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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

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  • $\begingroup$ 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? $\endgroup$
    – Gerry Myerson
    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

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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 $P^Tx=0$

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  • $\begingroup$ 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. $\endgroup$
    – Gerry Myerson
    Aug 28, 2014 at 22:59

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