Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In many textbooks of pattern recognition I have seen the following statement:

If a matrix $A$ in the quadratic form $F(\text{x}) = \text{x}^{T}A\text{x}$ is positive definite, then it follows that the surfaces of constant $F(\text{x})$ are hyperellipsoids, with principal axes having lengths proportional to $\lambda_{k}^{-1/2}$.

$A$ is a $n \times n$ matrix, $\lambda_k$ is an eigenvalue of $A$, $\text{x}$ is a non-zero vector and $k = 1, ..., n$.

My question is: "Why are the principal axes of hyperellipsoids proportional to the eigenvalues of $A$? Why are they proportional to $\lambda_{k}^{-1/2}$? And why are the surfaces of constant $F(\text{x})$ hyperellipsoids?"

Can someone give me a proof? Visualization perhaps? etc. Anything that would make me see that this indeed is the case :)

Hope my question is clear :) Thank you for any help!

share|improve this question

1 Answer 1

up vote 4 down vote accepted

If $A \in \def\R{\mathbb R}\R^{n\times n}$ is positive definite, there is an orthogonal $Q \in O(n)$ (that is a matrix whose columns form an orthonormal basis of $\R^{n}$) such that $QAQ^t = \mathrm{diag}(\lambda_1, \ldots, \lambda_n) =: \Lambda$ with the $\lambda_k$ positive. For $c > 0$ and $x \in \R^n$ we have \begin{align*} F(x) &= c\\ \iff x^t Ax &= c \\ \iff x^t Q^t QAQ^t Qx &= c\\ \iff (Qx)^t \Lambda Qx &= c\\ \iff \sum_k \lambda_k (Qx)_k^2 &= c\\ \iff \sum_k \frac{(Qx)_k^2}{(c^{1/2}\lambda_k^{-1/2})^2} &= 1 \end{align*} So in the transformed coordinates $y = Qx$ given by $Q$ we see that the hypersurface $\{F = c\}$ is given by $$ \frac{y_1^2}{(c^{1/2}\lambda_1^{-1/2})^2} + \cdots + \frac{y_n^2}{(c^{1/2}\lambda_n^{-1/2})^2} = 1 $$ that is it forms a hyperellipsoid with principal axes in direction $Qe_k$ (where $e_k$ denotes the $k$-th standard unit vecotr) having lengths $c^{1/2}\lambda_k^{-1/2}$.

share|improve this answer
    
+1 @martini Nice! thank you very much :) That helps :) –  jjepsuomi Jul 22 '13 at 10:25
1  
Clear derivation. (+1) For the sake of completeness, what is $e_k$? –  vesszabo Jul 22 '13 at 10:28
1  
@vesszabo Added. –  martini Jul 22 '13 at 10:29
    
@martini Thanks :-) –  vesszabo Jul 22 '13 at 10:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.