Question

Reduce the following quadratic form to standard form and find its principal axes:

4x^2-4xy+y^2

Answer 1

$4x^2-4xy+y^2\\ \begin{pmatrix} 4 & -2 \\ -2 & 1 \end{pmatrix}\\ \mathrm{The\:eigenvalues\:of}\:A\:\mathrm{are\:the\:roots\:of\:the\:characteristic\:equation}\:\det \left(A-λ\:I\right)=0\\ λ=5,\:λ=0\\$

Therefore the orthogonal canonical reduction of the quadratic form is

$𝑄 = (𝑥′ 𝑦′) \begin{pmatrix} 0 & 0 \\ 0 & 5 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = 5y'^2$

Principal axes at $λ=0\\$

The reduced row echelon form of the matrix is $\begin{bmatrix} 1 & 2\\ 0 & 0 \end{bmatrix}$

To find the null space, solve the matrix equation$\begin{bmatrix} 1 & 2\\ 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}= \begin{bmatrix} 0\\ 0 \end{bmatrix}$

If we take $x_ 2 =t x_2=t,$ then $x_1​=−2tx_1​=−2t$ .

$Thus, x⃗ =\begin{bmatrix} -2t \\ t \end{bmatrix}= \begin{bmatrix} -2 \\ 1 \end{bmatrix} t$

$|x|= \sqrt{(-2)^2+1^2}=\sqrt5$

$\frac{1}{\sqrt5}\begin{pmatrix} -2 \\ 1 \end{pmatrix}$

Principal axes at $λ=5$

The reduced row echelon form of the matrix is $\begin{bmatrix} 1 & -0.5\\ 0 & 0 \end{bmatrix}$

To find the null space, solve the matrix equation$\begin{bmatrix} 1 & -0.5\\ 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}= \begin{bmatrix} 0\\ 0 \end{bmatrix}$

If we take $x_ 2 =t x_2=t,$ then $x_1​=−2tx_1​=−2t$ .

$Thus, x⃗ =\begin{bmatrix} -0.5t \\ t \end{bmatrix}= \begin{bmatrix} -0.5 \\ 1 \end{bmatrix} t$

$|x|= \sqrt{(-0.5)^2+1^2}=\frac{\sqrt5}{2}$

$\frac{2}{\sqrt5}\begin{pmatrix} 0.5 \\ 1 \end{pmatrix}$