Practice: Quadratic Forms and Definiteness

(a) $Q=2x^2-4xy+5y^2$ is positive definite. Squared coefficients 2 and 5 on the diagonal; the cross coefficient $-4$ halved to $-2$:

Check. $Q(1,1)=2-4+5=3$, and $x^{\mathsf{T}}Ax$ at $(1,1)$ is $2-2-2+5=3$ ✓. $\mathrm{tr}A=7$, $detA=10-4=6$, so $p(\lambda )={\lambda }^2-7\lambda +6=(\lambda -6)(\lambda -1)$ and the eigenvalues are $6$ and $1$. Both positive, so positive definite. The negative cross term notwithstanding. The shortcut agrees: and . Axes. For $\lambda =6$: , so $-4x-2y=0$ gives $y=-2x$ and $v_1\propto (1,-2)$. For $\lambda =1$: , so $x=2y$ and $v_2\propto (2,1)$. Normalised,

perpendicular since $2-2=0$ ✓. In these coordinates $Q=6y_1^2+y_2^2$. Confirmation. $Q(v_1)=\frac{1}{5}[2(1)-4(1)(-2)+5(4)]=\frac{2+8+20}{5}=\frac{30}{5}=6={\lambda }_1$ ✓. (b) $R$ is positive semidefinite.

Block diagonal: the leading $2\times 2$ block has trace 2 and determinant 0, so eigenvalues $2$ and $0$; the isolated entry contributes $1$. The eigenvalues are 2, 1, 0. All nonnegative with one zero, so positive semidefinite, not definite. A vector where it vanishes. The zero eigenvalue's eigenvector is $(1,-1,0)$, and

with $(1,-1,0)\ne 0$. That single vector is what separates semidefinite from definite. Elsewhere the form is strictly positive: $R(1,1,0)=4$, $R(0,0,3)=9$. (c) The non-symmetric presentation. Expanding, $x^{\mathsf{T}}Mx=5x^2+4xy+0\cdot yx+5y^2=5x^2+4xy+5y^2$. So this is the form of the worked example, and its symmetric matrix is

with eigenvalues 7 and 3. Why $M$'s eigenvalues are the wrong ones. $M$ is triangular, so its eigenvalues are both 5. But $x^{\mathsf{T}}Mx$ is a scalar and therefore equals its own transpose $x^{\mathsf{T}}{M}^{\mathsf{T}}x$, so $M$, $M^{\mathsf{T}}$ and their average all define the same function. The antisymmetric part of $M$ contributes nothing to any value of the form, and any property computed from it, eigenvalues included, is an artefact of how the form was written rather than a fact about the form. Only the symmetric representative carries the properties the classification needs: real eigenvalues and an orthonormal eigenbasis, neither guaranteed for a general matrix. Here the verdict of positive definiteness would have come out right by luck, but the principal values 5 and 5 are wrong, and so would be any axis or level-set conclusion drawn from them. (d) Completing the square for $Q=2x^2-4xy+5y^2$.

Check at $(1,1)$: $2(0{)}^2+3=3$, matching $Q(1,1)=3$ ✓. At $(2,1)$: direct $8-8+5=5$; completed $2(1{)}^2+3=5$ ✓. What agrees. Both coefficients, 2 and 3, are positive, and so are both eigenvalues, 6 and 1. The count of positive, negative and zero coefficients matches, which is Sylvester's law of inertia, and that is enough to classify the form. What does not. The values differ: $\{2,3\}$ against $\{6,1\}$. The variables $x-y$ and $y$ are not orthogonal directions, so the change of variables is invertible but not a rotation. It removes the cross term at the cost of distorting lengths, and the numbers it produces are not the form's values along perpendicular axes. (e) What orthogonality preserves. An orthogonal change of variables $y=P^{\mathsf{T}}x$ satisfies $\Vert y\Vert =\Vert x\Vert$, so it preserves lengths, angles and areas. It is a rotation, possibly with a reflection. A general invertible change preserves only incidence: straight lines stay straight and the origin stays fixed, but distances and perpendicularity do not survive. For the level set $Q=1$ this is the whole difference. Under the orthogonal change, the set is a genuine ellipse whose axes point along $v_1$ and $v_2$ and whose semi-axis lengths are $1/\sqrt{{\lambda }_i}$, for part (a), $1/\sqrt{6}\approx 0.408$ along $(1,-2)/\sqrt{5}$ and $1$ along $(2,1)/\sqrt{5}$. Those are measurable facts about the curve in the plane. Under the completed-square change, the set becomes $2u^2+3v^2=1$, which is an ellipse in the $(u,v)$ coordinates but a sheared image of the true one. Its "axes" of length $1/\sqrt{2}$ and $1/\sqrt{3}$ describe the transformed picture, not the original curve, and the directions $x-y=\text{const}$ and $y=\text{const}$ are not perpendicular in the plane. So: completing the square is sufficient to classify, because signs are preserved by any invertible change. Only the spectral route describes, because only it preserves the geometry it is describing.