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Let $F(x,y,z)=2x^2+3y^2+5z^2-xy-xz-yz$ and

$A=\begin{pmatrix}2 & -\frac{1}{2} & -\frac{1}{2} \\-\frac{1}{2} & 3 & -\frac{1}{2} \\-\frac{1}{2} & -\frac{1}{2} & 5\end{pmatrix}$ be the associated matrix.

The inertia of $A$, denoted $\operatorname{in}(A)$, is defined as the triple

$\operatorname{in}(A):=(n_1,n_2,n_3)$

where $n_i, i=1,2,3$ denoted the number of positive, negative, and zero eigenvalues of $A$ respectively. Prove the following:

$i)$ If $\operatorname{in}(A) = (3,0,0)$ then the quadratic is an ellipsoid.

so this problem is saying it has 3 positive eigenvalues and no negative and no zero eigenvalues.

i know the general equation for an ellipsoid is $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ and that it only makes since for the eigenvalues to be all positive.

i'm not too sure how to actually "prove" it other than the fact if it was a negative it'd be a different shape

Michael Hardy
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Charlene
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  • You provide a lot of context, which is great. However, it's not totally clear what you're asking. Are you asking something like "how can I prove that a quadratic is an ellipsoid iff the associate matrix is positive definite"? Or, are you asking how to show that this particular quadratic is an ellipsoid (i.e. how to show that this matrix is positive definite)? – Ben Grossmann Aug 10 '15 at 16:44
  • yes that's what i'm trying to ask. – Charlene Aug 10 '15 at 16:47
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    The first one or the second one? – Ben Grossmann Aug 10 '15 at 16:47
  • the first one, sorry i didnt see the second part. – Charlene Aug 10 '15 at 16:49
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    For the second part: it's easy to see that this matrix is positive definite since it is symmetric and strictly diagonally dominant with positive diagonal entries. – Ben Grossmann Aug 10 '15 at 17:07
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    Related: http://math.stackexchange.com/questions/1388326/diagonal-matrix-a – Surb Aug 10 '15 at 17:13
  • same matrix with really good answer on how to send to congruent diagonal, typeset by another user http://math.stackexchange.com/questions/1388281/find-the-transitional-matrix-that-would-transform-this-form-to-a-diagonal-form – Will Jagy Aug 10 '15 at 18:48

1 Answers1

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Define $x = (x_1,x_2,x_3)^T$. Note that $F(x_1,x_2,x_3) = x^TAx$.

Now, by Sylvester's law of inertia: if $A$ is positive definite, there exists an invertible matrix $S$ such that $$ A = S^T\pmatrix{1&0&0\\0&1&0\\0&0&1} S = S^TS $$ Or in other words, we have $$ F(x_1,x_2,x_3) = x^TS^TSx = (Sx)^T(Sx) $$ Now, what does this mean geometrically? Let $s_{ij}$ denote the entries of $S$. Let $y = (y_1,y_2,y_3)^T = Sx$, so that $$ \pmatrix{y_1\\y_2\\y_3} := S \pmatrix{x_1\\x_2\\x_3} = \pmatrix{s_{11}x_1 + s_{12}x_2 + s_{13}x_3\\ s_{21}x_1 + s_{22}x_2 + s_{23}x_3\\ s_{31}x_1 + s_{32}x_2 + s_{33}x_3} $$ We can think of this as a change of variables from $x$ to $y(x)$. What we have, then, is that $$ F(x_1,x_2,x_3) = y^Ty = y_1^2 + y_2^2 + y_3^2 $$ That is, there is a linear change of variables that takes us from this quadratic to the sphere. That can only be true if we have an ellipsoid.

Ben Grossmann
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    really good presentation of *non-orthogonal* diagonalization of the same matrix (same person asked), sometimes called finding a congruent diagonal matrix (and answers the inertia question) http://math.stackexchange.com/questions/1388281/find-the-transitional-matrix-that-would-transform-this-form-to-a-diagonal-form/1388381#1388381 – Will Jagy Aug 10 '15 at 18:52
  • I typeset a different such problem, including matrices, at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr/1392111#1392111 – Will Jagy Aug 10 '15 at 18:57