Given any vector k $\epsilon$ $R^{3}$ consider k= $\sum_{j=1}^{3}$ $c_{j}u_{j}$ where $u_{1}$,$u_{2}$,$u_{3}$ are the orthonormal basis vectors (I don't know how to make them bold sorry about that, but they are vectors). This means in the basis B the vector k looks like ($c_{1}$,$c_{2}$,$c_{3}$). Rewrite the expression (Hk)$\cdot$ k= . $k^{T}$ Hk in terms of the basis vectors B
We are given that H= \begin{bmatrix} -3&1 &2 \\ 1&-3 &2 \\ 2& 2 & 0\end{bmatrix}
and B={$u_{1}$,$u_{2},$$u_{3}$} (u are vectors)
I have calculated $u_{1}$ $u_{2}$ $u_{3}$ but I'm not sure if this question needs them. I know this is quite a straightforward question of computation, but I have no idea how to rewrite the vector k in terms of B so I know k=$c_{1}$$u_{1}$+$c_{2}$$u_{2}$+$c_{3}$$u_{3}$. Does it mean that k=($c_{1}$,$c_{2}$,$c_{3}$)B? And how should I deal with the dot product?
In the basis $B=\{u_1, u_2, u_3\}$ we have $$ k=\left[ \begin{array}{c} c_1\\c_2\\c_3 \end{array}\right].$$ Hence $$ Hk=\left[ \begin{array}{rrr} -3 & 1 & 2\\ 1 & -3 & 2\\2 & 2 & 0 \end{array} \right] \left[ \begin{array}{c} c_1\\c_2\\c_3 \end{array}\right]=\left[ \begin{array}{c} -3c_1+c_2+2c_3\\c_1-3c_2+2c_3\\2c_1+2c_2 \end{array}\right] $$ and consequently $$ (Hk)\cdot k=k^{T}Hk=(-3c_1+c_2+2c_3)c_1+(c_1-3c_2+2c_3)c_2+(2c_1+2c_2)c_3=$$ $$=-3c_{1}^{2}-3c_{2}^{2}+2c_1c_2+4c_1c_3+4c_2c_3.$$