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Suppose that V is finite-dimensional and that the dimension of V is odd. Show that there does NOT exist a linear transformation T : V → V such that ker(T) = range(T).

I'm just getting thrown off with the fact that the dimension of V is odd. I believe that I have to use the formula that dimV = dim ker(T) + dim range(T) but I'm not really sure how to start...

Any help would be appreciated!

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    If two numbers are equal they will have the same parity. Is this possible if the sum is odd? – 2017-02-08

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By the rank-nullity theorem, $\dim V = \dim \ker T + \dim\operatorname{range}T$. If you know that the dimension of $V$ is odd, what can you deduce about the values of the summands $\dim \ker T$ and $\dim\operatorname{range}T$, and hence the existence of such a linear transformation?

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    So since dimV is odd, we can assume that dim Ker T is not equal to dim Range T. Is that enough to imply that there does not exist such a linear map? – 2017-02-08
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    @MichelleDrolet Effectively yes. It might help to assume such a map did exist and then see what that implies about the dimension of $V$. It seems like you're getting the idea. – 2017-02-08
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    Yep! I definitely understand it now. Thank you for your help! – 2017-02-09
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$\dim \ker T$ and $\dim $ range $T$ are two integers whose sum is odd. So, they can't be equal. If two spaces have different dimensions, they can't be the same space.