Linear systems and quadratic surfaces

Planes lead to linear systems; quadratic constraints lead to ellipsoids, hyperboloids and paraboloids. Matrix notation treats both systematically. Row reduction resolves compatible plane intersections, while orthogonal eigenvector coordinates reveal the principal axes of a symmetric quadratic form.

1 Matrices and intersections of planes

A system $Ax=b$ combines linear equations, one per row. Elementary row operations are reversible changes of equations, so they preserve the solution set when applied to the augmented matrix $[A\mid b]$. Pivots identify determined variables; free variables describe a line or higher-dimensional affine family. A zero-coefficient row with nonzero right-hand side proves inconsistency.

Worked three-plane intersection. Solve

$$x+y+z=6,\quad2x-y+z=3,\quad x+2y-z=2.$$

Subtract twice the first equation from the second to get $-3y-z=-9$; subtract the first from the third to get $y-2z=-4$. The latter gives $y=2z-4$, and substitution gives $z=3$, $y=2$, $x=1$. All three original equations verify the point. The coefficient determinant is 7, so there is a unique intersection.

A nonzero determinant is a square-system criterion. For a rectangular collection of plane constraints, rank and consistency are the appropriate language. Three equations can describe a line if one is redundant, or no point if one contradicts the others.

2 Quadratic forms and symmetric matrices

Every real quadratic expression can be written $x^TAx+b^Tx+c$, with symmetric $A$: a cross-term coefficient $2a_{ij}$ is shared between entries $a_{ij}$ and $a_{ji}$. An orthogonal eigenbasis gives $A=Q\Lambda Q^T$, so under $x=Qu$,

$$x^TAx=u^T\Lambda u=\sum_i\lambda_i u_i^2.$$

Orthogonal coordinates preserve lengths and angles. A general invertible coordinate substitution can diagonalize other ways but may distort geometry, so semiaxis lengths should be read only after the coordinate metric is understood.

For invertible $A$, completing the square with $z=x+\tfrac12A^{-1}b$ yields

$$x^TAx+b^Tx+c=z^TAz+c-\tfrac14b^TA^{-1}b.$$

Expand the right side to prove the identity. This translation locates the center, then orthogonal diagonalization locates principal axes. If $A$ is singular, a linear term in a zero-eigenvalue direction may produce a paraboloid; the inverse formula is unavailable.

3 Standard spatial objects

With positive constants $a,b,c$, the ellipsoid $u^2/a^2+v^2/b^2+w^2/c^2=1$ has semiaxes $a,b,c$. A positive-definite quadratic form equal to a positive constant has this shape after scaling its principal coordinates; equal to zero it gives only the center, and equal to a negative constant it gives no real points.

The surface $u^2/a^2+v^2/b^2-w^2/c^2=1$ is a one-sheet hyperboloid; with right-hand side $-1$ it becomes a two-sheet hyperboloid after rearrangement. With right-hand side zero it is a cone. Sign patterns matter more than a memorized name.

For singular forms, $u^2+v^2=1$ is a cylinder because $w$ is unrestricted. The equation $w=u^2+v^2$ is an elliptic paraboloid, while $w=u^2-v^2$ is a hyperbolic paraboloid. These cannot be reduced to a centered nondegenerate ellipsoid by simply dividing coefficients.

Worked rotated ellipsoid. For

$$2x^2+2xy+2y^2+z^2=6,$$

the $xy$ block has eigenvectors $(1,1)/\sqrt2$ and $(1,-1)/\sqrt2$ with eigenvalues 3 and 1. Set $u=(x+y)/\sqrt2$, $v=(x-y)/\sqrt2$, $w=z$. The equation is $3u^2+v^2+w^2=6$, an ellipsoid with semiaxes $\sqrt2,\sqrt6,\sqrt6$. The cross term describes a rotation of axes, not an additional fourth dimension or a failure of quadratic classification.

4 Tangent planes from implicit equations

Suppose $F$ is $C^1$ on an open neighborhood of $p$, $F(p)=0$, and $\nabla F(p)\ne0$. For its local level surface $F(x,y,z)=0$, the tangent plane is $\nabla F(p)\cdot(x-p)=0$. Differentiating $F(r(t))=0$ along any surface curve shows its tangent is orthogonal to the gradient, deriving the normal direction.

On $x^2+y^2+z^2=9$ at $(1,2,2)$, the gradient is $(2,4,4)$ and the plane is $x+2y+2z=9$. At the cone vertex of $x^2+y^2-z^2=0$, the gradient vanishes, so this regular tangent-plane theorem does not apply. One cannot infer a unique plane from the zero equation $0=0$.

Existence of principal axes and all tangent directions

The real symmetric spectral theorem used for principal axes has the following proof. Its maximization step is justified by compactness of the sphere, and its multiplier step is covered by the general one-constraint proof in the optimization chapter.

