given $\varphi((x_1, x_2)$,$(y_1, y_2)$) = $x_1y_1 − x_2y_2$ where $\varphi:\mathbb{R}^2 \times \mathbb{R}^2 \rightarrow \mathbb{R}$ how can i find the subspaces $W$ such that $W^\perp = W$ ?
Finding subspaces orthogonal to itself
linear algebraorthogonalityvector-spaces
Related Solutions
That basis is an orthnormal basis, and therefore the coordinates of $(u,v)$ with respect to that basis are $\bigl\langle(u,v),a\bigr\rangle(=ux_1+vx_2)$ and $\bigl\langle(u,v),b\bigr\rangle(=uy_1+vy_2)$. That is$$(u,v)=(ux_1+vx_2)a+(uy_1+vy_2)b.$$
Once you have found one requirement that fails, you are done. You do not need to check every requirement. A single failed requirement is enough (although $T(ax)=aT(x)$ might have been faster to check).
Also note that you don't need to do it in any kind of generality. You have used $(x_1,y_1)$ and $(x_2,y_2)$ here, but you could just as well have used $(1,0)$ and $(0,1)$. If a requirement that should hold for all vector pairs fails for even a single vector pair, it has still failed.
In fact, doing it generally leads you into a trap. You have shown that $$ T((x_1,y_1)+(x_2,y_2)) = T(x_1,y_1) + (0, x_1y_2+x_2y_1+x_2y_2)\\ =T(x_1,y_1)+T(x_2,y_2)+(0,x_1y_2+x_2y_1) $$ (note that I took the liberty of finishing the calculations you had; that last line is what we are really after, as it has $T(x_1,y_1)+T(x_2,y_2)$ in it). However, we are not done.
It is theoretically possible that that extraneous term $(0,x_1y_2+x_2y_1)$ that shouldn't be there is actually always $0$. In this case it turns out not to be, obviously, but there are plenty of obscure ways to write $0$. Suddenly you will find yourself with a transformation that truly is linear, but where you get a monstrous extra term that turns out to always be $0$. Or similarly for any other property you would want to check in the future.
At any rate, you must check that this extra term not only looks like it could be non-zero, but actually is non-zero, at least occasionally. Checking this by using concrete numbers to start with is the easiest in my opinion, although you may have bad luck when picking your numbers.
Best Answer
Hint: Find a non-zero vector $(a,b)$ such that $\varphi \bigl((a,b),(a,b)\bigr)=0$ and take $W=\{\lambda(a,b)\mid\lambda\in\mathbb R\}$.