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Find a basis $B$ for the orthogonal complement $V^{\bot}$ of the space V. What is the dimension of $V^{\bot}$?

Let $V$ be the subspace of all vectors in $R^6$, such that

$$x_1+x_2=x_3+x_4=x_5+x_6$$

I'm solving it this way: $$x_1+x_2=x_5+x_6$$ and $$x_3+x_4=x_5+x_6$$

Thus $$x_1+x_2-x_5-x_6=0$$ and $$x_3+x_4-x_5-x_6=0$$

So the matrix $A$ is:

$$\begin{bmatrix} 1&1&0&0&-1&-1\\ 0&0&1&1&-1&-1\\ \end{bmatrix}$$

So the rank of the matrix is $2$.

And the basis for $NulA=V$ are $\begin{bmatrix} -1\\ 1\\ 0\\ 0\\ 0\\ 0\\ \end{bmatrix}$, $\begin{bmatrix} 0\\ 0\\ -1\\ 1\\ 0\\ 0\\ \end{bmatrix}$, $\begin{bmatrix} 1\\ 0\\ 1\\ 0\\ 1\\ 0\\ \end{bmatrix}$, $\begin{bmatrix} 1\\ 0\\ 1\\ 0\\ 0\\ 1\\ \end{bmatrix}$.

I don't understand how to solve further. Any explanations about orthogonality would be appreciated (don't really understand this topic). Thank you in advance.

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1 Answer 1

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You almost have the answer to your question. You have shown that $V$ is the subspace of vectors which are orthogonal to $v_1:=\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \\ -1 \\ -1 \end{bmatrix}$ and $v_2:=\begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \\ -1 \\ -1 \end{bmatrix}$. Let $W$ be the subspace spanned by $v_1$ and $v_2$. You have $V = W^\perp$ so $W = (W^\perp)^\perp = V^\perp$. But $(v_1,v_2)$ clearly is a basis of $W$.

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  • $\begingroup$ As a remark, the fundamental result of orthogonality in finite dimension is the Gram-Schmidt process. A consequence is that if $E$ is an Euclidian space of dimension $n$ and $W$ is a subspace then $\dim W^\perp = n - \dim W$. Now you immediately have $W \subseteq (W^\perp)^\perp$ and, since those spaces have the same dimension, you get the equality $W = (W^\perp)^\perp$. $\endgroup$
    – BrL
    Commented May 10, 2016 at 13:46
  • $\begingroup$ Thank you! I guess I need to read a theory more carefully $\endgroup$ Commented May 10, 2016 at 13:47

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