# Orthogonal complement

1. Mar 17, 2014

### AwesomeTrains

1. The problem statement, all variables and given/known data
$G:=$
\begin{pmatrix}
1 & 0 & 0 & 1 \\
0 & 1 & 1 & 0 \\
0 & 1 & 1 & 0\\
1 & 0 & 0 & 1 \\
\end{pmatrix}

$B(x,y) = x^{T}Gy$

$B: \textit{R}^{4} X \textit{R}^{4} \rightarrow \textit{R}$

Find $(\textit{R}^{4})^{\bot}$

2. Relevant equations

$(\textit{R}^{4})^{\bot} = \left\{x | B(x,y)=0,\forall y\in\textit{R}^{4} \right\}$

3. The attempt at a solution
I think I have to solve a linear equation system but I don't know how to set it up.
I could only find examples where a subspace of the vektor space was given.
Any hints are much appreciated.

2. Mar 17, 2014

### Staff: Mentor

This is what I would do. Start by writing out what B(x, y) means.

$$B(x, y) = \begin{bmatrix} x_1 & x_2 & x_3 & x_4\end{bmatrix} \begin{bmatrix}1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} y_1 \\ y_2 \\ y_3 \\ y_4\end{bmatrix}$$

You want all vectors x in R4 such that B(x, y) = 0 for any y in R4.

3. Mar 17, 2014

### AwesomeTrains

Yes, but wouldn't that give me one equation with 8 unknowns.
$y_{1}(x_{1}+x_{4})+y_{2}(x_{2}+x_{3})+y_{3}(x_{2}+x_{3})+y_{4}(x_{1}+x_{4}) = 0$

4. Mar 17, 2014

### Staff: Mentor

But the expression on the left side has to be identically zero for any choices of the y values, so what conditions does that place on the four x values?

5. Mar 17, 2014

### AwesomeTrains

$x_{1}+x_{4}=0$ and $x_{2}+x_{3}=0$
Is that what you mean?

6. Mar 17, 2014

### Staff: Mentor

Yes.

7. Mar 18, 2014

### AwesomeTrains

Then $x_{1} = -x_{4} = -b$, $x_{2} := a$, $x_{3} = -x_{2} = -a$ and $x_{4} := b$
$\quad (R^{4})^{\bot}=a\begin{pmatrix} 0 \\ 1 \\ -1 \\ 0 \\ \end{pmatrix}+ b\begin{pmatrix} -1 \\ 0 \\ 0 \\ 1 \\ \end{pmatrix}$
Have I got it right?

8. Mar 18, 2014

Looks good.