Find the ON-base

Question

I can't figure out the b) question. I know that I am supposed to use the formula $ x^tAx$ but what does $ x^tAx$ mean? How can I figure out what the martix A is?

Question

Answer 1

Here $x$ is a 2-dimensional vector: $x = \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix}$. So $$x^\mathsf{T}Ax = \begin{bmatrix} x_1 & x_2 \\ \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} =a_{11}x_1^2 + (a_{12} + a_{21})x_1x_2 + a_{22}x_2. $$ That's definition of a quadratic form. In practice, we write $A$ as a symmetric matrix, so that calculations are easier. $$ ax_1^2 + bx_1x_2 + c x_2^2 \quad \rightsquigarrow \quad A = \begin{bmatrix} a & \frac b 2 \\ \frac b 2 & c \\ \end{bmatrix} $$ In that form, we can diagonalize $A$ with orthonormal change-of-basis matrix. That is for any symmetric matrix $A = A^\mathsf{T}$ there exist diagonal $J$ and invertable $P$ such that $A = P J P^\mathsf{T}$, and $P^{-1} = P^\mathsf{T}$. We want to find that $P$, because it will give us the diagonal form: $$ x^\mathsf{T}A x = x^\mathsf{T} P J P^\mathsf{T} x = (P^\mathsf{T}x)^\mathsf{T} J (P^\mathsf{T}x) = (P^{-1}x)^\mathsf{T} J (P^{-1}x) = u^\mathsf{T} J u. $$ Here $u$ is vector written in the new basis.

In our case we have $2x_1^2 + 2x_1x_2 + 2x_2$, so $a = 2, \frac b 2 = 1, c = 2$. We diagonalize $\begin{bmatrix} 2 & 1 \\ 1 & 2 \\ \end{bmatrix} $, and pick eigenvectors with norm 1. You should get someting similar to $$ \begin{bmatrix} 2 & 1 \\ 1 & 2 \\ \end{bmatrix} = \begin{bmatrix} -\frac 1 {\sqrt 2} & \frac 1 {\sqrt 2} \\ \frac 1 {\sqrt 2} & \frac 1 {\sqrt 2} \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 3 \\ \end{bmatrix} \begin{bmatrix} -\frac 1 {\sqrt 2} & \frac 1 {\sqrt 2} \\ \frac 1 {\sqrt 2} & \frac 1 {\sqrt 2} \\ \end{bmatrix}. $$ Now take $u = P^T x$ and get $u_1 = \frac{-x_1 + x_2}{\sqrt 2}, u_2 = \frac{x_1 + x_2}{\sqrt 2}$. Our quadratic form is now $u_1^2 + 3 u_2^2$. You can plug in definition of $u$ to see that it is correct.