Determinant of a 2x2 Quadratic Form

Suppose we have a 2$-d$ quadratic form $Q : \mathbb{R}^2 \to \mathbb{R}$. How can we compute its determinant?

This problem is a little under-specified. In fact, we can always pick a basis for $\RR^2$ so that our quadratic form is diagonal with diagonal entries $0$ or $\pm 1$. So there is always some basis where our quadratic form has determinant $0$ or $\pm 1$.

But usually, we have a choice of inner product on $\mathbb{R}^2$, and if we restrict to orthonormal bases with respect to this inner product then $Q$ does indeed have well-defined eigenvectors, and a well-defined trace and determinant. And once we have the eigenvectors, the trace and determinant are easy to evaluate. If $e_1, e_2$ are the eigenvectors with eigenvalues $\lambda_1, \lambda_2$, then the trace of $Q$ is given by $\tr Q = \lambda_1 + \lambda_2 = Q(e_1) + Q(e_2),$ and the determinant of $Q$ is given by $\det Q = \lambda_1 \lambda_2 = Q(e_1)Q(e_2).$

All this eigenvector business is actually not needed for computing the trace: we can pick any orthonormal vectors $X, Y \in \mathbb{R}^2$, and the trace is given by $Q(X) + Q(Y)$.

However, the determinant is trickier. In general, $\det Q$ is not equal to $Q(X)Q(Y)$ for a general pair of orthonormal vectors $X$ and $Y$. To see what goes wrong, we can write out $Q$ as a matrix in the $X, Y$ basis. The diagonal entries of this matrix are simply $Q(X)$ and $Q(Y)$. To obtain the off-diagonal entries, we can apply the polariziation identity (essentially a fancy name for the fact that $xy = \tfrac{1}{2}(x+y)^2 - \tfrac{1}{2}x^2 - \tfrac{1}{2}y^2$). Now that we have the matrix entries for $Q$, we can apply the usual formula that $\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = a d - b c$ and we find that $\det Q = Q(X) Q(Y) - \tfrac{1}{4}\left(Q(X+Y) - Q(X) - Q(Y)\right)^2.$

This formula simplifies a little if we define a new unit vector $Z := \tfrac{1}{\sqrt{2}}\left(X + Y\right)$ which lies half way in between $X$ and $Y$. Then we can do a little bit of algebra to find that $\begin{aligned} \det Q &= Q(X) Q(Y) - \left(Q(Z) - \tfrac{1}{2}\left(Q(X) + Q(Y)\right)\right)^2\\ &= Q(X)Q(Y) - Q(Z)^2 - \tfrac{1}{4}\left(Q(X)+Q(Y)\right)^2 + Q(Z)\left(Q(X) + Q(Y)\right)\\ &= \tfrac{1}{4}\left(Q(X) - Q(Y)\right)^2 + Q(Z) \left(Q(X) + Q(Y) - Q(Z)\right). \end{aligned}$

We can simplify this formula even more by introducing the complementary unit vector $W := \tfrac{1}{\sqrt{2}}\left(X - Y\right)$. Note that $Z, W$ also form an orthonormal basis for $\RR^2$. Since we can use the any orthonormal basis of $\RR^2$ to compute the trace of $Q$, we know that $Q(X) + Q(Y) = \tr Q = Q(Z) + Q(W)$, and therefore the expression $Q(X) + Q(Y) - Q(Z)$ in our formula is simply equal to $Q(W)$. So in the end, the determinant of $Q$ is given by $\det Q = \tfrac{1}{4}\left(Q(X) - Q(Y)\right)^2 + Q(Z) Q(W),$ where $X$ and $Y$ form one orthonormal basis, and $Z$ and $W$ form another orthonormal basis rotated by $45^\circ$.

As one last sanity check, note that if we take $Z$ and $W$ to be the eigenvectors $e_1$ and $e_2$, then $Q(X) = Q(Y)$, and we so find again that the determinant is given by $\det Q = Q(e_1) Q(e_2)$. But now we see that this is really a special case, and in general we need to add a correction term involving a pair of vectors rotated by $45^\circ$ from our starting vectors.