I've been given a very confusing homework problem that is as follows:

Let U be the set of all vectors u in $ℝ^4$ such that $2(u_1) + 3(u_3) - 2(u_4) = 0$ (i.e. U is the solution space of a given system).

(a) Find the dimension of U.

(b) Find a basis of U.

(c) Write U using the basis.

So firstly I'm not sure what $2(u_1) + 3(u_3) - 2(u_4) = 0$ . Is this vector the solution space of all other vectors in U?

If the dimension of a vector space Dim(U)=n then the dimension should be 4, no? Furthermore a basis of U should be a linear combination of any vector in the space, so would a linear combination of the given vector [2 0 3 -2] be sufficient?

I'm finding only abstractions of this problem in the book, my notes, and online.

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  • $\begingroup$ As I understand it, $u:$={$(u_1, u_3,u_4): 2u_1 +3u_3-2u_4=0$} if all of these where in $\mathbb R^3$, you would have a plane through the origin. $\endgroup$ – user99680 Jun 23 '14 at 22:17
  • $\begingroup$ If you have four variables and one equation, how many degrees of freedom do you have? $\endgroup$ – rlartiga Jun 23 '14 at 22:32
  • $\begingroup$ If I have four variables and one of them is arbitrary, I should be have to account for three variables, correct? $\endgroup$ – Csteele5 Jun 23 '14 at 22:47
  • $\begingroup$ @Csteele5 yes! :) $\endgroup$ – rlartiga Jun 24 '14 at 13:27

$U = \{ u \in \mathbb{R}^4 | 2u_1 + 3u_3 - 2u_4 = 0 \}$. That is, the set of four dimensional vectors with real components such that the constraint is satisfied.

Since $\dim \mathbb{R}^4 = 4$ and there is one constraint, we intuitively expect that $\dim U = 3$.

Some guesswork/experimentation can find a basis.

It should be clear that $b_1 = (0,1,0,0) \in U$.

Suppose $u_1 = 1, u_2, = u_3 =0$, then if we choose $u_4 = 1$, we see that $b_2=(1,0,0,1) \in U$.

Now suppose $u_1=0, u_2 = 0, u_3 = 2$, then if we choose $u_4 = 3$, we have $b_3=(0,0,2,3) \in U$.

It is easy to check explicitly that if $u \in U$, then we can write $u$ as a combination of the $b_k$.

Explicitly, $u = u_1 b_2 + u_2 b_1 + {1 \over 2 } u_3 b_3$. (Unfortunate numbering of basis elements.)

It is also straightforward to check that the $b_k$ are linearly independent.

If $\sum \alpha_k b_k = (\alpha_2, \alpha_1, 2 \alpha_3, \alpha_2+3 \alpha_3) = 0$, then we must have $\alpha_1 = \alpha_2 = \alpha_3 = 0$.

Hence $\dim U = 3$.

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  • $\begingroup$ I see why (0,1,0,0) is ∈ U, but I don't understand your reasoning behind the (1,0,0,1) vector. $\endgroup$ – Csteele5 Jun 23 '14 at 22:57
  • $\begingroup$ Well, you can view the constraint as $u_4 = {1 \over 2 } (2 u_1 + 3 u_3)$. So, if you assign $u_1,u_3$ (and $u_2$, although it has no effect here) then you can compute a $u_4$ such that $u=(u_1,...,u_4) \in U$. I just picked $(u_1,u_3)$ from $(1,0), (0,1)$. $\endgroup$ – copper.hat Jun 24 '14 at 2:15

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