Why $\dim U+\dim U^\perp=\dim V$? Let $V$ be a vector space. Given $U\le V$, define $U^\perp=\{f\in V^\ast\mid f(u)=0\,,\;\forall u\in U\}$ and given $W\le V^\ast$, define $W^\perp=\{u\in V\mid f(u)=0\;,\;\forall f\in W\}$. This defines a Galois connection between subspaces of $V$ and subspaces of $V^\ast$. If $V$ is finite dimensional, why is $\dim U+\dim U^\perp=\dim V$? Unfortunately, I am not very comfortable with linear algebra. Any hint will be appreciated.
 A: Suppose $V$ is finite dimensional, and let $n=dim(V)$, clearly also $V^*$ has dimension $n$ since it is its dual space, and let $m=dim(U)$. Fix a basis $\{v_1,...,v_m\}$ for $U$ and let $\mathcal{B}=\{v_1,..,v_m,..v_{m+1}...,v_n\}$ a basis for $V$. Denote with $\{v_1^*,...,v_n^*\}$ the dual basis for $V^*$; so: $v_j^*(v_i)=\delta_{ij}$ where $\delta_{ij}$ is the Kronecker's delta. 
Let $v\in V$, in coordinates $v=\sum_{i=1}^n a_iv_i$, and let $f\in V^*$; $f=\sum_{j=1}^n b_jv_j^*$; then
$$ f(v)=\sum_{i,j=1}^n a_ib_jv_j^*(v_i). \quad (1)$$
If $u\in U<V$ then $u=\sum_{i=1}^m a_iv_i+0v_{m+1}+...+0v_n$ (obviously); suppose now $f\in U^{\perp}$; then $f(u)=0$ for every $u\in U$, so by the equality above (1)
$$f(u)=\sum_{i,j=1}^n a_ib_jv_j^*(v_i)=\sum_{i,j=1}^m a_ib_jv_j^*(v_i)=0 \quad (2)$$
now; $v_j^*(v_i)=0$ if $j>m$ for every $i$, on the other hand, since $v_j^*(v_i)=\delta_{ij}$ when $j\le m$ and the equality (2) hold for every $u\in U$ this means that $b_j=0$ if $j\le m$; in other words
$$f=\sum_{j=m+1}^n b_jv_j^*.$$
Then $f\in U^{\perp}$ iff is a linear combination of $\{v_{m+1}^*,...,v_n^*\}$, this imply that the dimension of $U^{\perp}$ is $n-m$; then
$$dim(U)+dim(U^{\perp})=m+n-m=n=dim(V).$$
