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I am trying to understand how to find the orthogonal complement of a subspace $M$ of a vector space $V$.

From my understanding, $M^\perp$ is also a subspace of $V$ where all its vectors are perpendicular (orthogonal) to the columns of $M$, which would mean that the dot product of those vectors with each column of $M$ is $0$.

However, how do I find the $M^\perp$ subspace, when $M = \{ \vec{x} \in \mathbb{R}: 2x_2 + 3x_1 = 0\}$? In general, if there's a general approach, how does one attempt to find $M^\perp$ of an arbitrary $M$?

From my understanding, in this case, $M$ is the set of vectors in a plane that satisfy that equation, where $x_1$ and $x_2$ should respectively be the first and second components of the vector $\vec{x}$.

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For a finite dimensional vector space equipped with the standard dot product it's easy to find the orthogonal complement of the span of a given set of vectors:

Create a matrix with the given vectors as row vectors an then compute the kernel of that matrix.

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Orthogonal complement is defined as subspace $M^\perp = \{ v\in V\,|\, \langle v, m\rangle = 0,\forall m\in M\}$. This is really a subspace because of linearity of scalar product in the first argument. Also, it is easy to see that $M = (M^\perp)^\perp$ and that $M\dotplus M^\perp = V$ (in finite dimensional case).

So, in your example you have $$M = \{ (x,y)\in \mathbb R^2\,|\, 3x + 2y = 0\} = \{ (x,y)\in \mathbb R^2\,|\, \langle(x,y),(3,2)\rangle = 0\}$$

i.e. $M$ is orthogonal complement of subspace spanned by $(3,2)$. But, by $M = (M^\perp)^\perp$, we then have that $M^\perp = \operatorname{span}\{(3,2)\}$.

It would be overkill in this example, but in general if you can find basis for $M$, you can always find $M^\perp$. Let's say that $M=\operatorname{span}\{e_1,\ldots,e_m\}$. Then $$\langle x,e_i\rangle= 0,\quad i=1,\ldots,m$$ is a homogeneous system of $m$ linear equations which you can solve and solution subspace will be orthogonal complement.

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    $\begingroup$ Why exactly do you say that $M^{\perp} = span \{ (3, 2) \}$? $\langle (3, 2), (3, 2) \rangle \neq 0$... $\endgroup$ – nbro Feb 23 '17 at 22:28
  • $\begingroup$ @nbro, $3x + 2y = 0$ explicitly states that every vector in $M$ is orthogonal to $(3,2)$ and that all vectors orthogonal to $(3,2)$ are in $M$, i.e. $M = \operatorname{span}\{(3,2)\}^{\perp}$. $\endgroup$ – Ennar Feb 24 '17 at 5:35
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In the plane, it's easy. There's only one line through the origin that's perpendicular to any other given line.

So if the first line is generated by $(a,b)$, you can take the second (perpendicular) line to be generated by $(b,-a)$ since $(a,b)\cdot(b,-a) = ab-ba=0$.

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