1
$\begingroup$

Given a function $$q(x,y)=2x^2-2xy +2y^2$$.

Find the minimum point of the following function by first converting it to a matrix form and using the diagonalisation of the matrix to find its minimum point. (I know its easy just by using the usual method. But the point is that i want to learn how to use linear algebra using this question as an example.)

What i did was to express the function in matrix form $A$. $$A=\begin{pmatrix}2&-1\\-1&2 \end{pmatrix}.$$

Then using the diagonalisation formula in linear algebra where $$D=P^TAP$$ I find the eigenvector of A which is

$$P=\begin{pmatrix}1&-1\\1&1 \end{pmatrix}.$$

and do a matrix multiplication to find D. Im stuck at this From this point onwards. I know that after finding D, $$q(x,y)=x^TAx=x^TPDP^Tx=(P^Tx)^TDP^Tx=x'^TDx'^T$$ And i need to change it to a rotated coordinate system x'y' but how exactly to go about doing it from here. I know that after the rotation the expression becomes $$x^2+3y^2$$. But how to get there? Could anyone explain

$\endgroup$
6
  • $\begingroup$ Your matrix $P$ is not invertible. $\endgroup$
    – Lolman
    Aug 28, 2014 at 13:35
  • $\begingroup$ Sorry typo. I just edited $\endgroup$
    – ys wong
    Aug 28, 2014 at 13:37
  • $\begingroup$ You want the eigenvalues. If they are both positive, then the form is never negative. But evidently it does take on the value zero, so that's the minimum. A more interesting question is, what is the minimum value of the form when restricted to the unit circle, and where is that minimum attained? $\endgroup$ Aug 28, 2014 at 13:39
  • $\begingroup$ Still not a rotation... The determinant of $P$ isn't equal to one. And on the last line you got both $x'$ trasposed. So you should get $$x^2+3y^2$$ I think it is clear where is the minimum. $\endgroup$
    – Lolman
    Aug 28, 2014 at 13:42
  • $\begingroup$ Actually im more interested in knowing how to express q(x,y) in another coordinate using the formula $$q(x,y)=x^TAx=x^TPDP^Tx=(P^Tx)^TDP^Tx=x'^TDx'^T$$ $\endgroup$
    – ys wong
    Aug 28, 2014 at 13:44

1 Answer 1

1
$\begingroup$

First you define as you do a new variable:

$$x'=P^T x$$

In the new coordinate system you have:

$$q(x')=x'^TDx'=\begin{pmatrix}x' & y'\end{pmatrix}\begin{pmatrix}d_1&0\\0&d_2 \end{pmatrix}\begin{pmatrix}x'\\y' \end{pmatrix}=d_1x'^2+d_2y'^2$$

If $d_1$ and $d_2$ are positive the minimum is when $(x' y')=(0,0)$ therefore you have to solve when $P^Tx=0$

$\endgroup$
3
  • $\begingroup$ How do u evulate $d_1$ and $d_2$ $\endgroup$
    – ys wong
    Aug 28, 2014 at 13:57
  • $\begingroup$ Are the elements of the diagonal $\endgroup$
    – rlartiga
    Aug 28, 2014 at 14:05
  • $\begingroup$ $d_1$ and $d_2$ are the eigenvalues. $\endgroup$ Aug 28, 2014 at 22:59

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.