Linear Algebra: Find four unit vectors in $\mathbb{R}^3$ with the same angle between each. I need to determine the set of all angles $\theta$ such that there exists some four distinct unit vectors $\vec{v}_1, ..., \vec{v}_4 \in \mathbb{R}^3$ and the angle between any $\vec{v_i}$ and $\vec{v_j}$ is $\theta$ if $i\ne j$.  I cannot geometrically imagine an instance of this, although I can pretty easily see examples for three distinct vectors.  If you have two vectors, you can easily add a third such that its angle with the other two is equal to the angle of the first two.  But if you have three vectors of equal pair-wise angle, I can't see any way to rearrange or place a fourth so that they share the same angle.
The dot-product sounds relevant since any vector dotted with itself is 1, and dotted with any other vector is $\cos\theta$.  I've seen the trick enough times to suspect that I form the matrix $A^TA$ where the columns of $A$ are the vectors.  Then 
$$A^TA = \begin{pmatrix}
1 & \cos\theta & \cos\theta & ... & \cos\theta \\
\cos\theta & 1 & \cos \theta & ... & \cos\theta \\
... & ... & ... & ... & ... \\
\cos\theta & \cos\theta & ... & \cos\theta & 1 
\end{pmatrix}$$
That is to say, 1s down the main diagonal, $\cos\theta$ everywhere else. It's a symmetric matrix, but I'm not sure what to do with it.  I could take the determinant but I'm not sure what that would give me.  
 A: Suppose $v_1, v_2, v_3, v_4$ are such vectors. I claim that $v_2 - v_1, v_3 - v_1, v_4 - v_1$ form a basis for $\Bbb{R}^3$. It suffices to prove that they are linearly independent.
Suppose $a_2(v_2 - v_1) + a_3(v_3 - v_1) + a_4(v_4 - v_1) = 0$. Dot producting with $v_i$ (where $i = 2, 3, 4$), this becomes
$$a_i(1 - \cos(\theta)) = 0 \implies a_i = 0,$$
as clearly $\cos(\theta) = 1$ fails to make the vectors distinct. Therefore, the vectors are linearly independent in $\Bbb{R}^3$, and are thus a basis.
Let $w = v_1 + v_2 + v_3 + v_4$. I now claim that $w = 0$. Note first that,
$$w \cdot v_i = 1 + 3\cos(\theta),$$
and hence
$$w \cdot (v_i - v_j) = 0.$$
In particular, $w$ is perpendicular to all three basis vectors from before, proving $w = 0$ as required.
Since $0 = w \cdot v_1 = 1 + 3\cos(\theta)$, we must have $\cos(\theta) = -\frac{1}{3}$.
A: The approach you guessed actually works: As you noted yourself, an eigendecomposition of the inner product matrix $G=G_{\theta}$ (which is typically called Gramian matrix, by the way) is, with $x:=\cos(\theta)$, 
$$
G=
\begin{pmatrix}
1 &1 &0 &0\\
1 &-1 &1 &0\\
1 &0 &-1 &1 \\
1 &0  &0 &-1
\end{pmatrix}
\begin{pmatrix}
1+3x & 0 & 0 & 0\\
0 & 1-x & 0 & 0\\
0 & 0 & 1-x & 0 \\
0 & 0 & 0 & 1-x
\end{pmatrix}
\begin{pmatrix}
1 &1 &0 &0\\
1 &-1 &1 &0\\
1 &0 &-1 &1 \\
1 &0  &0 &-1
\end{pmatrix}^{-1}
$$
If there exists a matrix $A\in\mathbb{R}^{3\times 4}$ such that $G=A^{\top}A$, then $G$ can have at most rank $3$. Since the left and right matrices above are independent of $x$ and full rank, the middle matrix must have rank $3$. This can only happen if $x=1$ (which is a trivial case I leave to you) or when $x=-1/3$. This already proves the first bit:
If a solution exists, we must have $x=-1/3$ or $\theta=\arccos(-1/3)=109.5 ^{\circ} $.
To prove that there is a solution, i.e a matrix $A$ such that $G_{109.5}=A^{\top}A$, note that for $x=-1/3$ you have a positive semi-definite matrix. Such matrices have a decomposition as $A^{\top}A$ for $A\in \mathbb{R}^{3\times 4}$; you could use the singular value decomposition to prove this. $\Box$ 
Practically, you can find $A$ as follows:
Write the eigenvector matrix from above be
$$
(v_1|V_2)
$$
where $v_1=(1,1,1,1)$ and $V_2$ is the remaining $3\times 4$ part. Find an orthonormal basis of that part to get a new matrix $\tilde{V_2}$ of eigenvectors to the eigenvalue $1-x$. Since
$$
M=(v_1/2|\tilde{V_2})
$$
is an orthonormal basis, you have
$$
G=MDM^{-1}=MDM^{\top}.
$$
with $D$ the middle matrix from above. Since the first entry of $D$ is zero, this reduces to
$$
\tilde{V_2}\tilde{D}\tilde{V_2}^{\top}=(\tilde{V_2}\sqrt{\tilde{D}})(\sqrt{\tilde{D}}\tilde{V_2})^{\top},
$$
where you can read off the solution $A:=(\sqrt{\tilde{D}}\tilde{V_2})^{\top}$.
