I'm trying to prove that a given subset of a given vector space is an affine subspace.

Now I'm having some trouble with the definition of an affine subspace and I'm not sure whether I have a firm intuitive understanding of the concept. I have the following definition:

"A subset A of a vector space V is called affine subspace, if there exists a vector $v \in V$ and a subspace $U_A \subseteq V$ such that $$A=v + U_A= \{v+u : u \in U_A\}$$

So as far as I understand the definition, an affine subspace is simply a set of points that is created by shifting the subspace $U_A$ by $ v \in V$, i.e. by adding one vector of V to each element of $U_A$. Is this correct?

Now I have two example questions:

1) Let V be the vector space of all linear maps $ f:\mathbb{R}$ -> $\mathbb{R}$. Addition and scalar multiplication on V is defined by: $f(x+y) = f(x) + f(y)$, $f(\lambda x)= \lambda f(x)$. Let $x_0$ and $y_0$ be real numbers. Prove that the subset $V_{y_0} = \{f \in V : f(x_0) = y_0 \}$ is an affine subspace of V.

2) Let $L \in \mathbb{R}^n$ be the set of all n-tuple that solve a non-homogenous system of m linear equations with n unknowns. Prove that if such a system is solvable, then L is an affine subspace of $\mathbb{R}^n$.

Well in the first case I would have to find one linear map that maps from $\mathbb{R}$ to $\mathbb{R}$ and a subspace of V such that the set of all linear maps that yield $y_0 $ when $x_0$ is used. I'm not sure about the selection of $v$ and $U_A$...

For the second question I don't have any clue how to approach this really. I would guess that if a system is solvable there is either one solution or infinitely many. In the first case I guess I could somehow add the null space to the solution vector and in the second case I could add $\mathbb{R}^3 $ as $U_A$ to any vector?!

I would really appreciate any kind of help.

  • $\begingroup$ Ah yes I mixed the variable names up. I'll edit this. Could you please help me with the question? $\endgroup$ – eager2learn Dec 15 '13 at 15:30

You're right: You want one particular element $v\in V$, and then the subspace $U$ by which to translate it.

In your first example, assuming $x_0,y_0\ne 0$, you want a particular linear map $f_0$ with $f_0(x_0)=y_0$, say $f_0(x)=\dfrac{y_0}{x_0}x$, and then you want to find the subspace of all maps $f$ with $f(x_0)=0$. Since you're considering the case of linear maps $\Bbb R\to\Bbb R$, this subspace consists only of the zero map. It would be more interesting if, for example, you were considering the vector space $V$ of all linear maps $\Bbb R^n\to\Bbb R$. I'll let you work out the details that now we can take $$f_0(\mathbf x) = \dfrac{y_0}{\|\mathbf x_0\|^2}(\mathbf x\cdot\mathbf x_0) \qquad\text{and}\qquad U = \{f\in V: f(\mathbf x_0)=0\}\,.$$ If you choose a basis for $\Bbb R^n$ consisting of $\{\mathbf x_0,\mathbf v_2,\dots,\mathbf v_n\}$, then $U$ has as a basis the linear maps $f_i$, $i=2,\dots,n$, defined by $f_i(\mathbf v_i) = 1$ and $f_i(\mathbf x_0) = f_i(\mathbf v_j) = 0$ for all $j\ne i$.

Similarly, in your second example, for an $m\times n$ matrix $A$, the set $L$ of solutions of $A\mathbf x = \mathbf b$ is found by taking a particular solution $\mathbf x_0$ (you said the system had a solution) and adding to it the subspace of all solutions of the homogeneous system $A\mathbf x =\mathbf 0$.

If you've taken a differential equations course, you see the same story there with homogeneous and inhomogeneous linear differential equations.

  • $\begingroup$ Thanks so much, I really appreciate your answer. I have one question about the answer to the second example. I don't really understand why can obtain all solutions to $Ax=b$ by taking a particular solution and then adding the solutions of the homogenous system to it. Could you explain that part please? $\endgroup$ – eager2learn Dec 15 '13 at 16:41
  • $\begingroup$ Have you learned how to solve systems of linear equations by row reduction (Gaussian elimination)? $\endgroup$ – Ted Shifrin Dec 15 '13 at 16:49
  • $\begingroup$ Yes I did, but I can' think of anything that would help me answer this question. $\endgroup$ – eager2learn Dec 15 '13 at 16:58
  • $\begingroup$ Well, your solution, in vector form, is the sum of a vector of constants (this is a particular solution) and a linear combination of the vectors, one for each free variable, that are solutions of the homogeneous system. $\endgroup$ – Ted Shifrin Dec 15 '13 at 17:29

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