# Find a necessary and sufficient condition for all vectors to be eigenvectors.

Let $$\mathscr{A} : V \to V$$ - linear operator.

And we know that for every $$v \in V$$ $$\mathscr{A}v = \lambda v$$ (Every vector are eigenvector).

The question is about finding necessary and sufficient conditions to $$\mathscr{A}$$ for beeing this statement true.

I found 2 operators (they are really trivial):

For the identity operator, all nonzero vectors of the space are eigenvectors (with an eigenvalue equal to one).

For a zero operator, all nonzero vectors of the space are eigenvalues ​​(with an eigenvalue equal to zero).

But what other conditions we can find for this question?

• Consider what happens when you add 2 eigenvectors. Show that the eigenvalue is the same everywhere.
– Eric
Apr 4, 2021 at 17:49
• Try and show that $A$ is a multiple of the identity. Also, I assume you mean for any $v\in V$ there exists $\lambda \in \mathbb{F}$ (which may depend on $v$) such that $Av=\lambda v$ Apr 4, 2021 at 17:54

I think that you can see that you application is $$\lambda*I$$ for some $$\lambda$$ using the matrix expresion:
$$f$$ is determinated by how it goes in a basis so select $$\{v_1,\dots v_n\}$$ basis of $$V$$. By our hypothesis $$f(v_1)=\lambda_1 v_1$$,. . .,$$f(v_n)=\lambda_n v$$ So the matrix of $$f$$ respective to this basis is: $$B=diag(\lambda_1,. . .,\lambda_n)$$
Now select a vector $$v=\alpha_1 v_1+. . .+\alpha_n v_n$$ Then $$f(v)=\lambda_1 \alpha_1 v_1+. . .+ \lambda_n \alpha_n v_n$$ so As $$v$$ needs to be an eigenvector then it is needed that $$\lambda_1=. . .=\lambda_n$$ (In other to have $$\lambda=\lambda_i$$ as a common factor) And then $$f=\lambda_1 I=\lambda I$$