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I started reviewing linear algebra, from a different textbook (Axler's), after taking a fast paced summer class. Unfortunately, I've become confused with a concept that is introduced at the end of chapter one. That is, sum of subspaces.

Axler's text defines the sum of subspaces as follows.

Let $U_1,U_2,...,U_m$ be subspaces of a vectorspace $V$. Then we say $U_1+U_2+...+U_m=\{u_1+...+u_m:u_1\in U_1,...,u_m\in U_m\}$

I thought I understood this concept, but I'm afraid I don't because I am having trouble answering the following assertions he asks us to verify.

First that if we let $U_1,U_2,...,U_m$ be subspaces of a vectorspace $V$, then the sum of those subspaces is a subspace of $V$.

Also, this is what really had me tricked to thinking I understood it. Let $U=\{(x,0,0)\in \mathbb R^3: x\in \mathbb R\}$ and $W=\{(0,y,0)\in \mathbb R^3:y\in \mathbb R\}$, then $U+W= \{(x,y,0):x,y\in \mathbb R\}$. So this example made me think it was pretty straight forward and that I understood it, but in the next few lines he says let $Z= \{(y,y,0)\in \mathbb R^3:y\in\mathbb R\}$. Then $U+W=U+Z$ (which I am asked to verify).

Could someone please help me understand the definition and the verifications?

EDIT: I currently see the definition to say that when we take a collection of sets that are subspaces the sum of the sets is a set which consists of the sum of all their elements. However when I say that it seems to me the summed set consists of just one elements (the total sum of all the elements).

Thank You

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    $\begingroup$ What do you understand the definition to mean currently? It's sort of hard to help you understand it without knowing what you currently think of it. $\endgroup$
    – qaphla
    Aug 7, 2014 at 16:55
  • $\begingroup$ Okay give me a moment I will edit it to be more clear. My apologies. $\endgroup$
    – Valentino
    Aug 7, 2014 at 16:57
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    $\begingroup$ No need to apologize -- everyone is new to this sometime. :) $\endgroup$
    – qaphla
    Aug 7, 2014 at 16:57
  • $\begingroup$ It would be clearer if written: $U=\{(u_1,0,0): u_1\in \mathbb R\}$, $W=\{(0,w_2,0):w_2\in \mathbb R\}$, $S=U+W= \{(s_1,s_2,0):s_1,s_2\in \mathbb R\}$ $\endgroup$
    – mins
    Jan 16 at 16:24

4 Answers 4

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A point that might confuse you is that the same letters are used, let us rewrite this.

  1. If you have some element from $U+W$ then it is of the form $(x,y,0)$ for some $x,y$.

  2. If you have some element from $U+Z$ then it is of the form $(v+w,w,0)$ for some $v,w$.

Now the claim is that these two really yield the same collection of elements. I assume you can see that each element of the latter form is of the former. More specifically if we pick a generic element $ (v+w, w,0) $ in $U+Z$, we can equivalently express the element in the form $ (x,y,0) $ by setting $ x = v + w$ and $ y = w $. This shows that $ U + Z \subset U + W$.

For the other direction we can select an arbitrary element $(x, y, 0) $ of $ U + W $ and re-express it in the form $(v+w,w,0)$ by setting $ v = x - y $ and $ w = y $. The latter shows $ U + W \subset U + Z$.

The two subset relations $U+Z \subset U+W $ and $ U + W \subset U + Z$ imply that $U+Z = U+W$.


Added following the edit: a single element of the sum $U+W$ is one element of $U$ plus one element of $W$. And $U+W$ is the set of all these elements together.

So, you have since $(19,0,0)$ in $U$ and $(0,-3,0)$ in $W$ that $(19,0,0)+(0,-3,0)$ in $U+W$. However, you can of course evaluate that sum $(19,0,0)+(0,-3,0)= (19,-3, 0)$.

