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By definition, $U + V = \{\mathbf{u} + \mathbf{v} : \mathbf{u} \in U\ \; \& \; \mathbf{v} \in V\}$.
Let $U = \{ \; u_1 = (1, 1, 0, \color{green}0), u_2 = (-3, 7, 2, \color{green}1) \;\}, V = \{ \; \mathbf{v} = (x_1,x_2,x_3,0) : x_i \in \mathbb{R} \; \}$.
Find $dim(U + V)$. (Source)

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My work: $U + V = \{\; \mathbf{u_1 + v} \; , \; \mathbf{u_2 + v} \; \} = \{\; (1 + x_1, 1 + x_2, x_3, \color{green}{0}) \; , \; (-3 + x_1,7 + x_2, 2 + x_3, \color{green}{1}) \; \}$.
Since the last entry of $\mathbf{u_1 + v} $ is $ \color{green}{0}$ and that of $\mathbf{u_2 + v} $ is $ \color{green}{1}$, thus $U + V$ is lin-indept.
Then how and why would you divine/previse $U + V = \mathbb{R^4}$? Then what's being done?

I also tried to determine the $\operatorname{span}(U + V)$ by solving for $a_1$ and $a_2$, in terms of the $b_i$s, in: $a_1(\mathbf{u_1 + v}) + a_2(\mathbf{u_2 + v}) = (b_1, b_2, b_3, b_4) \implies$ $\left[ \begin{array}{cc|c} 1 + x_1 & -3 + x_1 & b_1\\ 1 + x_2 & 7 + x_2 & b_2 \\ x_3 & 2 + x_3 & b_3 \\ 0 & 1 & b_4 \end{array} \right] $. Too involved?

This question precedes dimensions/theorems of the 4 subspaces, Orthogonality, Determinants, eigenvalues and eigenvectors, and linear transformations. Please omit them.


2nd Supplementary to Pete L. Clark's Answer :

$\Large{{2.1.}}$ Doesn't $U+V = \mathbb{R}^4$ straightaway imply: $\{\mathbf{e_i}\}_{1 \le i \le 4}$ is a basis for $U + V$?
What's the answer doing? Why bother to note that $\mathbf{e_4} = \cdots $ is a linear combination of $ \cdots $?

$\Large{{3.1.}}$ What's wrong with my work intended for finding a basis for $(U + V)$?

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  • $\begingroup$ Your definition of $U+V$ might be misleading, since it is the subspace spanned by $u_1,u_2$ and $V = \langle (x_1,x_2,x_3,0) : x_i \in \Bbb R \rangle$. So to prove that it has dimension $4$ you have to show that the standard basis of $\Bbb R^4$, $e_1,\dots,e_4$ is in $U+V$, which the author does. $\endgroup$ – BIS HD Nov 26 '13 at 14:42
  • $\begingroup$ By the way, this is not an instance of "direct sum": we say that $V+W$ is a direct sum when $V \cap W = \{0\}$. $\endgroup$ – Pete L. Clark Nov 27 '13 at 11:36
  • $\begingroup$ Thank you. I fixed the title. Please feel free to edit it. $\endgroup$ – NNOX Apps Dec 2 '13 at 8:30
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The subspace $V$ has dimension $3$ inside $\mathbb{R}^4$ (thus it is a hyperplane). Since $V \subseteq V+U \subseteq \mathbb{R}^4$, the first subspace has dimension $3$ and the last subspace has dimension $4$, we must have either that $\operatorname{dim} V+U = 3$ -- and thus $V+U = V$ -- or $\operatorname{dim} V+U = 4$ -- and thus $V+U = \mathbb{R}^4$. Moreover, $V+U = V$ holds if and only if $U \subset V$.

So it all boils down to whether $U$ is contained in $V$. This is immediate to check: does every vector in $U$ have last coordinate equal to $0$? Clearly not -- look at $u_2$ -- so $U+V = \mathbb{R}^4$.

Added: With regard to the "supplementary": 1) Previously I had a "$W"$ in my answer. That was a typo: all instances of it have been changed to $U$. 2) Yes, if $U+V = \mathbb{R}^4$ then $\{e_1,e_2,e_3,e_4\}$ is a basis for $U+V$. You asked how one would know that $U+V = \mathbb{R}^4$, so my answer gave a more detailed and motivated argument for that. Your work doesn't make much sense to me, since you start by writing that the set $U+V$ has just two elements $s_1$ and $s_2$. Clearly this is not the case. Nor is it clear what $s_1$ and $s_2$ mean, since you have unspecified variables $x_1$, $x_2$, $x_3$ in them.

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  • $\begingroup$ Thank you for your answer. What's $W$? Also, could you please respond to my supplementary in my OP and in your answer (not as a comment)? $\endgroup$ – NNOX Apps Dec 2 '13 at 8:45
  • $\begingroup$ Thank you very much again for your answer. +1. I've clarified the notation in my 2nd supplementary. Could you please respond to the new one in my OP in your answer? $\endgroup$ – NNOX Apps Dec 7 '13 at 8:11
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You know a basis of $V$: $\{e_1,e_2,e_3\}$; thus a spanning set for $U+V$ is $$ \{u_1,u_2,e_1,e_2,e_3\} $$ Perform Gaussian elimination on $\begin{bmatrix} 1&-3&1&0&0\\ 1&7&0&1&0\\ 0&2&0&0&1\\ 0&1&0&0&0 \end{bmatrix} $ and you'll find your basis: $RREF = \begin{bmatrix} 1 & ~ & ~ & 1 & ~ \\ ~ & 1 & ~ & ~ & ~ \\ ~ & ~ & 1 & -1 & ~ \\ ~ & ~ & ~ & ~ & 1 \\ \end{bmatrix}$

Of course, the specific case could be managed in a different way, since the subspace $U$ is clearly not contained in $V$, so by dimensionality reasons it must be $U+V=\mathbb{R}^4$. The method with Gaussian elimination works in any case: all you need is spanning sets for $U$ and $V$.

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  • $\begingroup$ Thanks. +1. Are the "dimensionality reasons" those explained by Pete L. Clark? If not, could you please dilate on them? $\endgroup$ – NNOX Apps Dec 7 '13 at 8:13
  • $\begingroup$ @LePressentiment Yes, they're the same: the vector $u_2$ doesn't belong to $V$ (it has a nonzero last coordinate) and $V$ has dimension $3$. $\endgroup$ – egreg Dec 7 '13 at 10:42

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