what is the dimension of a subspace of $\mathcal{L}(V,W)$ I need to find the dimension of $U$ in terms of $\dim(V)$ and $\dim(W)$ where $U=\{f\in\mathcal{L}(V,W)\mid f(v_o)=0,v_o \text{ fixed}\}$ where $V,W$ are vector spaces and $\mathcal{L}(V,W)=\{f:V\rightarrow W\mid f \text{ is linear}\}$ is a vector space with pointwise operations.
Does anyone know how I would go about doing this?
Thank you
 A: I suppose $V$ and $W$ are finite dimensional, say $m$ and $n$ respectively. $\mathcal L(V,W)$ has dimension $mn$. If $v_0\ne 0$, the vector equation $f(v_0)=0$ translates into $n$ independent linear equations in the coefficients of the matrix associated to $f$. Hence the dimension of the subspace of the linear maps from $V$ $W$ that vanish at $v_0$ is equal to $mn-n=(m-1)n$.
If $v_0=0, \enspace U=\mathcal L(V,W)$, and its dimension is $mn$.
A: If $v_0=0$, you have no problem, as $U=\mathcal{L}(V,W)$. Thus we can assume $v_0\ne0$; extend it to a basis $\mathscr{B}=\{v_1=v_0,v_2,\dots,v_m\}$ of $V$, where $m=\dim V$. Fix also a basis $\mathscr{C}$ of $W$.
The map that associates to $f\in\mathcal{L}(V,W)$ its representing matrix $M(f)$ with respect to the bases $\mathscr{B}$ and $\mathscr{C}$ is an isomorphism between $\mathcal{L}(V,W)$ and the vector space of $n\times m$ matrices (where $n=\dim W$).
The image of $U$ under this isomorphism consists of the matrices with a zero first column, which is clearly a subspace of dimension $mn-n$.
An abstract version of this proof uses the homomorphism theorem. A map $f\in U$ induces a unique linear map $V/\langle v_0\rangle\to W$ and, conversely, a linear map $V/\langle v_0\rangle\to W$ induces a linear map belonging to $U$, by composing with the canonical projection $V\to V/\langle v_0\rangle$.
So this induces an isomorphism $U\to\mathcal{L}(V/\langle v_0\rangle, W)$. If $v_0\ne0$, then $\dim V/\langle v_0\rangle=\dim V-1$.
A: Hint :
If you take $W=\mathbb{R}$ so $U$ is a Hyperplane (dim of $U$ will be $n-1$), then  deduce for general $W$
