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My prof made an offhand remark about this in class and didn't go into detail. I've took a bit of stats before and to me these two statements seem identical, I don't see how they're different.

I also found this question; is this what my prof is referring to:

Set Addition vs. Set Union

If it is, I'm not sure how does Minkowski addition fits into linear algebra.

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    $\begingroup$ $U\cup W$ is not a vector space, in general. $\endgroup$ – Thomas Andrews Oct 5 '17 at 1:12
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    $\begingroup$ Consider $U$ the $x$-axis in $\Bbb R^2$ and $W$ the $y$-axis in $\Bbb R^2$. Then $U\cup W$ are the axes and only the axes, i.e. have at least one coordinate zero. $U+W$ however is the entirety of $\Bbb R^2$. $\endgroup$ – JMoravitz Oct 5 '17 at 1:15
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The definitions couldn't be more different : Let $U,W$ be vector spaces.

$U \cup W = \{x \in U \textrm{ or } x \in W\}$.

$U + W = \{x + y : x \in U, y \in W\}$.

An vector is in $U \cup W$ if and only if it is either in $U$ or in $W$ or in both.

A vector is in $U + W$ if and only if it is the sum of two (not necessarily distinct or unique) vectors from $U$ and $W$ respectively.

To give you an example, let's consider the vector subspaces of $\mathbb R^2$, given by $U = \{(x,y) : x = 2y\}$, and $W = \{(x,y) : x = 3y\}$. You can clearly see that these are vector spaces : they are closed under addition and scalar multiplication.

What is an element of $U \cup W$? As sets in $\mathbb R^2$, $U$ and $W$ are just lines passing through the origin. So as a set, $U \cup W$ is just two lines passing through the origin. From this image, we conclude that it cannot be a subspace either.

But what is $U + W$? This is the set of all possible sums of elements from $U$ and $W$. That is, if $(2a,a) \in U$ and $(3b,b) \in W$, then an element of $U+W$ is $(2a+3b,a+b)$.

We claim that $U+W = \mathbb R^2$. To see this, let $(x,y) \in \mathbb R^2$. We want $(x,y) = (2a+3b,a+b)$ for some $a,b$. It's easy to see that if we choose $a = 3y-x, b = x-2y$, then infact this holds true. In other words, if we take the point $(2a,a) \in U$ and $(3b,b) \in V$, then their sum is $(x,y)$. Since $x,y$ were arbitrary, we conclude that $U + W = \mathbb R^2$.

Hence, clearly $U \cup W$ and $U + W$ are different sets. In fact, more often than not they will be very different from each other. For one, $U \cup W$ is not a subspace most of the time,while $U + W$ is always a subspace.

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    $\begingroup$ It's worth noting that they are not completely unrelated sets: certainly $U \cup W \subseteq U + W$, and in fact $U+W$ is the subspace generated by $U \cup W$. $\endgroup$ – Bungo Oct 5 '17 at 1:19
  • $\begingroup$ Yes, that is correct. Thank you for the comment. $\endgroup$ – астон вілла олоф мэллбэрг Oct 5 '17 at 1:20
  • $\begingroup$ @Bungo: I'm not seeing how U + W can be generate from U∪W $\endgroup$ – L to the V Oct 5 '17 at 1:27
  • $\begingroup$ @LtotheV Every element of $U+W$ is of the form $u + w$ where $u \in U$ and $w \in W$. As $U \subseteq U \cup W$ and $W \subseteq U \cup W$, this means that an arbitrary element of $U+W$ is the sum of two elements contained in $U\cup W$. $\endgroup$ – Bungo Oct 5 '17 at 1:31
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For example, let $$ U=\{(x,0): \ x\in\mathbb{R}\}, W=\{(0,y): \ y\in\mathbb{R}\}. $$ Then $$ U\cup W=\{(x,y): \ x=0\text{ or }y=0\}, U+W=\{(x,y): \ x,y\in\mathbb{R}\}=\mathbb{R}^2. $$ You see the difference.

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