3 number of a function

[nhrock3]

3)
$$q:R^{3}->R$$ is defined by $$q(x,y,z)=4(xy+yz+zx)$$
find the minimal $$M\in R$$ so $$q(x,y,z)\leq M(x^{2}+y^{2}+z^{2})$$
?
why in the solution the calculate the caracteristic polinomial
?
why if (t+2) is in power 2 then we have -2 in two members of
the formula q(v) ??
our polinomial doesn't separated into different lenear member
so in order to find its jordan form we need to find the minimal polinomial
etc..
but in the solution they said it straight forward why??
why did they coose in the end to put eigen vaule iside?
why its minimal?
we could put a vector which not is 0.5
?
is it true that the as the eigen vectors would diagonolise A
so is their orthonormal basis whould show Q as in sum of squares
correct?

 

[susskind_leon]

You want your bilinear form $B(x,y,z)\equiv M(x^2+y^2+z^2)-q(x,y,z)$ to be positive definite. This is equivalent to all eigenvalues being positive. Therefore you need the characteristic polynomial. So you actually need to compute
$$P(\lambda)=|B-\lambda id_3|$$
B is given by
$$\begin{pmatrix} M & -2 & -2\\ -2 & M & -2 \\ -2 & -2& M \end{pmatrix}$$
So, you compute the characteristic polynomial of
$$\begin{pmatrix} M-\lambda & -2 & -2\\ -2 & M-\lambda & -2 \\ -2 & -2& M-\lambda \end{pmatrix}$$
Now, let's call $t\equiv M-\lambda$
Then you get the characteristic polynomial
$$(t-4)(t+2)^2$$
which has solutions $t=4,t=-2$, which means
$$\lambda=M-4, \lambda=M+2$$
Now, you go figure out for which M all eigenvalues are positive ;)

 

[Mark44]

nhrock,
The image in your first post (at http://i42.tinypic.com/2a8hlpg.png) was way too large, and had a huge amount of whitespace at the bottom. Please edit the image using Paint or another image editing tool so that it is no larger than about 1200 x 800 pixels.