0
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $v_1, v_2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.