2
$\begingroup$

Is the the following Proof Correct?

Given a vector space $V$ such that $\dim V=1$ and $T\in\mathcal{L}(V,V)$ where $\mathcal{L}=\{H|H:V\to V\}$ then there exists $\lambda\in\mathbf{F}$ such that $Tv=\lambda v$ $\forall v\in V$.

Proof. Assume that $T\in\mathcal{L}(V,V)$ let $w$ be some vector in $V$ such that $w\neq 0$, evidently $w$ is basis for $V$ consequently $$\forall u\in V\exists\tag{1}\lambda\in \mathbf{F}(u=\lambda w)$$ Now let $v$ be an arbitrary vector in $V$ since $T:V\to V$ it follows that $Tv\in V$ thus by $(1)$ it follows that for some $\beta\in \mathbf{F}$. $Tv=\beta w$

$\blacksquare$

$\endgroup$
2
$\begingroup$

No, you haven't shown what was asked. You were asked to show that there exists $\lambda \in \mathbb{F}$ such that for all $v \in V$ we have $Tv = \lambda v$ (note the order of quantifiers). In particular, $\lambda$ shouldn't depend on $v$.

You have shown instead that for every $v \in V$ there exists $\beta \in \mathbb{F}$ (which possibly depends on $v$) such that $Tv = \beta w$. This is not what was asked. Note that your proof doesn't use the linearity of $T$ and without the linearity of $T$, the result is not true.

$\endgroup$
  • $\begingroup$ Thank you for your answer yes now that i have looked at the statement carefully you are quite right $\endgroup$ – Atif Farooq Aug 1 '17 at 16:51
1
$\begingroup$

Yep, looks fine to me. Alternatively you could have written the $1 \times 1$ matrix representing the linear transformation and concluded directly from there.

$\endgroup$
1
$\begingroup$

You have to be careful since in (1) the scalar $\lambda$ depends on $v$, but in your statement, it doesn't.

The ansatz to choose a basis $w\neq 0$ of $V$ is very good. Now use that $T$ is linear!

Define $u:=Tw$. There exists $\lambda\in\mathbb F$ such that $u=\lambda w$.

(Consider that $\lambda$ depends on $u$ and $w$!)

We get $Tw=\lambda w$. Now let be $v\in V$ arbitrary and $\mu\in\mathbb F$ such that $v=\mu w$.

(Here $\mu$ depends on $v$ and $w$)

Since $T$ is linear, we conclude $$Tv=T(\mu w)=\mu Tw=\mu\lambda w=\lambda\mu w=\lambda v.$$Since $\lambda$ depends on $w$ and $u=Tw$ but not on $v$ we are done.

$\endgroup$

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.