Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Question: Let V,W be vector spaces over field F. We mark L(V,W) as the vector space of linear transformations from V to W. Let $v_0 \ne 0$. We define a transformation: $\Psi: L(V,W) \to W$ that sends a linear transformation $T \in L(V,W)$ to $T(v_0) \in W$

What is the image of $\Psi$ ? Evaluate $dim(Ker \Psi)$

What I did: I said that $dim(V)=m, dim(W)=n$, therefore $dim L(V,W)=mn$ I see that $\Psi$ can't be injective because the dimensions of L and W are not equal. $Im\Psi=\Psi(T)=T(v_0)$ Can't figure how should I calculate the dim of the Kernel. A thought I had is that since dimW < dimL then if $\Psi$ is surjective then $dimKer\Psi = mn-m$, but I don't know what if it's not surjective.

Thanks in advance

share|cite|improve this question

Well, if I'm not missing something, $\Psi$ is always surjective (for any $V, W$), at least assuming $dim(V)$ and $dim(W)$ are finite, as you implicitely seem to do when you write $dim(V)=m$ and $dim(W)=n$. To show this, complete $v_{0}$ to a base of $V$, let's say $(v_{0},\dots,v_{m-1})$ (a single non zero vector is linearly independent so you can do this). For any $w\in W$ consider the linear transformation $T\in L(V,W)$ such that $T(v_{0})=w$ and $T(v_{i})=0$ for any $i=1,\dots,m-1$. Obviously $\Psi (T)=w$ so that $Im (\Psi)=W$ and then one gets $dim(Ker(\Psi))=mn-n=dim(L(V,W))-dim(Im(\Psi))$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.