0
$\begingroup$

Prove that if $T[V]=\ker(T)$, then $n$ is even, where $\dim(V)=n$

We know that from the dimension thereom:

rank(T) + nullity(T) = n

So we can say that $n=\dim(T[V])+\dim(\ker(T))$

But I don;t know how to go on from here. What exactly does it mean when it says: $T[V]=\ker(T)$

I am aware there is a similar question to this already on MSE, but the answer didn't provide details and so I'm finding it hard to understand:

Let $V$ be $n$ dimensional real vector space. Show that of $T[V]=\ker(T)$, then $n$ is even

$\endgroup$
1
$\begingroup$

The image is the same as the kernel so in particular the rank and nullity of $T$ are equal.

$\endgroup$
  • $\begingroup$ but why does this imply $n$ even $\endgroup$ – K Split X Oct 25 '17 at 23:54
  • $\begingroup$ @KSplitX: $x + x = 2x$ $\endgroup$ – Joe Johnson 126 Oct 25 '17 at 23:55
  • $\begingroup$ In that way, okay thank you $\endgroup$ – K Split X Oct 25 '17 at 23:58
1
$\begingroup$

Combine your $n=\dim(T[V])+\dim(\ker(T))$ and $T[V]=\ker(T)$ to get $$n=\dim(T[V])+\dim(T[V])=2\dim(T(V))$$

so $n$ is even.

$\endgroup$

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.