- a) Prove that $n$ is even:
I figure that since $n$ is the dimension of the vector space,
By the rank-nullity theorem:
$n = \dim(\mathrm{im}(T)) + \dim(\ker(T))$
But since image and kernel are identical, by calling $x = dim(ker(T))$ we have:
$n = 2x$ so n can only be even.
Am I right with this reasoning?
- b) Give an example of such linear transformation
What example comes to your mind? I don't know if I should think at a certain example or generalize the answer.
Thank you
Best Answer
a) You are right.
b) A general example (I mean, for all $n$) would be$$f(x_1,x_2,\dots,x_{2n-1},x_{2n})=(x_2,0,x_4,0,\ldots,x_{2n},0).$$Then$$\operatorname{Im}f=\ker f=\{(x_1,x_2,\ldots,x_{2n})\,|\;x_2=x_4=\cdots=x_{2n}=0\}.$$