Show that a positive definite (not necessarily symmetric) matrix induces a hyperellipse

Question

Consider $A\in M_n(\mathbb{R})$ a positive definite matrix and a matrix $B\in M_{n \times p}(\mathbb{R})$, with $n\geq p$ and $rank(B)=p$.

i) Show that $C=B^TAB$ is positive definite.

ii) Show that the equation $x^TAx=1$ is a hyperellipse in $\mathbb{R}^n$.

For i), $$ x\neq 0 \Rightarrow Bx\neq 0. $$

So $$ \langle B^tABx,x\rangle = \langle A(Bx),Bx \rangle > 0.$$

But for ii) I have no idea.

Actually, before posting an answer, I should have clarified: by positive definite, do you mean symmetric positive definite? — artificial_moonlet, Apr 25 '15 at 12:49
In fact the exercise does not say that the matrix is symmetric. — Roland, Apr 25 '15 at 12:56
In some texts, positive definite is a priori defined to include symmetric. You're sure this is not the case? — artificial_moonlet, Apr 25 '15 at 12:57
This exercise is of an ancient race, so I can not guarantee you that the matrix is symmetric fact. There is not any other way to show this? — Roland, Apr 25 '15 at 12:59

artificial_moonlet · Accepted Answer · 2015-04-25T14:02:39.830

2

1) The following works for $A$ symmetric.

We can diagonalize $A$: $$A = Q^TDQ,$$ where $Q$ is unitary and $D$ is diagonal. Such a factorization exists since $A$ is symmetric positive definite and therefore normal. Then $$x^TAx = x^TQ^TDQx = (Qx)^TDQx.$$ Let $y = Qx$. Then you have $$ y^TDy = 1.$$

Working things out in components, you should see that this is precisely the equation for a hyperellipse.

2) When $A$ is not symmetric, the situation is messier, but still doable.

Define $B:= \frac{1}{2}(A+A^T)$. $B$ is symmetric positive definite, so by part 1), $B$ gives rise to a hyperellipse. I claim $B$ and $A$ define the same hyperellipse.

Indeed, $$ \frac{1}{2}x^TA^Tx = \frac{1}{2}x^TAx ,$$ which you can see by working things out component-wise.

Then $$x^TBx:= \frac{1}{2}(x^TAx+x^TA^Tx) = x^TAx.$$

edited Apr 25 '15 at 14:02

answered Apr 25 '15 at 12:48

artificial_moonlet

957

How would you do the proof of i)?. – Roland Apr 25 '15 at 14:08
You have the proof for i) in the problem statement... – artificial_moonlet Apr 25 '15 at 14:10
In the proof you supposes that the matrix is symmetric, D is the eigenvalues of matrix? – Roland Apr 25 '15 at 14:12
Yes. $Q^TDQ$ is the eigenvalue decomposition of $A$. This is a standard result. – artificial_moonlet Apr 25 '15 at 14:13
One last thing, the product of two positive-defned matrix is always positive definite? – Roland Apr 25 '15 at 14:16
That's another question entirely, whose answer again depends on how you define positive definite. See http://math.stackexchange.com/questions/113842/product-of-symmetric-positive-semidefinite-matrices-is-positive-definite?rq=1. You do not need to multiply two positive definite matrices in this proof, though. – artificial_moonlet Apr 25 '15 at 14:22
I'm just a little in doubt this last part $\langle A(Bx),Bx \rangle > 0$ – Roland Apr 25 '15 at 14:26
1

$Bx$ is a vector in $\mathbb{R}^n$. Applying the fact that $A$ is positive definite gives you the result, without looking at $AB$. – artificial_moonlet Apr 25 '15 at 14:34

Show that a positive definite (not necessarily symmetric) matrix induces a hyperellipse

1 Answers1