Showing set of tensored states span a space I have the four states 
$$
\lvert1\rangle \lvert1\rangle - \lvert0\rangle \lvert0\rangle \\
i\lvert1\rangle \lvert1\rangle + i\lvert0\rangle\lvert0\rangle \\
\lvert0\rangle \lvert1\rangle + \lvert1\rangle \lvert0\rangle \\
\lvert0\rangle \lvert1\rangle - \lvert1\rangle \lvert0\rangle 
$$
What is a quick and easy way to show they are an orthogonal basis in $\mathbb{C^2}\otimes \mathbb{C^2}$. I've got the simple orthogonal bit down which I guess proves independence, but how can I show the span of the set?
 A: $\Bbb C^2\otimes \Bbb C^2$ is four dimensional, and you have four linearly independent vectors. They have to span the whole space because of dimensionality.
Just think what would happen if they didn't span the space. You would pick another vector outside of their span, and that would have to be linearly independent with the first four vectors. But that would mean the dimension of the space was at least 5 (an absurdity.)
In general, any $n$ linearly independent vectors of an $n$ dimensional space must span the space. Conversely, any $n$ vectors that span an $n$ dimensional vector space must be linearly independent.
A: $$
\begin{eqnarray}
|1\rangle|1\rangle-|0\rangle|0\rangle \\
i|1\rangle|1\rangle+i|0\rangle|0\rangle \\
|0\rangle|1\rangle+|1\rangle|0\rangle \\
|0\rangle|1\rangle-|1\rangle|0\rangle \\
\end{eqnarray}
$$
forms an orthogonal basis, since the (scaled) matrix
$$
U=\frac{1}{\sqrt{2}}\pmatrix{
-1 &0&0&1\\
i&0&0&i\\
0&1&1&0\\
0&1&-1&0
}
$$
which transforms the standard basis to your (entangled) basis, is unitary (check $U^\dagger U\overset{!}{=}\bf 1$) and this preserves orthogonality.
