How to solve this basic type of quadratic form question?

Question

Here's a problem from a previous exam.

1. Find the rank and signature of the following quadratic form on $\mathbb R^3)$ $$q(x)=x_1^2-6x_1x_2+4x_1x_3+7x_2^2-10x_2x_3+8x_3^2$$
2. Find the symmetric bilinear form $f(x,y)$ corresponding to $q$
3. Prove the form below is an inner product on $\mathbb R^3$, where $ \left\langle \cdot,\cdot \right\rangle$ is the standard inner product on $\mathbb R^3$. $$h(x,y)=x_1y_1-3x_1y_2-3x_2y_1+2x_1y_3+2x_3y_1+7x_2y_2-5x_2y_3-5x_3y_2+8x_3y_3+ \left\langle x,y \right\rangle $$

I don't understand the theoretical material at all and am hoping to understand it by getting help in solving this thing step by step..

Answer 1

It helps to start by answering 2. Note that we can rewrite $q$ by splitting any cross-products into two. In particular, $$ q(x) = x_1x_1 - 3x_1x_2 + 2x_1 x_3\\ -3x_2x_1 + 7x_2x_2 - 5 x_2x_3\\ +2x_3x_1 - 5x_3x_2 + 8 x_3x_3 $$ If we define $$ f(x,y) = x_1y_1 - 3x_1y_2 + 2x_1 y_3\\ -3x_2y_1 + 7x_2y_2 - 5 x_2y_3\\ +2x_3y_1 - 5x_3y_2 + 8 x_3y_3 $$ then it's clear that $q(x) = f(x,x)$. Moreover, $f$ is symmetric since $f(x,y) = f(y,x)$. So, we say that $f$ is the symmetric bilinear form corresponding to $q$.

Importantly, we can write $f$ using matrices. Note that if $x$ and $y$ are column-vectors, then $f(x,y) = x^TAy$, where $A$ is the matrix $$ A = \pmatrix{1&-3&2\\-3&7&-5\\2&-5&8} $$ thus, we call $A$ the matrix corresponding to the quadratic form $q$ and the (symmetric) bilinear form $f$. Notably, the matrix corresponding to a quadratic form is always symmetric.

To answer 1, find the rank of the matrix $A$, and find out how many positive eigenvalues $A$ has. Notably, there are ways to do this without explicitly computing the eigenvalues of $A$.

To answer 3, note that $A + I$ (the matrix corresponding to $h$) has only positive eigenvalues.

Answer 2

Write the quadratic form as a sum of squares, usingwhatever method you want. I shall use here Lagrange's (using $\;x,y,z\;$ intead of the annoying (for me) $\;x_1,...,x_3$):

$$q\left(\bf x\right) =x^2-6xy+4xz+7y^2-10yz+8z^2=$$

$$=(x-(3y-2z))^2-9y^2+12yz-4z^2+7y^2-10yz+8z^2=$$

$$=(x-(3y-2z))^2-2y^2+2yz+4z^2=(x-(3y-2z))^2-2\left(y-\frac12z\right)^2+\frac92z^2$$

and there you have the signature: $\;(p,n)=(2,1)\;$ and rank three.

Other way: with matrices:

$$q\left(\bf x\right)=(x\;y\;z)\begin{pmatrix}1&\!-3&2\\\!-3&7&\!-5\\2&\!-5&8\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}$$

Now bring the matrix to diagonal fgorm (why is it always possible with matrics of quadratic forms?) by row/column elementary operations:

$$\begin{pmatrix}1&\!-3&2\\\!-3&7&\!-5\\2&\!-5&8\end{pmatrix}\stackrel{R_2+3R_1,\,R_3-2R_1}\rightarrow \begin{pmatrix}1&\!-3&2\\0&\!-2&1\\0&1&4\end{pmatrix}\stackrel{C_2+3C_1,\,C_3-2C_1}\rightarrow \begin{pmatrix}1&0&0\\0&\!-2&1\\0&1&4\end{pmatrix}\stackrel{R_3+\frac12R_2}\rightarrow$$$${}$$

$$\begin{pmatrix}1&0&0\\0&\!-2&1\\0&0&\frac92\end{pmatrix}\stackrel{C_3+\frac12C_2}\rightarrow\begin{pmatrix}1&0&0\\0&\!-2&0\\0&0&\frac92\end{pmatrix}$$

and once again you get the same signature and the same rank, ofcourse.