Find the rank and signature of the following

Question

Find the rank and signature of the quadratic forms: (a)$x_1x_2+x_2x_3+....+x_{2k-1}x_{2k} $

(b)$\sum_{i,j=1}^{n} (x_i - x_j)^2$

I am stuck with both of these. In (a) I was thinking of turning each $x_ix_{i+1} $ into $\frac{1}{4}[(x_i +x_{i+1})^2-(x_i - x_{i+1})^2]$ but what to do after that?

Answer 1

Here is an answer for the first quadratic form.

Two proofs :

a) A simple proof, inspired by this answer.

Let us work on this equivalent form:

$$Q=4x_1x_2+4x_2x_3+4x_3x_4+4x_4x_5+\cdots+4x_{2k-1}x_{2k}$$

Let us group the constituent terms of $Q$ under the following form (with the notable exception of the last term, left alone)

$$Q=4(x_1x_2+x_2x_3)+4(x_3x_4+x_4x_5)+\cdots+4x_{2k-1}x_{2k}$$

Let us factor the common terms variables :

$$Q=4(x_1+x_3)x_2+4(x_3+x_5)x_4+\cdots+4x_{2k-1}x_{2k}$$

Then using repeatedly identity $4ab=(a+b)^2-(a-b)^2$ :

$$Q=\{\color{red}{+}(x_1+x_3+x_2)^2\color{red}{-}(x_1+x_3-x_2)^2\}+\{\color{red}{+}(x_3+x_5+x_4)^2\color{red}{-}(x_3+x_5-x_4)^2\}+\cdots+\{\color{red}{+}(x_{2k-1}+x_{2k})^2\color{red}{-}(x_{2k-1}-x_{2k})^2\}$$

Therefore, each pair of terms between curly brackets $\{ \ \ \}$ "contributes" to a $\color{red}{+}$ and a $\color{red}{-}$ sign in the signature of the quadratic form. Previous expression having $2k$ terms, one can conclude that :

The signature of $Q$ has $k$ times "+" and $k$ times "-"

Remark : in a fully rigorous proof, one has to verify that all the linear forms inside the squares $(x_1+x_3+x_2), \ (x_1+x_3-x_2)$, etc. are independent ; I leave it to you...

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b) Second proof based on eigenvalues. Though looking indirect compared to the previous one, this proof is rather efficient and permits to reach the signature without having to find a Gauss decomposition, which can be very tedious in some cases.

let us now consider the equivalent form :

$$2x_1x_2+2x_2x_3+....+2x_{n-1}x_{n} \ \ \text{with} \ \ n=2k$$

You should know that the signature of a quadratic form can be obtained by counting the number (+,-,z) of resp. positive, negative and zero eigenvalues of the associated matrix (https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia).

Here the associated $n \times n$ matrix is :

$$\begin{pmatrix}0&1&0&0&\cdots&0&0\\ 1&0&1&0&\cdots&0&0\\ 0&1&0&1&\cdots&0&0\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\\ 0&0&0&0&\cdots&0&1\\ 0&0&0&0&\cdots&1&0 \end{pmatrix}$$

with a superdiagonal and a subdiagonal of $1$s.

It happens that this tridiagonal Toeplitz matrix has rather simple eigenvalues

$$2\cos(k\pi/(n+1)) \ \ for \ \ k=1,2,...n$$

(see "Toeplitz case" in https://en.wikipedia.org/wiki/Tridiagonal_matrix#Eigenvalues).

From there it is not difficult to conclude concerning the signs of these eigenvalues :

• if $n=2k$ (we are in this even case), the signature of the quadratic form is $(+,-,z)=(k,k,0)$ ; (as many positive and negative eigenvalues ; no zero eigenvalue).

Moreover, as there are no zero eigenvalues, the rank is $n=2k$ (full rank).

Had we been in the case $n=2k+1$, the signature would have been $(+,-,z)=(m,m,1)$ (presence of a single zero eigenvalue).

Remark: The explicit decomposition we have obtained in the first method is by no means unique. For example, in the particular case of

$$4(x_1x_2+x_2x_3+x_3x_4)$$

one has also the following decomposition

$$(x_1+x_2+x_3+x_4)^2-(-x_1+x_2-x_3+x_4)^2-(x_1+x_4)^2+(x_1-x_4)^2$$