What is the intersection of the following two subspaces and why?

$U \cap W$ with $U=\langle 2-e^{3x},4x+e^{3x}\rangle$ and $W=\langle 1,e^x, e^{3x}\rangle$. I thought too easily: $e^x$ can't be in the union, because nothing in $U$ can be written as $\lambda e^x$ and $x$ can't be in the union because it won't fit in $W$.

  • $\begingroup$ Do you really mean union or do you mean sum, that is, the smallest subspace containing $U$ and $W$? $\endgroup$ – lhf Jan 20 '12 at 13:01
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    $\begingroup$ union? union of sub-spaces is not a sub-space unless one is contained in the other. Do you mean sum? Or intersection? $\endgroup$ – Dennis Gulko Jan 20 '12 at 13:01
  • $\begingroup$ Sorry for that! I meant intersection ;). $\endgroup$ – www.data-blogger.com Jan 20 '12 at 13:08
  • $\begingroup$ Please use \langle and \rangle for the angle brackets in TeX. The symbols < and > not only look different, they also give different spacing because they are relation symbols. $\endgroup$ – Harald Hanche-Olsen Jan 20 '12 at 13:24
  • $\begingroup$ Thanks, you are right. Looks now more it looks like in my book :). $\endgroup$ – www.data-blogger.com Jan 21 '12 at 8:04

I would not call that union, but instead an intersection. Let's do it from the very definition : the intersection is made of the element of W which are also in U, i.e. those elements of the form $a + b e^x + c e^{3x}$ ($a,b,c\in\mathbb R$) such that there exists $u$ and $v$ (also real numbers) with $$a + b e^x + c e^{3x} = u(2-e^{3x}) + v (4x + e^{3x}) \qquad \forall x \in \mathbb R$$ Re-arrange this: $$a + b e^x + c e^{3x} = 2 u + 4 v x + (v - u) e^{3x} \qquad \forall x \in \mathbb R$$ For this to be satisfied, you need (because the functions which map $x$ to $1$, $x$, $e^{x}$ and $e^{3x}$ are all linearly independent): $$ \begin{cases} a = 2 u\\ b = 0\\ 4 v = 0\\ c = v - u \end{cases} $$ which is equivalent to $$ \begin{cases} a = - 2 c\\ b = 0\\ v = 0\\ u = -c \end{cases} $$ You thus need $(a,b,c) = (-2C, 0, C)$ for some real number $C$, and then indeed there exists $(u,v)$ (namely $(-C, 0)$) such that the system is satisfied. This means that the elements of the intersection are all the multiples of $-2 + e^{3x}$.


One possible way: since $U\not\subseteq W$ and $dim U=2$, $dimW=3$, you know that $dim(U\cap W)$ is either $0$ or $1$ (why?). You can see that $2-e^{3x}$ is an element of both $U$ and $W$ (why?), hence is an element of $U\cap W$. This implies that $U\cap W=\langle2-e^{3x}\rangle$ (how?).


Ah, found a proof myself :), is this correct?

Take $x \in U \cap W$. Since $x \in U$ it holds that $x = \lambda (2-e^{3x})+ \mu (4x + e^3x)$. $x \in W \Rightarrow x = \alpha \cdot 1 + \beta e^x + \gamma e^{3x}$. For any $\gamma, \mu, \alpha, \beta, \gamma$. Since $x = x$ we can subtract those two expressions which gives: $0=1 \cdot (2 \lambda - \alpha) + x(4 \mu) - e^x \cdot \beta + e^{3x} (\mu -\lambda - \gamma)$ and since $1, x, e^x$ and $e^{3x}$ are linear independent (I will not proof that), it follows that $\beta = 0, \mu = 0, 2 \lambda = \alpha, -\lambda=\gamma$. Now choose $\epsilon = \lambda$, then $\gamma = -\epsilon$ and $\alpha = 2 \epsilon$. Filling this in for $x$ gives $x = \epsilon (2 - e^{3x})$ and in the other equation: $x = 2\epsilon - \epsilon e^{3x} = \epsilon(2-e^{3x})$. And since $\lambda$ is a free variable, so is $\epsilon$, so $x \in U \cap W \Leftrightarrow x \in <2 - e^{3x}>$.


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