Indexes of inertia for the function

[smile1]

Hello everyone

Here is the question

Find positive and negative indexes of inertia for the function $q(x)=TrX^2$ on the space $M_n(R)$

I did some work, first I suppose $X$ as a n by n matrix, then $TrX^2=a_{11}^2 +...+a_{nn}^2+2(a_{ij}a_{ji})$

It seems like that all terms are positive, unless $a_{ji}=-a_{ji}$, hence the positive index will be $3n$ and the negative index is $0$.

Am I right?

Thanks.

 

[Sudharaka]

Hello everyone

Here is the question

Find positive and negative indexes of inertia for the function $q(x)=TrX^2$ on the space $M_n(R)$

I did some work, first I suppose $X$ as a n by n matrix, then $TrX^2=a_{11}^2 +...+a_{nn}^2+2(a_{ij}a_{ji})$

It seems like that all terms are positive, unless $a_{ji}=-a_{ji}$, hence the positive index will be $3n$ and the negative index is $0$.

Am I right?

Thanks.

Hi simile, :)

By definition (read >>this<<) the moment of inertia is the number of positive or negative terms in the standard from of a quadratic form. That is positive moment of inertia stands for the number of positive terms in the standard form and negative moment of inertia stands for the number of negative terms in the standard form.

So you have to first reduce your quadratic form into it's standard from. Notice that,

\begin{eqnarray}

q(x)&=&\mbox{Tr }X^2\\

&=&\sum_{i=1}^{n}\sum_{j=1}^{n}x_{ij}x_{ji}\\

&=&\sum_{i=1}^{n}x_{ii}^2+\sum_{i>j}^{n}\sum_{j=1}^{n}\frac{1}{2}[(x_{ij}+x_{ji})^2-(x_{ij}-x_{ji})^2]\\

\end{eqnarray}

This is in fact the standard form of \(q\). We have written each non-square term \(2x_{ij}x_{ji}\) where \(i\neq j\) as \(\frac{1}{2}[(x_{ij}+x_{ji})^2-(x_{ij}-x_{ji})^2]\). So if you count the number of negative terms you'll get, \(\frac{n^2-n}{2}\) and the number of positive terms is \(\frac{n^2+n}{2}\).

There's another method of doing this which is explained >>here<<.