Finding an inner product Question:
Given 2 vector spaces $U=sp(1,1), W=sp(2,0)$.
How do I find an inner product in $\Bbb R^2$ s.t. $U=W^{+}$ (orthogonal)
I would love an explanation for the algorithm really, more than this specific question.
Thanks.
 A: This is of course the same argument as the one given by Franklin.vp, whose answer appeared while I was too long at typing.
Given $u_1,\ldots,u_n$ in a  finite-dimensional inner product space $(V, (\cdot,\cdot))$, the following are equivalent:
1- there exists another inner product on $V$ such that $\{u_1,\ldots,u_n\}$ become an orthonormal set.
2- $u_1,\ldots,u_n$ are linearly independent.
It is clear that 1 implies 2. In the other direction, complete the set into a basis $\{u_1,\ldots,u_m\}$ of $V$ and take any orthonormal basis $\{v_1,\ldots,v_m\}$ for the old inner product. Then consider the linear map $A\in L(V)$ defined by $Au_j:=v_j$ for every $j$. Since it sends a basis onto a basis, $A$ is invertible. Just check that
$$
(x,y)':=(Ax,Ay)\qquad \forall x,y\in V
$$
defines a new inner product on $V$ for which $\{u_1,\ldots,u_m\}$ is now an orthonormal basis. This is a standard renorming procedure, in the particular case of a norm coming form an inner product.
So in your particular case, just consider the matrices
$$
B:=\pmatrix{1&2\\1&0}\qquad A=B^{-1}=\pmatrix{0&1\\\frac{1}{2}&-\frac{1}{2}}
$$
so that $B$ sends the canonical orthonormal basis of $\mathbb{R}^2$ (for $(x,y)=x_1y_1+x_2y_2=x^Ty$), namely $V_1=(1,0)$, $v_2=(0,1)$ to the vectors $u=(1,1)$ and $w=(2,0)$ which span $U$ and $W$ respectively. Then the inverse matrix $A=B^{-1}$ does what we want for the new inner product
$$
(x,y)':=(Ax,Ay)=\frac{1}{4}(x_1y_1+2x_2y_2-x_1y_2-x_2y_1)
$$
to turn $\{u,w\}$ into an orthonormal set, so that their spans $U$ and $W$ be orthogonal, whence $U=W^\perp$ and $W=U^\perp$.
A: Given two non-collinear vectors $u$ and $v$ you can get an inner product to make them orthogonal in the following way. 
Find first a linear transformation that sends $u$ to $(1,0)$ and $v$ to $(0,1)$. If you already have their coordinates in this basis (the standard basis) then the transformation is the inverse $A$ of the matrix $[uv]$ formed by putting both vectors as columns. So you get $Au=(1,0)$ and $Av=(0,1)$. Now the standard inner product makes $(1,0)$ and $(0,1)$ orthogonal. So, if you define
$$<x,y>:= (Ax)\cdot(Ay),$$
where $\cdot$ denotes the standard inner product. You get an inner product such that $<u,v>=0$.
A: If a set of vectors form a basis, you can take co-ordinates with respect to that basis – you can write an arbitrary vector as a linear combination of the basis vectors $v_i$. Then you can define an inner product as, if $x = \sum x_i v_i$ and $y = \sum y_i v_i$,
\[x\cdot y = \sum x_i y_i\]
Clearly the $v_i$ will automatically be orthogonal wrt this inner product.
This is the standard way of making the standard inner product from the standard basis, but of course any basis is as good as any other, so it works just as well in this case, too!
