Graph the curve on the $2D$ plane defined by $3x^2+3y^2-2xy-2=0$
Let $q(X)=3x^2+3y^2-2xy , X=(x,y)^T$. The matrix associate with q is $A=\begin{bmatrix}3&-1\\-1&3 \end{bmatrix}$. Since its symmetric,$\exists P$ orthogonal such that $P^TAP=Λ$.
By finding A's eigenvalues, we see that this matrix is similar to $Λ =\begin{bmatrix}2&0\\0&4 \end{bmatrix} $ which is the associate matrix of $q'(Z) = 2z_1^2+4z_2^2$, where $Z=PX$. That gives me the equation $2z_1^2+4z_2^2=2 \implies z_1^2+2z_2^2=1$, which is an ellipse. The initial equation describes a rotated elliptic cylinder, while the ellipse is not rotated.
My question is, shouldn't the graph of the ellipse be a projection of the graph of the initial equation in the $2D$ plane?
Your $z_1$-$z_2$ coordinate system is rotated relative to the original one. That’s kind of the point of diagonalizing $A$: to shift into a coordinate system in which the principal axes of the quadric—the eigenspaces of $A$—are the coordinate axes.