Changing co-ordinate systems quadratic forms A = $$\begin{bmatrix}1 & 4 \\\4 & -4\end{bmatrix}$$ and $$q(\mathbf{x})= \mathbf{x}\cdot A \mathbf{x}, $$ then what is? $$q(\mathbf{e_1})= $$ $$q(\mathbf{e_2})=$$ $$$$ $$$$ $$$$ Im really struggling this this. I have ascertained that $$q(\mathbf{x}) = x^2+8xy-4y^2$$ Also the Eigenvalue $$λ_1=\frac{\sqrt{89}-3}{2}$$ $$λ_2=\frac{-\sqrt{89}-3}{2}$$ Eigenvectors $$\begin{bmatrix}\frac{\sqrt{89}+5}{8}\\\1\end{bmatrix}$$ $$\begin{bmatrix}\frac{-\sqrt{89}+5}{8}\\\1\end{bmatrix}$$

Why don't you just use the definition? $$ q(e_1)=e_1\cdot Ae_1=e_1\cdot\begin{pmatrix}1\\\4\end{pmatrix}=1 $$ and $$ q(e_2)=e_2\cdot Ae_2=e_2\cdot\begin{pmatrix}4\\\\-4\end{pmatrix}=-4. $$ You don't need the eigenvalues and eigenvectors of $A$ to compute the value of the quadratic form at some points.

Otherwise you computed $$ q(x,y)=x^2+8xy-4y^2. $$ You can also use this to compute $$ q(e_1)=q(1,0)=1^2+8\cdot1\cdot0-4\cdot 0^2=1 $$ and $$ q(e_2)=q(0,1)=0^2+8\cdot 0\cdot 1-4\cdot 1^2=-4. $$