Find $a$ in equation so that it can define a scalar product I have to determine which values for $a$ where
$g(u,v) = u_1v_1 + 5u_2v_2 + au_3v_3+ 2(u_1v_2+u_2v_1)+3(u_1v_3+u_3v_1)+4(u_2v_3+u_3v_2)$
defines a scalar product in $\mathbb{R}^3$.
I need tips to begin. 
I have tried solving it by setting $g(u,u) \geq 0$
which gives 
$u_1^2+5u_2^2+au_3^2+4u_1u_2+6u_1u_3+8u_2u_3\geq 0$
But even in this case $a$ is not solvable.
 A: A wide hint: your $g(u,v)$ can be written as $u^TAv$ (or $\langle u, AV\rangle$) for a suitable matrix $A$.  Find that matrix $A$, and find its eigenvalues (which will be functions of $a$); by diagonalization, your condition for positive semidefiniteness of the product is that all those eigenvalues are $\geq 0$.
A: First, a brief digression. Let $$B=\begin{bmatrix}b_{11} & b_{12} & b_{13}\\b_{21} & b_{22} & b_{23}\\b_{31} & b_{32} & b_{33}\end{bmatrix},$$ and note that $$\begin{align}u^TBv &=[x_1\:x_2\:x_3]\left[\begin{array}{c}b_{11}y_1+b_{12}y_2+b_{13}y_3\\ b_{21}y_1+b_{22}y_2+b_{23}y_3\\ b_{31}y_1+b_{32}y_2+b_{33}y_3\end{array}\right]\\ &= (b_{11}y_1+b_{12}y_2+b_{13}y_3)x_1+(b_{21}y_1+b_{22}y_2+b_{23}y_3)x_2+(b_{31}y_1+b_{32}y_2+b_{33}y_3)x_3\\ &= \sum_{i=1}^3\sum_{j=1}^3b_{ij}x_iy_j\end{align}$$
This gives us a (potentially) more convenient (or at least compact) form in which to write $g.$ Namely, $g(u,v)=u^TBv,$ where $$B=\begin{bmatrix}1 & 2 & 3\\2 & 5 & 4\\3 & 4 & a\end{bmatrix}.$$
In order for $g$ to represent a scalar product, it is necessary and sufficient that the following be positive:


*

*$\det[1]$

*$\det\begin{bmatrix}1 & 2\\2 & 5\end{bmatrix}$

*$\det B$

