Find values of $\lambda$ for which quadratic form is defined positively Given quadratic form is:
$$2x_1^2+8x_2^2+x_3^2+2\lambda x_1x_2-2x_1x_3+2x_2x_3$$
$\textbf{What I was thinking about}:$
1.) As far as I can understand we have to use Sylvester's criterion to determine its sign.
2.) To check its sign we have to write down the matrix of сoefficients taken from  the quadric form
3.) Check if leading principles minors are positive or negative, it shows us the sign of the quadratic form.

$\textbf{Questions:}$
What should be in matrix of coefficients? (I mean I do not know what  to write as a row and what as a column)?
How do I find the values of  $\lambda$ if I know that quadratic form is defined positively?
 A: The matrix is$$\begin{pmatrix}2&\lambda&-1\\\lambda&8&1\\-1&1&1\end{pmatrix}.$$By Sylvester's criterion, the quadratic form is positive definite if and anly if$$2>0\text{, }\det\begin{pmatrix}2&\lambda\\\lambda&8\end{pmatrix}>0\text{, and}\det\begin{pmatrix}2&\lambda&-1\\\lambda&8&1\\-1&1&1\end{pmatrix}>0.$$The first condition is automatic, the second one is equivalent to $16-\lambda^2>0$ and the third one is equivalent to $6-2\lambda-\lambda^2>0$. These conditions hold if and only if $-1-\sqrt7<\lambda<-1+\sqrt7$.
A: $$2x_1^2+8x_2^2+x_3^2+2\lambda x_1x_2-2x_1x_3+2x_2x_3=\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix}^T\color{blue}{\begin{pmatrix}2 & \lambda &-1\\\lambda & 8 & 1\\-1 & 1 & 1\end{pmatrix}}\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix}$$
and the characteristic polynomial of the blue matrix is 
$$ \color{blue}{p(x)}= x^3-11x^2+(24-\lambda^2)x+(\lambda^2+2\lambda-6).$$
We know that this polynomial always has real roots since the spectrum of a symmetric matrix belongs to $\mathbb{R}$ by the spectral theorem. The negativity of $p(0)$ is required to ensure the positive-definiteness of the given quadratic form, and $p(0)$ is negative iff $\color{blue}{-\sqrt{7}-1<\lambda<\sqrt{7}-1}$. In such a case $16-\lambda^2>0$ holds and Sylvester's criterion ensures that the blue matrix is positive definite.
A: Putting it simply, coefficient of $x_ix_j$ is $a_{ij}$. But since $x_ix_j = x_jx_i$, coefficients of mixed terms go divided by $2$, as in $$\begin{pmatrix} 2 & \lambda & -1 \\ \lambda & 8 & 1 \\ -1 & 1 & 1\end{pmatrix}$$From here on your guess is correct, it is a direct application of Sylvester's Criterion. Can you proceed?
A: Let $x_1=a$, $x_2=b$ and $x_3=c$.
Hence, we need 
$$c^2-2(a-b)c+2a^2+8b^2+2\lambda ab\geq0$$
for all reals $a$, $b$ and $c$, which says
$$(a-b)^2-(2a^2+8b^2+2\lambda ab)\leq0$$ or
$$a^2+2(\lambda+1)ab+7b^2\geq0,$$ which gives
$$(\lambda+1)^2-7\leq0$$ or
$$-\sqrt7-1\leq\lambda\leq\sqrt7-1.$$
Done!
I think that our form defined positively and not  strictly defined positively, 
which says that it should be $\geq0$ and not $>$. 
