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?

  • $\begingroup$ Consider what happens when you add 2 eigenvectors. Show that the eigenvalue is the same everywhere. $\endgroup$
    – Eric
    Apr 4, 2021 at 17:49
  • $\begingroup$ 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 $Av=\lambda v$ $\endgroup$
    – Matthew H.
    Apr 4, 2021 at 17:54

1 Answer 1


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 $f=\lambda_1 I=\lambda I$


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