Question

Find the rank of the quadratic form 2x2+2y2+3z2-xy+4yz +5xz .

Answer 1

Answer on Question#40158, Math, Linear Algebra

Find the rank of the quadratic form $2x^{2} + 2y^{2} + 3z^{2} + xy + 4yz + 5xz$.

Solution.

Find a canonical expression of the quadratic form $(x,y,z) = 2x^{2} + 2y^{2} + 3z^{2} + xy + 4yz + 5xz$, by using Lagrange's Reduction.

$$\begin{array}{l} f(x, y, z) = \left(2x^{2} + \frac{1}{8}y^{2} + \frac{25}{8}z^{2} - xy + 5xz - \frac{5}{4}yz - \frac{1}{8}y^{2} - \frac{25}{8}z^{2} + \frac{5}{4}yz\right) + 2y^{2} + 3z^{2} \\ + 4yz = \\ = \left(\sqrt{2}x - \frac{1}{2\sqrt{2}}y + \frac{5}{2\sqrt{2}}z\right)^{2} + \frac{15}{8}y^{2} - \frac{1}{8}z^{2} + \frac{25}{4}yz. \end{array}$$

The quadratic form $\frac{15}{8}y^{2} - \frac{1}{8}z^{2} + \frac{25}{4}yz$ is in two variables $y$ and $z$ and does not depend on $x$. Now we repeat the above described procedure for this quadratic form:

$$\begin{array}{l} f(x, y, z) = \left(\sqrt{2}x - \frac{1}{2\sqrt{2}}y + \frac{5}{2\sqrt{2}}z\right)^{2} + \frac{15}{8}y^{2} - \frac{1}{8}z^{2} + \frac{25}{4}yz = \\ = \left(\sqrt{2}x - \frac{1}{2\sqrt{2}}y + \frac{5}{2\sqrt{2}}z\right)^{2} + \frac{15}{8}\left(y^{2} + \frac{10}{3}yz\right) - \frac{1}{8}z^{2} = \\ = \left(\sqrt{2}x - \frac{1}{2\sqrt{2}}y + \frac{5}{2\sqrt{2}}z\right)^{2} + \frac{15}{8}\left(y^{2} + \frac{10}{3}yz + \frac{25}{9}z^{2} - \frac{25}{9}z^{2}\right) - \frac{1}{8}z^{2} = \\ = \left(\sqrt{2}x - \frac{1}{2\sqrt{2}}y + \frac{5}{2\sqrt{2}}z\right)^{2} + \frac{15}{8}\left(y + \frac{5}{3}z\right)^{2} - \frac{209}{72}z^{2}. \end{array}$$

Putting

$$\left\{ \begin{array}{l} t_{1} = \sqrt{2}x - \frac{1}{2\sqrt{2}}y + \frac{5}{2\sqrt{2}}z \\ t_{2} = y + \frac{5}{3}z \\ t_{3} = z \end{array} \right.$$

So, the canonical form is

$$g(t_{1}, t_{2}, t_{3}) = t_{1}^{2} + \frac{15}{8}t_{2}^{2} - \frac{209}{72}t_{3}^{2}$$

The rank of the quadratic form in the canonical form is equal to the total number of square terms. The rank of $f(x, y, z) =$ the rank of $g(t_{1}, t_{2}, t_{3}) = 3$.

Answer: 3