Quadratic Forms

Portfolio variance isn't just a sum of each asset's individual variance — it also depends on how every pair of assets moves together. Writing that out term by term for even a modest portfolio produces a sprawling expression full of covariance terms. A quadratic form is the compact way to package all of that: one matrix, one vector, one expression, and — crucially — a guarantee that the result behaves sensibly (never negative, for a real variance) that falls straight out of the matrix's eigenvalues.

An analogy: a bowl versus a saddle

Picture the graph of a function of two variables as a landscape. Some landscapes are bowls — every direction you walk from the bottom, you go up. Some are saddles — walk one way and you go up, walk another and you go down. A quadratic form is the algebraic recipe behind that landscape's shape, and whether it's a bowl (always positive), an upside-down bowl (always negative), or a saddle (mixed) is fixed entirely by the eigenvalues of the underlying matrix — no need to trace out the whole landscape to know which one you have.

The math, one symbol at a time

For a symmetric matrix $A$ and a vector $x$, the quadratic form is the scalar

$$Q(x) = x^\top A x = \sum_{i,j} A_{ij} x_i x_j.$$

In words: multiply every pair of entries $x_i, x_j$ by the matrix entry $A_{ij}$ that links them, and add everything up. When $A$ is a covariance matrix and $x$ is a vector of portfolio weights, $Q(x)$ is exactly the portfolio's variance — diagonal terms $A_{ii}x_i^2$ are each asset's own variance contribution, off-diagonal terms $A_{ij}x_ix_j$ capture how pairs move together.

The matrix's eigenvalues classify the shape: if every eigenvalue is positive, $A$ is positive definite and $Q(x) > 0$ for every nonzero $x$ (a bowl); if eigenvalues are mixed in sign, $Q(x)$ can go either way (a saddle); if some eigenvalue is exactly zero, $Q(x)$ can be zero for a nonzero $x$ (positive semidefinite — the flat direction of a valley floor).

Worked example 1: portfolio variance by hand

Two assets with variances $\sigma_1^2 = 0.04$, $\sigma_2^2 = 0.09$ and covariance $0.03$ give covariance matrix $A = \begin{pmatrix}0.04 & 0.03\\0.03 & 0.09\end{pmatrix}$. For weights $x = (0.6, 0.4)$:

$$Q(x) = 0.04(0.6)^2 + 0.09(0.4)^2 + 2(0.03)(0.6)(0.4) = 0.0144 + 0.0144 + 0.0144 = 0.0432.$$

Portfolio variance is 0.0432, so volatility is $\sqrt{0.0432} \approx 20.8\%$. Every piece of that number came directly from the quadratic form — no separate variance-of-a-sum derivation needed.

Worked example 2: checking positive definiteness

Is $B = \begin{pmatrix}2 & 3\\3 & 2\end{pmatrix}$ a valid covariance-style matrix? Its eigenvalues solve $(2-\lambda)^2 - 9 = 0$, giving $\lambda = 5$ and $\lambda = -1$. One eigenvalue is negative, so $B$ is not positive definite — indeed $x = (1,-1)$ gives $Q(x) = 2(1) + 2(1) + 2(3)(1)(-1) = 4 - 6 = -2$, a negative "variance." A matrix like this could never legitimately arise as a covariance matrix; if you ever compute one and it looks like $B$, it signals a data or estimation error, not a real risk structure.

Adjust the coefficient and watch the parabola flip from a bowl to an upside-down bowl as the leading sign changes — the one-dimensional version of what eigenvalue sign does to a quadratic form in many dimensions.

What this means in practice

Portfolio variance, quadratic penalty terms in regularized regression, and the second-order (curvature) term in a Taylor expansion of any risk or loss function are all quadratic forms. Optimizers rely on positive definiteness to guarantee a unique minimum exists; risk systems rely on it to guarantee variance estimates are never nonsensically negative. When a covariance matrix is estimated from limited historical data, it can come out only positive semidefinite or even mildly indefinite due to estimation noise — a problem addressed by the The Nearest Correlation Matrix Problem repair technique.