4.3 Symmetry gives orthogonal eigenvectors

For real symmetric $A$ and real vectors $u,v$, if $Au=\lambda u$, $Av=\mu v$, then

$$\lambda u^Tv=(Au)^Tv=u^TA^Tv=u^TAv=\mu u^Tv.$$

Distinct eigenvalues therefore have orthogonal eigenspaces. The stronger spectral theorem states that every real symmetric $n\times n$ matrix, where $n\ge1$, has an orthonormal eigenbasis, hence $A=QDQ^T$ with $Q^TQ=I$ and real diagonal $D$.

Here is a proof using earlier chapters. The continuous function $x^TAx$ attains a maximum on the unit sphere, which is closed and bounded. At a maximizing unit vector $v$, the regular constraint $x^Tx=1$ has nonzero gradient $2v$. The multiplier condition gives $2Av=2\lambda v$, so $v$ is a real unit eigenvector. Its perpendicular subspace is invariant: if $v^Tw=0$, then $v^TAw=(Av)^Tw=\lambda v^Tw=0$. Restrict $A$ to that $(n-1)$-dimensional subspace in an orthonormal basis; the restriction remains symmetric. Induction, beginning with dimension one, supplies an orthonormal eigenbasis there. Together with $v$, it proves the theorem. An orthonormal basis for the perpendicular subspace is constructed by the Gram–Schmidt method.

The orthonormal-complement construction and the tangent plane

To make the induction constructive, start with the unit eigenvector $v$ and extend it to a linear basis $v,w_2,\ldots,w_n$ by selecting standard coordinate vectors outside the span until the span is all of $\mathbb R^n$. Set $u_1=v$; at step $j$, subtract $\sum_{i<j}(w_j\cdot u_i)u_i$ from $w_j$. The result is orthogonal to the earlier $u_i$ and is nonzero, since otherwise $w_j$ lay in the previous span. Divide by its positive norm. The resulting $u_2,\ldots,u_n$ form an orthonormal basis of $v^\perp$. In that basis the restriction matrix has entries $u_i^TAu_j=u_j^TAu_i$, so it is symmetric and the induction applies. Dimension one begins the induction, and dimension zero is the empty complement.

For the level surface, assume explicitly that $F$ is $C^1$ on an open neighborhood of $p$, $F(p)=0$, and $\nabla F(p)\ne0$. Choose the last coordinate so $F_z(p)\ne0$. The general implicit theorem proved in the optimization chapter gives the graph $z=\varphi(x,y)$ near $p$. Its tangent vectors are $(u_1,u_2,D\varphi(p')u)$, and differentiating the identity gives $D\varphi=-F_z^{-1}(F_x,F_y)$. Hence they are exactly all vectors orthogonal to $\nabla F(p)$: any orthogonal vector has its last coordinate uniquely forced by the first two, and the graph curve $s\mapsto(p'+su,\varphi(p'+su))$ realizes it. Thus the tangent plane is precisely $\nabla F(p)\cdot(x-p)=0$. Existence of partial derivatives at one point alone would not justify this graph or tangent-plane argument.

5 Graded practice

E1 Foundation. Solve $x+y+z=3$, $x-y+z=1$, $z=1$, and interpret uniqueness.

E2 Translation. Classify $x^2+y^2+z^2-2x+4y=4$ by completing squares.

E3 Rotation. Remove the cross term in $3x^2-2xy+3y^2+2z^2=8$ and find its principal semiaxes.

E4 Degeneracy. Classify $x^2+y^2-z=0$ and find its tangent plane at $(1,0,1)$.

6 Full solutions

S1. $z=1$, so $x+y=2$ and $x-y=0$, giving $(1,1,1)$. All three variables are determined, and the independent plane constraints intersect in one point.

S2. $(x-1)^2+(y+2)^2+z^2=9$. It is a sphere of radius 3 centered at $(1,-2,0)$. The constant changes by the completed-square offsets, not just by moving linear terms across the equality.

S3. With $u=(x+y)/\sqrt2$, $v=(x-y)/\sqrt2$, the $xy$ eigenvalues are 2 and 4. The equation becomes $2u^2+4v^2+2z^2=8$, giving semiaxes $2,\sqrt2,2$ along $u,v,z$ respectively.

S4. It is an elliptic paraboloid $z=x^2+y^2$. The gradient of $F=x^2+y^2-z$ at the point is $(2,0,-1)$, so $2(x-1)-(z-1)=0$, or $z=2x-1$.

Summary

Row reduction solves linear spatial constraints without relying on drawings. Symmetric matrices organize quadratic expressions; translations locate centers and orthogonal eigenvectors reveal principal axes. Singular quadratic parts and vanishing level gradients require separate treatment rather than an unjustified inverse or tangent-plane formula.