Positive-definite quadratic criteria with an equality budget

4.4 Quadratic forms and a global constrained formula

If $A=QDQ^T$ and $z=Q^Tx$, then $x^TAx=z^TDz=\sum_i\lambda_iz_i^2$. Consequently $A$ is positive definite exactly when every eigenvalue is positive. Negative eigenvalues produce negative directions, and zero eigenvalues produce nontrivial flat directions. This general-size criterion extends Chapter 2's two-variable determinant test.

Worked example 2: diagonalize a quadratic criterion. For $\Sigma=\begin{pmatrix}5&2\\2&2\end{pmatrix}$, eigenvectors $(2,1)^T$ and $(1,-2)^T$ have eigenvalues six and one. Dividing by $\sqrt5$ gives an orthogonal matrix $Q$. In its coordinates,

$$w^T\Sigma w=6z_1^2+z_2^2,$$

proving positive definiteness and exposing unequal curvature in perpendicular directions.

For a symmetric positive-definite $\Sigma$ and constraint $\mathbf1^Tw=1$, multiplier equations yield

$$w_* =\frac{\Sigma^{-1}\mathbf1}{\mathbf1^T\Sigma^{-1}\mathbf1},\qquad V_* =\frac1{\mathbf1^T\Sigma^{-1}\mathbf1}.$$

The denominator is positive because the inverse is positive definite. To prove global optimality, write another feasible vector as $w_*+h$ with $\mathbf1^Th=0$. The cross term vanishes since $\Sigma w_*$ is a multiple of $\mathbf1$, leaving $V(w_*+h)=V_*+h^T\Sigma h>V_*$ for $h\ne0$. Nonnegative-weight constraints, if present, still need checking separately; the equality-only formula can produce negative components in other examples.

Exercises

P06. Verify the eigenvectors and diagonal quadratic expression for $\Sigma$ in Worked example 2.

P09. Use the equality-constrained formula for $\Sigma=\operatorname{diag}(2,8)$ and calculate the minimizing two weights and value. Verify directly by substituting $w_2=1-w_1$.

Full solutions

P06. $\Sigma(2,1)^T=(12,6)^T=6(2,1)^T$ and $\Sigma(1,-2)^T=(1,-2)^T$. The normalized vectors are orthogonal, so $Q^T\Sigma Q=\operatorname{diag}(6,1)$ and the quadratic becomes $6z_1^2+z_2^2$.

P09. $\Sigma^{-1}\mathbf1=(1/2,1/8)^T$ has sum $5/8$, so weights are $(4/5,1/5)$ and value $8/5$. Directly, $2x^2+8(1-x)^2=10(x-4/5)^2+8/5$, proving the same unique global minimum.