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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.

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    $\begingroup$ That's what $x, y \in \mathbb{F}$ means in set-builder notation here. $\endgroup$ Jan 27, 2022 at 4:00
  • $\begingroup$ 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. $\endgroup$
    – blamocur
    Jan 27, 2022 at 4:02

1 Answer 1

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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?

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