Let be the follow bilinear form on $\mathbb{R}^4$ given by the matrix:

\begin{pmatrix}
 1 & 1 & 1 & 1 \\
 1 & 0 & 0 & 0 \\
 1 & 0 & 1 & 1 \\
 1 & 0 & 1 & 0
 \end{pmatrix}

Consider the subspace $F$, generated by $ w\_1 = (1, 1, 0, 1) $ and $ w\_2 = (0, −1, 1, 1)$. Prove that $F$ is non-singular and find a basis for $F^{\perp}$. Find the orthogonal projection of $ v = (1, 0, −1, 1) $ over $F$.

Is there an easier way to show that $F$ is non-singular? My attemp:

$F$ non-singular if $rad(F) = \{ {0\_{\mathbb{R}^4}} \} $.

We calculate $F^{\perp}$ = $\langle (-1/2,1,1/4,0),(0,0,-1/2,1) \rangle $

$ rad(F) = F \cap F^{\perp} = \{ {0\_{\mathbb{R}^4}} \} $

Then, I have several problems to find the projection of $v$ over $F$. I know that I need an orthogonal basis for $F$, let be $W=(w\_1, w\_2)$, and then we can do something like:

$ p^r\_F (v) = \frac{\langle v,w\_1 \rangle}{\langle w\_1,w\_1 \rangle } w\_1 + \frac{\langle v,w\_2 \rangle}{\langle w\_2,w\_2 \rangle } w\_2 $

But, well, I'm not sure at all about that last step... Any help for that questions?? Thanks!!