# Finding minimum point of a function using linear algebra

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

• Your matrix $P$ is not invertible. Aug 28, 2014 at 13:35
• Sorry typo. I just edited Aug 28, 2014 at 13:37
• 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? Aug 28, 2014 at 13:39
• 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. Aug 28, 2014 at 13:42
• Actually im more interested in knowing how to express q(x,y) in another coordinate using the formula $$q(x,y)=x^TAx=x^TPDP^Tx=(P^Tx)^TDP^Tx=x'^TDx'^T$$ Aug 28, 2014 at 13:44

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 $P^Tx=0$

• How do u evulate $d_1$ and $d_2$ Aug 28, 2014 at 13:57
• Are the elements of the diagonal Aug 28, 2014 at 14:05
• $d_1$ and $d_2$ are the eigenvalues. Aug 28, 2014 at 22:59