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I am trying to learn about vector spaces, and I came across this question: If I have an equation which represents the kernel of a linear form $c$: $$2x_1+4x_2+2x_3+3x_4+6x_5+x_6=0$$

How do I find a basis for that kernel? I know I need to find vectors of $\mathbb{R^{6 \times 1}}$ for which this equation holds, but I do not know any consistend ways of finding a basis for it. I know how to do this the other way around, however. If I have a basis of a vector space, then I know how to find the basis of the annihilator space, or how to find a set of equations that every vector of my subspace fulfills.

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    Solve the equation for $x_1$ and let the other variables be free. You'll be able to generate a basis. – 2017-01-23
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    You are basically asking how to find the solutions to a linear equation. This is probably chapter 1 in your linear algebra course. I find it curious that when people learn about vector spaces and some more terminology, they forget how to solve basic equations. Notice that you have 6 variables and one (independent) equation, hence there are 5 free variables that can be used to find 5 independent solutions. – 2017-01-23
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    @Mathematician42 you are totally right, sometimes you just study so much that you forget the simplest things and especially when you are stressed it can be quite frustrating which was the case for me, I was just overthinking it way too much! – 2017-01-23

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Finding a basis of kernel of linear operator is equivalent to solving appropriate linear system. In your case, you are solving linear system $$2x_1+4x_2+2x_3+3x_4+6x_5+x_6 = 0$$

From the general theory we know that this subspace will be $5$-dimensional ($6-1=5$). Perhaps it will look more familiar to you if we write full matrix

$$\begin{pmatrix} 2 & 4 &2 &3 &6 &1\\ 0& 0 &0 &0 &0 &0\\ 0& 0 &0 &0 &0 &0\\ 0& 0 &0 &0 &0 &0\\ 0& 0 &0 &0 &0 &0\\ 0& 0 &0 &0 &0 &0\\ \end{pmatrix} \sim \begin{pmatrix} 1 & 2 &1 &3/2 &3 &1/2\\ 0& 0 &0 &0 &0 &0\\ 0& 0 &0 &0 &0 &0\\ 0& 0 &0 &0 &0 &0\\ 0& 0 &0 &0 &0 &0\\ 0& 0 &0 &0 &0 &0\\ \end{pmatrix}$$

and thus the basis will be $$\{(-2,1,0,0,0,0), (-1,0,1,0,0,0),(-3/2,0,0,1,0,0),(-3,0,0,0,1,0),(-1/2,0,0,0,0,1)\}$$

You can easily check that these are solutions of the system and that they are linearly independent.

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Your equation is satisfied by all vectors $(x_{1}, x_{2}, \ldots, x_{6})$ that are orthogonal to the vector $$ \hat{n} = (2, 4, 2, 3, 6, 1). $$ So, start with $\hat{n}$ and complete it to an orthogonal basis (e.g., using the Gram-Schmidt process). Then drop $\hat{n}$ from that basis, and what remains is the desired basis for the kernel.

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    Are you saying this to annoy the OP? Surely it works, but it is certainly not an efficient way. – 2017-01-23
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    @Mathematician42, No, I am not. Why don't you let the OP decide what annoys them and, if you have something more "efficient", why not present it as an answer? – 2017-01-23
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    The other answer does exactly what everyone learned as a standard (and more efficient) method, no need to repeat it. – 2017-01-23
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    @Mathematician42, the other answer gives no indication as to the most important, geometric, aspect of the problem: how would the student themselves have come up with it? It's great for helping someone in a jam to submit a homework assignment, but teaches little. Which of these is the main goal of this forum? – 2017-01-23
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    So you want to teach students to solve homogeneous systems using Gram-Schmidt? And what if the system is not homogeneous? When you have algebraic-geometric duality, you use it to improve understanding, not obfuscate a simple algorithm everyone knows from elementary school in its basic form. – 2017-01-23
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    @Ennar, yes, I criticized your answer (and, indirectly, OP's use of this website). In my answer, I clearly stated that the goal was finding an orthogonal complementary basis, and that Gram-Schmidt was only one possible way of doing it. (I chose it because it gives an unambiguous computational procedure.) I really don't know what you mean by "algebraic-geometric duality" or what you refer to as "a simple algorithm". If it is an algorithm, would you be able to give generic pseudocode? Besides, how do you know what "everyone knows from elementary school"? This website is international. – 2017-01-24
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    I really don't see how the concept of a "kernel" of a linear functional can be detached from orthogonality in a Euclidean (or Hilbert) space. Neither did the author of the https://en.wikipedia.org/wiki/Riesz_representation_theorem. Nothing obfuscates linear algebra more than approaching it with coordinates and matrices instead of as the **geometry of linear spaces**, a la Halmos. – 2017-01-24
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    The algorithm behind Ennar's answer is very easy. Take **any** system of linear equations. Put it in matrix form for simplicity. Row reduce the system. Once you have the row reduced form, declare variables which do not correspond to leading ones in the row reduced form as being free. Put all free variables equal to zero except for one and solve the system, to obtain a solution, repeat the last step for all free variables to obtain a (by definition) linearly independent set of solutions. Simple, elegant and efficient. – 2017-01-24
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    I don't get why you are trying to argue. The question is about solving the easiest type of equation. I'm fairly sure people were able to do it long before the Riesz-Fréchet theorem even existed. Also, the Gauss-algorithm also has a geometric interpretation, the easiest way to see this is two take two lines in $\mathbb{R}^2$ and calculate the intersection using row reductions. In each step draw the lines corresponding to the system of equations and notice that you are actually rotating the lines about the intersection point. This procedure much resembles what the Gramm-schmidt procedure does. – 2017-01-24
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    @avs if you are trying to insinuate that [Gaussian elimination](https://en.wikipedia.org/wiki/Gaussian_elimination) is somehow ambiguous, perhaps you should revise your linear algebra. Do you really solve linear systems using Gram-Schmidt? Why don't you demonstrate in your answer just how much more clear it is. – 2017-01-24