Ellipsoids From Positive Definite Forms

From a Quadratic Form to an Ellipsoid

A positive definite quadratic form assigns a positive value to every nonzero vector. Its level sets therefore describe how far vectors lie from the origin, but generally not in the same way in every direction. The eigenvectors of the matrix identify the directions in which the form has a particularly simple expression; the corresponding eigenvalues determine how quickly the form grows along those directions. Together, they give the orientation and dimensions of the ellipsoids described by equations of the form \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}=c\).

Throughout, \(\mathbf{x}\in\mathbb{R}^n\), and \(\mathbf{A}\in\mathbb{R}^{n\times n}\) is real and symmetric. The matrix is positive definite if \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}>0\) for every \(\mathbf{x}\ne\mathbf{0}\). The level set at \(c\in\mathbb{R}\) is the set of all \(\mathbf{x}\) satisfying \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}=c\). We use the earlier result on orthogonal diagonalization of real symmetric matrices: there is an orthonormal basis of eigenvectors of \(\mathbf{A}\).

The Eigenvector Coordinates and the Axis Formula

Let \(\mathbf{q}_1,\ldots,\mathbf{q}_n\) be an orthonormal eigenbasis, with \(\mathbf{A}\mathbf{q}_i=\lambda_i\mathbf{q}_i\). Write \(\mathbf{Q}=[\mathbf{q}_1\ \cdots\ \mathbf{q}_n]\). Orthonormality means \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\), and orthogonal diagonalization gives \(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)\). For coordinates \(\mathbf{y}=\mathbf{Q}^{\top}\mathbf{x}\), we have \(\mathbf{x}=\mathbf{Q}\mathbf{y}\), so

Positive definiteness forces every \(\lambda_i\) to be positive: for a unit eigenvector \(\mathbf{q}_i\), \(\lambda_i=\mathbf{q}_i^{\top}\mathbf{A}\mathbf{q}_i>0\). Thus, when \(c>0\), dividing by \(c\) expresses the level set as a sum of squared coordinates equal to \(1\). This is the standard ellipsoid equation in orthonormal coordinates.

Proof. Assume the stated hypotheses. Since the eigenvectors form an orthonormal basis, every \(\mathbf{x}\in\mathbb{R}^n\) has coordinates \(y_i=\mathbf{q}_i^{\top}\mathbf{x}\) and equals \(\sum_i y_i\mathbf{q}_i\). Orthogonal diagonalization gives \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}=\sum_i\lambda_i y_i^2\). Because \(c>0\) and every \(\lambda_i>0\), the equation \(\sum_i\lambda_i y_i^2=c\) is equivalent, after dividing by \(c\), to \(\sum_i y_i^2/(c/\lambda_i)=1\). This proves the displayed description in both directions: each point of the level set satisfies the sum equation, and any coordinates satisfying the sum equation give a point of the level set. On the \(i\)th principal direction, all coordinates except \(y_i\) are zero, so the equation becomes \(\lambda_i y_i^2=c\). Its two solutions are \(y_i=\pm\sqrt{c/\lambda_i}\), giving the two axis endpoints and the stated semiaxis length. \(\square\)

The formula also explains how to compare the axes. For fixed \(c>0\), smaller eigenvalues give longer semiaxes, because \(c/\lambda_i\) is larger when \(\lambda_i\) is smaller. If an eigenvalue is repeated, its eigenspace gives the corresponding principal directions; an orthonormal basis within that eigenspace can be chosen in more than one way, but all those directions have the same semiaxis length.

Worked Axis Calculations

What the Value of \(c\) Changes

The ellipsoid description requires \(c>0\). The other values have straightforward but different outcomes, and positive definiteness is essential to each conclusion.

Proof. If \(\mathbf{x}\ne\mathbf{0}\), positive definiteness gives \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}>0\), while at \(\mathbf{x}=\mathbf{0}\) the value is \(0\). Therefore no vector can have a negative value, proving emptiness for \(c<0\). When \(c=0\), the zero vector is a solution, and no nonzero vector can be a solution, so the level set is exactly \(\{\mathbf{0}\}\). When \(c>0\), the ellipsoid axes theorem applies and gives the ellipsoid description. \(\square\)

As \(c\) increases through positive values, every semiaxis length \(\sqrt{c/\lambda_i}\) scales by the same factor. For example, replacing \(c\) by \(4c\) doubles each semiaxis length without changing its principal direction. The level set itself is the ellipsoid’s surface; the solid it encloses is described by \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}\le c\). The same eigenvector coordinates show why: the inequality becomes \(\sum_i y_i^2/(c/\lambda_i)\le1\).

A Common Axis-Length Mistake

A common error is to report \(c/\lambda_i\) as the semiaxis length. That ratio is the square of the semiaxis length, because the standard equation uses squared coordinates. In the first worked example, the mistaken calculation would report a length \(4/1=4\) along \(\mathbf{e}_1\). But the proposed endpoint \((4,0)^{\top}\) gives \((4,0)\mathbf{A}(4,0)^{\top}=16\), not \(4\). The correction is to take the square root: the semiaxis length is \(\sqrt{4/1}=2\), and the endpoint \((2,0)^{\top}\) gives the required value \(4\).

It is also important not to confuse an eigenvalue with a length. A large eigenvalue means the quadratic form grows quickly along its eigenvector, so the ellipsoid must reach a smaller distance in that direction to stay at the same value \(c\). Thus the longest semiaxis corresponds to the smallest eigenvalue, and the shortest corresponds to the largest. This direction is the reverse of what a direct comparison of eigenvalue sizes might suggest.