"Q$(\left[ {\begin{array}{c} x \\ y \\ \end{array} } \right] )$ $ = x^2 - xy + y^2$
State the symmetric matrix that belongs to Q.
"Let $B = \{ \frac 1 {\sqrt{2}}\left[ {\begin{array}{c} 1 \\ 1 \\ \end{array} } \right], \frac 1 {\sqrt{2}}\left[ {\begin{array}{c} -1 \\ 1 \\ \end{array} } \right] \}$ State the matrix that belongs to Q with respect to the basis B. "
I know the answer to question 1 which is: $ M= \left[ {\begin{array}{cc} 1 & -\frac 1{2} \\ -\frac 1{2} & 1 \\ \end{array} } \right] $
However, I'm completely blank when it comes to question 2. Where do I start?
Suppose you have a vector represented in the standard basis $[x,y]^T$. In your new basis, this same vector is $$\frac{1}{\sqrt{2}}\left[\begin{matrix}1 & -1 \\ 1 & 1\end{matrix}\right]^T \left[\begin{matrix} x \\ y \end{matrix}\right]$$ So you want $$\left[\begin{matrix} x \\ y \end{matrix}\right]^TQ_{\text{standard basis}}\left[\begin{matrix} x \\ y \end{matrix}\right]=\left(\frac{1}{\sqrt{2}}\left[\begin{matrix}1 & -1 \\ 1 & 1\end{matrix}\right]^T \left[\begin{matrix} x \\ y \end{matrix}\right]\right)^T Q_{\text{new basis}} \frac{1}{\sqrt{2}}\left[\begin{matrix}1 & -1 \\ 1 & 1\end{matrix}\right]^T \left[\begin{matrix} x \\ y \end{matrix}\right]$$ $$\left[\begin{matrix} x \\ y \end{matrix}\right]^TQ_{\text{standard basis}}\left[\begin{matrix} x \\ y \end{matrix}\right]=\frac{1}{2} \left[\begin{matrix} x \\ y \end{matrix}\right]^T\left[\begin{matrix}1 & -1 \\ 1 & 1\end{matrix}\right] Q_{\text{new basis}} \left[\begin{matrix}1 & -1 \\ 1 & 1\end{matrix}\right]^T \left[\begin{matrix} x \\ y \end{matrix}\right]$$ $$Q_{\text{standard basis}}=\frac{1}{2} \left[\begin{matrix}1 & -1 \\ 1 & 1\end{matrix}\right] Q_{\text{new basis}} \left[\begin{matrix}1 & -1 \\ 1 & 1\end{matrix}\right]^T $$ $$2 \left[\begin{matrix}1 & -1 \\ 1 & 1\end{matrix}\right]^{-1} Q_{\text{standard basis}} \left[\begin{matrix}1 & -1 \\ 1 & 1\end{matrix}\right]^{-T}= Q_{\text{new basis}} $$