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I've been doing some work in Linear Algebra for my course at school. I just want to be clear about how to find the orthogonal complement of a subspace. The basis for the subspace, W, is shown below, composed of 3 vectors:$$W = \begin{Bmatrix}\begin{bmatrix}1\\2\\3\\4 \end{bmatrix} \begin{bmatrix}-3\\4\\2\\6\end{bmatrix} \begin{bmatrix}2\\-2\\3\\5\end{bmatrix}\end{Bmatrix}$$

I would like to know if one simply sets $W*W^T$$=0$ and takes the columns of the resulting matrix as the basis of the orthogonal complement of W, provided that row reduction has been performed to make sure the remaining columns are linearly independent.

Please note that I have created an arbitrary set of vectors above that are not orthogonal, and so if they need to be for $W*W^T$$=0$ please point that out. Thank you.

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Think of a system of linear equations you want to solve that will say your vector $\mathbf x$ is orthogonal to $W$.

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  • $\begingroup$ I think I now better understand. x being orthogonal to W, as I just reviewed, means the inner product of x with all vectors in a set spanning W must be zero. So doing a little rearranging, I found that the transpose of W times x must also equal zero. So is it incorrect for me to say W*W^T = 0 when Row W = perp(Nul W^T)? $\endgroup$ – equalsign Apr 24 '13 at 6:00
  • $\begingroup$ EDIT: In above comment, I meant to ask "Therefore is it incorrect for one to say W*W^T = 0 when Col W = perp(Nul W^T)?" (NOT Row W) $\endgroup$ – equalsign Apr 24 '13 at 6:13
  • $\begingroup$ Yes, $\text{Col}(A)^\perp = N(A^T)$. Your $W$ is a subspace, not a matrix, so what you wrote doesn't quite make sense. $\endgroup$ – Ted Shifrin Apr 24 '13 at 11:13
  • $\begingroup$ No, you forget that a matrix can be comprised of vectors, and the subspace above is a basis of vectors, which can, in turn, be considered a matrix. I'm afraid you haven't really answered my question, which is specifically leading to "how can one find a vector perpendicular to a set of vectors?" And no, I am not looking for a simple answer such as "come up with a vector with the same number of elements as the vectors in the set whose inner product with each of the set's vectors is zero". I'm afraid it has to be more elegant than that. $\endgroup$ – equalsign Apr 25 '13 at 4:31
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I'd rather put the matrix in this way, here I just give a very simple example, you can solve your matrix in the same way: $$W = \begin{bmatrix} \begin{matrix}1\\0\\0 \end{matrix} \begin{matrix}0\\1\\0\end{matrix} \begin{matrix}0\\0\\1\end{matrix} \begin{matrix}1\\1\\1\end{matrix}\end{bmatrix} $$

Then find the null space of W by solving $$Wx = 0$$

You will get the basis for the nullspace: $$v = \begin{bmatrix} \begin{matrix}-1\\-1\\-1\\1 \end{matrix} \end{bmatrix}$$

the null space is spared by v, $$null(W) = span(v)$$ you can easily find subspace spanned by W is orthogonal to subspace spanned by v, because every basis (each row of W) is orthogonal to v.

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