understand better now, thanks. May 14, 2018 at 7:02

To wrap it up, we can show that if the whole space is made of eigenvectors of $T$, then $T$ is an homothetic transformation. With quantifiers, we have: $$\forall x\in V, \exists\lambda_x\in \mathbb{K},T(x)=\lambda_x x$$ and we want to show that: $$\exists\lambda\in \mathbb{K},\forall x\in V,T(x)=\lambda x$$ Then, all we have to do is to show that $\lambda_x=\lambda_y$ for all $x$ and $y$ in $V$.
• If $(x,y)$ is linearly dependent, $\exists \mu\in \mathbb{K}, x=\mu y$. Then $T(x)=\lambda_x x = \lambda_x \mu y$ but $T(x)=\mu T(y) = \mu \lambda_y y$. So $\lambda_x=\lambda_y$.
• If $(x,y)$ is linearly independent, then $T(x+y)=\lambda_{x+y} (x+y) = \lambda_{x+y} x + \lambda_{x+y} y$ and $T(x+y)=T(x)+T(y)=\lambda_x x + \lambda_y y$ thus $\lambda_x =\lambda_{x+y} =\lambda_y$.
To come back on what you wrote on two endomorphisms sharing the same eigenvectors, your intuition is right. That works because, since there is a basis of eigenvectors for each, both $T$ and $G$ are diagonalizable (both in the same basis, whose matrix we will denote by $P\in \mathcal{GL}_n(\mathbb{K})$).
Then $TG =PD_TP^{-1} \times PD_GP^{-1} = PD_TD_GP^{-1}= PD_GD_TP^{-1} =GT$ because two diagonal matrices commute.