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Let $E$ be a subspace of $\mathbb{R}^4$ that is composed of vectors that support the following system :

$$\begin{cases} x - 2y + z + t = 0 \\2x+y-2z-t = 0 \end{cases}$$

So, I've used Gaussian elimination so I've found the following basis for $E$: $$ B = \{(3,4,5,0), (1,3,0,5)\}. $$

Well now i defined $W= \{ w \in W \mid \langle v, w \rangle = 0 \}$ . What do I have to do next? I used the following system :

$$ \begin{cases}3x_1 + 4x_2 + 5x_3 = 0 \\ 1x_1+3x_2+5x_4 = 0 \end{cases} $$

by using the inner product, where $x_1,x_2,x_3,x_4$ are the elements of an arbitrary vector in $W$. Is this approach correct? If i now find the basis $W$ from this system will i find the orthogonal complement basis of $E$?

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Your approach is correct, but you probably mean inner product instead of cross product?

There's also a much easier way to see this. Note that if you take $$ v_1 = \left( \begin{matrix} 1 \\ -2 \\ 1 \\ 1\end{matrix} \right), v_2 = \left( \begin{matrix} 2 \\ 1 \\ -2 \\ -1\end{matrix} \right), $$ ( the components of the vectors are simply the coefficients in the system of equations determining $E$) we have $w \in E$ if and only if $\langle w, v_1 \rangle =0$ and $\langle w, v_2 \rangle = 0$. So in fact $E = \text{span}(v_1,v_2)^\perp$. But then the orthogonal complement is easily found to be $E^\perp = \text{span}(v_1,v_2)$. So you don't actually have to calculate or do any Gaussian elimination at all.

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  • $\begingroup$ Thank you for your answer! $\endgroup$ – Eduard6421 Jun 14 '17 at 18:51
  • $\begingroup$ @Eduard6421 You're welcome! $\endgroup$ – Demophilus Jun 14 '17 at 19:02

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