# Thread: Orthogonal Change of Variables

1. ## Orthogonal Change of Variables

Find an orthogonal change of variables X=PY such that:

–2x1^2 + 2x1x2 - 2x2^2

takes the form:

Ay1^2 + By2^2

Sorry if this may be confusing. The numbers next to the variables should be subscripts. For instance x1^2 should be x subscript 1 to the power of 2.

2. Write the quadratic form as $\displaystyle -2x_1^2+2x_1x_2-2x_2^2 = \begin{bmatrix}x_1&x_2\end{bmatrix} \begin{bmatrix}-2&1\\1&-2\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix}$. Then find the eigenvalues and eigenvectors of the matrix and diagonalise it.

3. ## confused about the same type of problem

my problem is find an orthogonal change in variables x=py such that
4x1^2 + 10x1x2 + 4x2^2
takes the form ?x1^2 + ?x2^2
p=?
the question marks are blanks for answers, there should be subscripts and ^2 is squared

so i think i understand how to find p... you diagonalize and find the eigenvectors, but i dont know how to answer the x1^2 + x2^2 part.

4. Originally Posted by luccasaurus
my problem is find an orthogonal change in variables x=py such that
4x1^2 + 10x1x2 + 4x2^2
takes the form ?x1^2 + ?x2^2
p=?
the question marks are blanks for answers, there should be subscripts and ^2 is squared

so i think i understand how to find p... you diagonalize and find the eigenvectors, but i dont know how to answer the x1^2 + x2^2 part.
Step 1: Write the quadratic form as $\displaystyle \mathbf{x}^{\textsc{t}}A\mathbf{x}$, where $\displaystyle \mathbf{x} = \begin{bmatrix}x_1\\x_2\end{bmatrix}$ and $\displaystyle A = \begin{bmatrix}4&5\\5&4\end{bmatrix}$.

Step 2: Diagonalise A, to get $\displaystyle A = P^{\textsc{t}}DP$, where D is the diagonal matrix $\displaystyle \begin{bmatrix}-1&0\\0&9\end{bmatrix}$ (-1 and 9 being the eigenvalues of A), and P is the orthogonal matrix whose columns are the corresponding normalised eigenvectors.

Step 3: Then the quadratic form is $\displaystyle \mathbf{x}^{\textsc{t}}P^{\textsc{t}}DP\mathbf{x} = \mathbf{y}^{\textsc{t}}D\mathbf{y} = -y_1^2 + 9y_2^2$, where $\displaystyle \mathbf{y} = \begin{bmatrix}y_1\\y_2\end{bmatrix} = P\mathbf{x}$.

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