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    $\begingroup$ I followed your numerical example and you were definitely correct in saying that the same letters being used somewhat seemed to confuse me. Also, I believe I have a much stronger confidence in the definition and what is being asked of me to verify but I am having trouble following the solutions people have provided here. Thank you very much for the help. $\endgroup$
    – Valentino
    Aug 8, 2014 at 2:29
  • $\begingroup$ Reading over it again just now I feel that I understand what you are saying up until you until after the word "former," everything after that seems unclear to me. $\endgroup$
    – Valentino
    Aug 8, 2014 at 2:48
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First, you can show that the sum of subspaces is also a subspace directly from definition. For example if $$ u_{1}+...+u_{m}\in U_{1}+...+U_{m} $$

where $u_{i}\in U_{i}$ and $$ u_{1}'+...+u_{m}'\in U_{1}+...+U_{m} $$

where $u_{i}'\in U_{i}$ then $$ u_{1}+...+u_{m}+u_{1}'+...+u_{m}'=(u_{1}+u_{1}')+...+(u_{n}+u_{n}') $$

and since $U_{i}$ is a subspace it is closed under addition and so $u_{i}+u_{i}'\in U_{i}$ and thus the above sum is in $U_{1}+...+U_{m}$.

Regarding your second example: $U+W=\{(x,y,0)|\, x,y\in\mathbb{R}\}$ is as straightforward as it seems, the part that looks a bit strange is to find $U+Z$ and lets work by definition to understand what it is:

$$ U=\{(x,0,0)|\, x\in\mathbb{R}\} $$ $$ Z=\{(y,y,0)|\, y\in\mathbb{R}\} $$

If $$ u+z=(a,b,c) $$

then clearly $c=0$. We claim that if $(x_{0},y_{0})\in\mathbb{R}^{2}$ then there are some $u\in U,z\in Z$ s.t $a=x_{0},b=y_{0}$- Indeed $u+z$ is of the form $$ (x,0,0)+(y,y,0)=(x+y,y,0)=(x_{0},y_{0},0) $$

then we have the solution $$ y=y_{0} $$ $$ x=x_{0}-y_{0} $$

and if you wish to verify $$ (x_{0}-y_{0},0,0)+(y_{0},y_{0},0)=(x_{0},y_{0},0) $$

So $$ \{(x,y,0)|x,y\in\mathbb{R}\}\subseteq U+Z $$

and the other containment is clear and so the two sets are in fact equal.

I hope this makes things clear, please comment if not

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The sum of $U_1$ and $U_2$ contains anything that can be obtained by adding something from $U_1$ to something from $U_2$.

You say you understand why $U+W$ in this example is equal to $\{(x,y,0) : x,y\in \Bbb R\}$, so it seems that you're puzzled about why $U+Z$ is also equal to this.

The sum of $U$ and $Z$ is anything that can be obtained by adding something from $U$ to something from $Z$. The claim is that this is equal to $U+W$. To be equal, the two spaces must contain the same vectors. So the claim is that we can get any vector $(x,y,0)$ from $U+W$ by adding something from $U$ to something from $Z$, and also that we can only get vectors of that form.

So you need to show:

  1. If you add something from $U$ to something from $Z$, the sum must have the form $(x,y,0)$, which will show that it is in $U+W$.
  2. Any vector $(x,y,0)$ that is in $U+W$ can also be obtained by adding something from $U$ to something from $Z$.
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I'm also reading from the same book, this is what I understood so far.

Let's say you have an element $u \in U$, this means that it has the form of $(x, 0, 0) \in \mathbb R^3$ for some $x \in \mathbb R$. For example $(1,0,0)$ is an element in $U$ since it satisfies this definition.

Similarly, $W$ is defined as the set of all elements that satisfy $(0,y,0)\in \mathbb R^3$ for some $y\in \mathbb R$. For example $(0,\sqrt{2},0) \in W$

Now to vertify the sum $U + W$, you want to find the 'general form' an element should have in order to be in $U + W$

Let's say we add $(1,0,0)$ and $(0,\sqrt{2},0)$ to get $(1,\sqrt{2},0)$

Since this is arbitrary, we can say that the general form of the elements in the set $U + W$ will always be $(x, y, 0) \in \mathbb R^3$ for some $x, y \in \mathbb R$

Similary for $U + Z$ we can take an example: $(0.5, 0.5, 0) \in Z$ such that the sum $(1,0,0)$ and $(0.5, 0.5, 0)$ will always give me a general form of $(x, y, 0) \in \mathbb R^3$ for some $x, y \in \mathbb R$

Since the sum of two subspaces is the set of all the possible sums of elements of these subspaces, the sets $U+W$ and $U+Z$ must contain the same elements, therefore equal.

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