Prove that $F$ is non-singular and find the orthogonal projection of $ v$

Question

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!!

Answer 1

Let be $W_F = (w_1,w_2)$ an orthogonal basis on $F$. For instance, $ w_1 = (1,1,0,1) $ $ w_2 = (-2,7,5,3) $.

Then,

$ M ( \langle \ ,\ \rangle , W_F ) = \begin{pmatrix} 5 & 0 \\ 0 & 55 \\ \end{pmatrix} $

So we finally find,

$ p^r_F (v) = \frac{2}{5} w_1 + \frac{-4}{55} w_2 = (\frac{6}{11},\frac{10}{11},\frac{-4}{11},\frac{2}{11}) $

Answer 2

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$$ P^T H P = D $$

$$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ - 1 & 1 & 0 & 0 \\ 0 & - 1 & 1 & 0 \\ 0 & 0 & - 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - 1 & 0 & 0 \\ 0 & 1 & - 1 & 0 \\ 0 & 0 & 1 & - 1 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & - 1 \\ \end{array} \right) $$ $$ $$

$$ Q^T D Q = H $$

$$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & - 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 0 \\ \end{array} \right) $$

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