Answer to Question #154594 in Linear Algebra for Ashweta Padhan

Question

Let q(x₁, x₂)= 3x₁² - 6x₁x₂ +11x₂². Then find an orthonormal change of coordinate that diagonalizes the above quadratic form q

Answer 1

 q(x₁, x₂)= 3x₁² - 6x₁x₂ +11x₂²

matrix of the quadratic equation is $A=\begin{pmatrix} 3 & -3 \\ -3 & 11 \end{pmatrix}$

first we find the orthonormal diagonalization

$det(A -\lambda\Iota) = 0$

$\Iota = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\lambda\Iota = \begin{bmatrix} \lambda\ & 0 \\ 0 & \lambda \end{bmatrix}$

$\because A -\lambda\Iota = A=\begin{bmatrix} 3 & -3 \\ -3 & 11 \end{bmatrix} - \begin{bmatrix} \lambda\ & 0 \\ 0 & \lambda \end{bmatrix} = \begin{bmatrix} 3-\lambda\ & -3 \\ -3 & 11-\lambda \end{bmatrix}$

$det(A -\lambda\Iota) = [(3-\lambda)(11-\lambda)] - 9 =0$

$33 -3\lambda-11\lambda + \lambda^2 -9= 0$

$\lambda^2 - 14\lambda +24= 0$

$\lambda = 2 , \lambda =12$ (eigen values)

when $\lambda=2: N(A-(2)\Iota)$

$\begin{bmatrix} 1 & -3 \\ -3 & 9 \end{bmatrix} \to$ $\begin{bmatrix} 1 & -3 \\ 0 & 0 \end{bmatrix}$ eigen space : span $\begin{bmatrix} 3 \\ 1 \end{bmatrix}$

we normalize the vectors to get orthonormal vectors ${1 \over \sqrt{1 + 9}} \begin{bmatrix} 3 \\ 1 \end{bmatrix}$

$\begin{bmatrix} {3 \over \sqrt{10 }} \\ {1 \over \sqrt{10}} \end{bmatrix}$

when $\lambda=12: N(A-(12)\Iota)$

$\begin{bmatrix} -9 & -3 \\ -3 & -1 \end{bmatrix} \to$ $\begin{bmatrix} -9 & -3 \\ 0 & 0 \end{bmatrix}$ eigen space : span$\begin{bmatrix} 1 \\ -3 \end{bmatrix}$

we normalize the vectors to get orthonormal vectors ${1 \over \sqrt{1 + 9}} \begin{bmatrix} 1 \\ -3 \end{bmatrix}$

$\begin{bmatrix} {1 \over \sqrt{10 }} \\ {-3 \over \sqrt{10}} \end{bmatrix}$

orthonormal diagonalization:

D= $\begin{bmatrix} 2 & 0 \\ 0 & 12 \end{bmatrix},P= \begin{bmatrix} {3 \over \sqrt{10 }} & {1 \over \sqrt{10 }} \\ {1 \over \sqrt{10 }} & -{3 \over \sqrt{10 }} \end{bmatrix}$

Now we make a change of coordinates: Before it was q(y₁,y₂)= P^-1x = P^Tx Now it will be q(x₁,x₂)= P^T(y₁,y₂)

(x₁,x₂) = $\begin{bmatrix} {3 \over \sqrt{10 }} & {1 \over \sqrt{10 }} \\ {1 \over \sqrt{10 }} & -{3 \over \sqrt{10 }} \end{bmatrix}$ (y₁,y₂)