# If $T \in \cal{L}(V)$ and every vector in $V$ is an eigenvector of $T$ then $T=cI$ for some $c \in \cal{F}$

Suppose $T$ is an operator on a vector space $V$ (i.e. a linear map from $V$ to itself) such that every vector $v \in V$ is an eigenvector of $T$. To show that $T$ is a scalar multiple of the identity operator.

Let be the underlying field of $V$. From the construction it is clear that for all $v \in V$, there exists $\lambda_v \in \cal{F}$ such that $Tv=\lambda_vv$

But I have to show that, as I understand it, $Tv=cv$, for all $v \in V$, for some $c \in \cal{F}$, i.e. the same $c$ works for all $v \in V$. How can I show that $\lambda_v$ is independent of $v$?

Any help would be much appreciated.

Choose a basis $B$ of $V$ and show that two members of $B$ have the same Eigenvalue: If $v_1,v_2\in B$ and $\lambda_1,\lambda_2$ are their Eigenvalues it follows: $T(v_1+v_2) = \lambda_1v_1 + \lambda_2v_2$ On the other hand $v_1+v_2$ ist an Eigenvector too. Also an Eigenvalue $\eta$ exists with $T(v_1+v_2)=\eta(v_1+v_2)$. Hence you get $\lambda_1=\eta$ and $\lambda_2=\eta\,$ form thel inear independence of $v_1$ and $v_2$. So you see that only one Eigenvalue exists and you are done.
Suppose every vector in $V$ is an eigenvector. For contradiction, assume that $v_1, v_2$ are (non-zero) eigenvectors whose corresponding eigenvalues $\lambda_1, \lambda_2$ are distinct.
Can you show that $v_1, v_2$ are linearly independent?
But $v_1 + v_2$ is an eigenvector too. Show that this contradicts the linear independence of $v_1, v_2$.