Finding a minimal scalar $m \in R$ such that $q(x,y,z) \leq m(x^2 + y^2 + z^2), \forall x,y,z \in R$ for $q$ quadratic form Given a quadratic form: 
$$
q(x,y,z) = 2zx + 4yz -2xy
$$
$q$ is given in the standard base of $V = R^3$
Find a minimal scalar $m \in R$ such that: 
$$
q(x,y,z) \leq m(x^2 + y^2 + z^2), \forall x,y,z \in R
$$
The idea seems to be orthogonal diagonalization of $q$ and looking at the diagonalizing matrix with coordinate vector with the norm as in the standard base and $m$ is the max eigenvalue of him... and so on but i dont realy understand whats going on in this way of solution. m is the eigenvalue of the vector of the coordinate???
Can someone make it more clear? 
Thanks. 
 A: The problem is equivalent to find the minimum saclar $m\in \mathbb R$ such that
$$
f_m(x,y,z) = m(x^2+y^2+z^2)-2xz-4yz+2xy \geq 0\quad \forall x,y,z\in \mathbb R
$$
Using linear algebra, the idea is to find the canonical form of the quadratic form $f_m$ in function of $m$. Since there is not the degree zero term in $f_m$ we have only some possibility for the canonical form:
\begin{gather}
X^2\\
X^2+Y^2\\
X^2+Y^2+Z^2\\
X^2-Y^2\\
X^2+Y^2-Z^2
\end{gather}
up to signs. This means that, after a linear change of coordinates, your quadratic form appears as one of the form written above. It is known that the quadratic form is completely determined by the eingenvalues of the corresponding matrix of the quadratic form $f_m$:
$$
\mathcal M (f_m)=\left[
\begin{matrix}
m & 1 & -1\\
1 & m & -2\\
-1&-2 & m
\end{matrix}
\right]
$$
Hence, in order to have $f_m\geq 0$, it's necessary and sufficient that all the eingenvalues of this matrix are more or equal than zero.
After some computations, we have that the characteristic polynomial of $\mathcal M(f_m)$ is:
$$
p_m(t) = (t-(m-2))(t-(m+1+\sqrt 3))(t-(m+1-\sqrt 3))
$$
and the eingenvalues are:
$$
\lambda_1 = m-2,\qquad \lambda_2 = m+1+\sqrt 3,\qquad \lambda_3 = m+1-\sqrt 3
$$
Imposing now all the eingenvalues more or equal than $0$ we obtain $m\geq 2$. Since we want the minimum value of $m$ we have to take $m=2$ that is the solution.

In order to convince yourself that all we did is correct, let's take $m=2$. We can observe that:
$$
f_2(x,y,z) = \frac{1}{2}\left(2x+y-z\right)^2 + \frac{3}{2}\left(y-z\right)^2
$$
so $f_2\geq 0$ for all $x,y,z \in \mathbb R$. Hence $m=2$ is a good value for the property. 
Moreover, $m=2$ is the minimum value: take $n=m-\varepsilon$ with $\varepsilon >0$. We have:
$$
f_{n}(x,y,z)=f_{2-\varepsilon}(x,y,z)=f_2(x,y,z) - \varepsilon(x^2+y^2+z^2)
$$
If we evaluate $f_n$ in $(0,1,1)$, we have:
$$
f_n(0,1,1)=f_2(0,1,1) - \varepsilon(0+1+1) = -2\varepsilon < 0
$$
Hence $f_n$ does not sotisfy the property and $m=2$ is the minimum.
