# The Sum of Subspaces in Linear Algebra

I'm having some trouble understanding the sum of subspaces in Axler's Linear Algebra Done Right. Here is the question that is giving me a headache:

Suppose that $$U = \left\{ (x, x, y, y) \in F^4 : x, y \in F \right\}$$ and $$W = \left\{(x, x, x, y) \in F^4 : x, y \in F \right\}$$. Then:

$$U + W = \left\{(x, x, y, z) \in F^4 : x, y, z \in F \right\}$$.

My confusion comes from the variables. Are $$x, y$$, and $$z$$ arbitrary numbers in $$F$$ (complex and real numbers)? How would you verify this? Thank you.

• That's what $x, y \in \mathbb{F}$ means in set-builder notation here. Jan 27, 2022 at 4:00
• Yes, x, y and z are arbitrary. $U$ consists of vectors in which the first two coordinates are equal and so are the last two. $W$ consists. of vectors in which the first 3 coordinates are the same and the last can be anything. Their sum consists of vectors in which the first two coordinates are the same and the other two are arbitrary. Jan 27, 2022 at 4:02

HINT

According to the definition of sum, $$w\in U + V$$ iff $$w = u + v$$, where $$u\in U$$ and $$v\in V$$.

Consequently, it results that \begin{align*} w & = (a,a,b,b) + (c,c,c,d) = a(1,1,0,0) + b(0,0,1,1) + c(1,1,1,0) + d(0,0,0,1) \end{align*}

This relation means that \begin{align*} U + V = \operatorname{span}\{(1,1,0,0),(0,0,1,1),(1,1,1,0),(0,0,0,1)\} \end{align*}

Now it remains to remove redundant directions.

In order to do so, observe that

\begin{align*} (1,1,1,0) = (1,1,0,0) + (0,0,1,1) - (0,0,0,1) \end{align*}

Can you take it from here?