Question

Let a quadratic form has expression 6x2+5xy+10y2 with respect to the standard basis of R2. Find its expression with respect to the new basis { (3,1) , (-2,1) } of R2.

Answer 1

Answer on Question #40021, Math, Linear Algebra

Let a quadratic form has expression $6x^2 + 5xy + 10y^2$ with respect to the standard basis of R2. Find its expression with respect to the new basis $\{\{3,1\}, \{-2,1\}\}$ of R2.

Solution.

$$Q = 6x^2 + 5xy + 10y^2$$

Standard basis in $R^2$ is:

$$\{\binom{1}{0}, \binom{0}{1}\}$$

New basis:

$$\{\binom{3}{1}, \binom{-2}{1}\}$$

So to write $Q$ with respect to new basis we should go to another terms:

$$x = 3x' + y'$$ $$y = -2x' + y'$$

Therefore,

$$\begin{aligned} Q' &= 6(3x' + y')^2 + 5(3x' + y')(-2x' + y') + 10(-2x' + y')^2 = \\ &= 6(9x'^2 + 6x'y' + y'^2) + 5(-6x'^2 + x'y' + y'^2) + 10(4x'^2 - 4x'y' + y'^2) = \\ &= 54x'^2 + 36x'y' + 6y'^2 - 30x'^2 + 5x'y' + 5y'^2 + 40x'^2 - 40x'y' + 10y'^2 = \\ &= 64x'^2 + x'y' + 21y'^2 = Q' \end{aligned}$$

So $Q'$ is a quadratic form in new basis $\{\binom{3}{1}, \binom{-2}{1}\}$.

Answer: $Q' = 64x'^2 + x'y' + 21y'^2$