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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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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. –  qaphla Aug 7 at 16:55
Okay give me a moment I will edit it to be more clear. My apologies. –  Valentino Aug 7 at 16:57
No need to apologize -- everyone is new to this sometime. :) –  qaphla Aug 7 at 16:57

3 Answers 3

up vote 3 down vote accepted

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 elment 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, like given $v_0,w_0$ just set $x_0 = v_0$ nad $y_0= v_0 + w_0$.

For the other direction the task is this: given $x_0,y_0$ find $v$ and $w$ such that $x_0 = v$ and $y_0 = v +w$. So you need to solve a (simple) system of equations. The first gives you $v= x_0$ and then you get $w= y_0 - x_0$.

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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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. –  Valentino Aug 8 at 2:29
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. –  Valentino Aug 8 at 2:48

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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