1
$\begingroup$

UNDERGRAD LINEAR ALGEBRA~

I'm really bad at this alternate coordinate stuff because I missed the last bit of classes and that's when we did change of bases. I know for a vector in standard basis $\bar{x}$ we can find a representation for the same thing in a base $B$, denoted $[\bar{x}]_B$ by finding $\bar{C}[\bar{x}]_b = \bar{x}$. So far practice problems have been simple manipulations of matrices, so I'm not sure how to approach this problem:

A conic in the xy coordinate system is $5x^2 -2\sqrt{3}xy+7y^2 = 16$. Find a rotation of coordinates that bring it to standard form. What is the matrix of this rotation? What is the cosine of the angle of the rotation? (this part I can probably do after getting started).

$\endgroup$
1
$\begingroup$

A general rotation matrix has the form $$R=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix},$$ while the equation of the conic can be written as $$\begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix}5&-\sqrt3\\-\sqrt3&7\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}=17.$$ Substiuting $(x,y)^T=R^{-1}(x',y')^T$ into the left-hand side of this equation and using $R^{-1}=R^T$ gives $$\begin{bmatrix}x'&y'\end{bmatrix} R\begin{bmatrix}5&-\sqrt3\\-\sqrt3&7\end{bmatrix} R^T \begin{bmatrix}x'\\y'\end{bmatrix} = \begin{bmatrix}x'&y'\end{bmatrix} \begin{bmatrix}6-\cos2\theta+\sqrt3\sin2\theta & -\sqrt3\cos2\theta-\sin2\theta \\ -\sqrt3\cos2\theta-\sin2\theta & 6+\cos2\theta-\sqrt3\sin2\theta\end{bmatrix} \begin{bmatrix}x'\\y'\end{bmatrix}.$$ We want the $x'y'$ term to vanish, so we must have $\sqrt3\cos2\theta+\sin2\theta=0.$ Solve for $\theta$. Note that the rows of the resulting rotation matrix will be eigenvectors of the symmetric matrix associated with the conic, but there’s no need to go through the usual eigenvector/eigenvalue computation to solve this problem.

$\endgroup$
1
$\begingroup$

One straightforward way of doing this is to find a rotation matrix, $\ U\ $, such that the matrix $$ D=U^\top\pmatrix{5&-\sqrt{3}\\-\sqrt{3}&7}U $$ is diagonal. The equation of your conic can be written as $$ \pmatrix{x&y} \pmatrix{5&-\sqrt{3}\\-\sqrt{3}&7} \pmatrix{x\\y}=16\ , $$ so if you define new coordinates $\ x', y'\ $ by $$ \pmatrix{x',y'}=U^{-1} \pmatrix{x\\y}\ , $$ —that is, with respect to axes subjected to the rotation determined by $\ U\ $—then the equation in the new coordinates is \begin{align} 16&= \pmatrix{x'&y'}U^\top \pmatrix{5&-\sqrt{3}\\-\sqrt{3}&7} U\pmatrix{x'\\y'}\\ &= \pmatrix{x'&y'}D \pmatrix{x'\\y'}\\ &= d_1x'^2+d_2y'^2\ , \end{align} where $\ d_1, d_2\ $ are the (diagonal) entries of $\ D\ $.

The columns of $\ U\ $ need to be chosen as normalised eigenvectors of the matrix $\ \pmatrix{5&-\sqrt{3}\\-\sqrt{3}&7}\ $, and $\ d_2, d_2\ $ will then be the corresponding eigenvalues.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.