If V is the vector space of all polynomials of one variable over complex numbers. If the endomorphism T: V -> V is surjective, is it also injective?
I have the following problem, so I thought that as it is surjective the dimension of the image is equals to the dimension of V. So, dim V = dim Ker(T) + dim Im(T), then dim Ker(T) = 0 so it is injective.
But as it is reflected in the answer, being surjective don't make it also injective. So I've thought that is because V has infinite dimension, is it correct? So if instead of V we have W that is the vector space of the polynomials of one variable with grade leq to 7 , my assumption would be true and then if it's surjective, it's also injective.
Moreover, what happens if the problem were :
If V is the vector space of all polynomials of one variable over complex numbers. If the endomorphism T: V -> V is injective, is it also surjective?
It would be false again?
Thanks!
Best Answer
Yes your thoughts are exactly right. For finite dimensional vector spaces injectivity and surjectivity is equivalent. However, for vector spaces of infinite dimension this is not true. For example take the map $\psi:\mathbb{C}[X] \to \mathbb{C}[X], f\mapsto f\cdot X$, then clearly $\psi$ is injective and $\mathbb{C}$-linear, but not surjective. Another more abstract example is: consider $\ell^1(\mathbb{R})=\ell^1=\{(a_n)_{n\in \mathbb{N}}:\sum_{n=1}^\infty\vert a_n\vert <\infty\}$ then $\ell^1$ becomes an $\mathbb{R}$-vector space by point-wise addition and scalar multiplication. Consider the map $$ s:\ell^1\to \ell^1,\quad (a_1,a_2,\ldots)\mapsto (0,a_1,a_2,\ldots)$$ then $s$ is linear and injective but not surjective. If you are interested in infinite dimensional vector spaces then functional analysis may be an interesting subject for you.
Another example is $$r:\ell^1\to \ell^1,\quad (a_1,a_2,\ldots)\mapsto (a_2,a_3,\ldots)$$ then $r$ is linear, surjective but not injective.