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


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.


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 $W$.

  • $\begingroup$ 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 May 10 '16 at 13:46
  • $\begingroup$ Thank you! I guess I need to read a theory more carefully $\endgroup$ – krszyoscezio May 10 '16 at 13:47

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