An ellipsoid centered at the origin is defined by the solutions $\mathbf{x}$ to the equation $\mathbf{x}^TM\mathbf{x} = 1$, where M is a positive definite matrix.

How can I see why M needs to be positive definite, based on the equation of an ellipse $Ax^2 + Bxy + Cy^2 = 1$ where $B-4AC < 0$? It looks like the idea is to make $B-4AC < 0$ equate to the requirement that all eigenvalues of $M$ are positive for a 2x2 matrix, but I can't seem to make it work.

Also, what other shapes can we represent with $\mathbf{x}^TM\mathbf{x} = 1$ when $M$ is not positive definite?

  • 1
    $\begingroup$ How about other conics? Parabolas, hyperbolas, double lines... $\endgroup$ Commented Aug 19, 2015 at 16:38
  • $\begingroup$ Could you explain what requirements on A are needed to represent those conics? $\endgroup$
    – scip
    Commented Aug 19, 2015 at 16:41
  • 1
    $\begingroup$ en.wikipedia.org/wiki/Conic_section#Discriminant_classification $\endgroup$ Commented Aug 19, 2015 at 16:42
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    $\begingroup$ Well, first it would help to know that if $A=\begin{bmatrix} a & b \\ b & c \end{bmatrix}$ (i.e. $A$ is $2 \times 2$ and symmetric) and $z=\begin{bmatrix} x \\ y \end{bmatrix}$ then $z^T A z = ax^2+2bxy+cy^2$. So $a$ matches your $A$, $2b$ matches your $B$, and $c$ matches your $C$. Now $B^2-4AC=4b^2-4ac$ has the opposite sign of the determinant of $A$. So $B^2-4AC<0$ is equivalent to $\text{det}(A)>0$. Finally the determinant is the product of eigenvalues, so knowing its sign tells you things about the sign of the eigenvalues. Can you work from there? $\endgroup$
    – Ian
    Commented Aug 19, 2015 at 16:51
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    $\begingroup$ Eh, I can be more explicit. The determinant is the product of the eigenvalues; the trace is the sum of the eigenvalues. Therefore if the determinant is zero, then you have one zero eigenvalue, and the other eigenvalue is the trace. If the determinant is negative, then you have one strictly positive eigenvalue and one strictly negative eigenvalue. If the determinant is positive, then both eigenvalues have the same sign, which is the same sign as the trace. So a symmetric $A$ is positive definite iff $a+c>0$ and $ac-b^2>0$. $\endgroup$
    – Ian
    Commented Aug 19, 2015 at 17:03

1 Answer 1


Consider the factorization of the matrix $A$ which is always possible since $A$ is a real symmetric matrix, and the : $A=P\Lambda P^{-1} =P\Lambda P^T$, where the matrix $P$ is orthogonal because $A$ is symmetric, and the diagonal matrix $\Lambda$ contains the eigenvalues of $A$. Then the ellipsoid can be rewritten as: $$x^TP\Lambda P^Tx = (P^Tx)^T\Lambda(P^Tx) = y^T\Lambda y = 1$$ where, of course, $y$ is derived from the linear map $\varphi:\mathbb{R}^n \rightarrow \mathbb{R}^n$, defined as $y = \varphi(x) = P^Tx$. Now, $P$ represents a pure rotation (since it is orthogonal, it preserves the norm of a vector) so that this new expression describes an ellipsoid having its axes aligned with the canonical base axes $e_1, ..., e_n$. Since the matrix is diagonal, the quadratic form gives: $$y^T\Lambda y =\lambda_1y_1^2+...+\lambda_ny_n^2 = 1$$ Of course, you could recognise the canonical form of an ellipsoid: $$\frac{y_1^2}{c_1^2} +...+ \frac{y_n^2}{c_n^2}=1$$ where $c_i$ is the length of the $i$-th semiaxis But, if the matrix is not positive definite, then the eigenvalues could assume also zero or even negative values.

In the first case, $c_i^2$ cannot exist since it should be obtained as $\frac{1}{\lambda_i}$, but dividing by zero is not allowed. In a suggestive way of thinking, it is like the ellipsoid were infinitely extended along that direction $e_i$, then it cannot be an ellipsoid.

In the second case, the $c_i$ become complex numbers, because they are the root of a negative number, and if all are negative, the ellipsoid is said 'imaginary' for trivial reasons.


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