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

  • $\begingroup$ Why didn't you accept the answer? $\endgroup$ – Viktor Glombik Dec 16 '18 at 10:39

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 $dim(Ker(\Psi))=mn-n=dim(L(V,W))-dim(Im(\Psi))$.


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