Question

Find the normal cononial form of the quadratic form 2xy+2yz-x2-y2-z2 . Hence compute its signature.

Answer 1

Answer on Question #40024, Math, Linear Algebra

Find the normal canonical form of the quadratic form $2xy + 2yz - x^2 - y^2 - z^2$. Hence, compute its signature.

Solution.

We should use Lagrange's Reduction. The reduction of a quadratic form to canonical form can be carried out by a procedure known as Lagrange's Reduction, which consists essentially of repeated completing of the square.

$$Q = 2xy + 2yz - x^2 - y^2 - z^2 =$$

rewrite it

$$= -x^2 + 2xy - y^2 + 2yz - z^2 =$$

First we should complete the square of terms with $x$:

$$= -(x^2 - 2xy + y^2) - z^2 + 2yz = -(x - y)^2 - z^2 + 2yz = -(x - y)^2 + Q_1$$

Do it with $Q_1$:

$$Q_1 = -z^2 + 2yz = -z^2 + 2yz - y^2 + y^2 = -(z^2 - 2yz + y^2) + y^2 = -(z - y)^2 + y^2$$

Therefore,

$$Q = -(x - y)^2 + y^2 - (z - y)^2$$

Inspection of this last expression for $Q$ shows those substitutions that will reduce $Q$ to the canonical form:

$$x' = x - y$$ $$y' = y$$ $$z' = z - y$$

Substituting $x', y'$ and $z'$ into the last expression for $Q$ gives

$$Q = -x'^2 + y'^2 - z'^2$$

which is of the canonical form, where $Q$ is expressed in terms of the new variables $x', y'$ and $z'$.

The signature of the quadratic form is the number $\mathcal{P}$ of positive squared terms in the reduced form.

So the signature of $Q$ is 1.

Answer:

$$Q = -x'^2 + y'^2 - z'^2$$ $$\text{Signature} = 1$$