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This is an example from Axler's "Linear Algebra Done Right":

Suppose $U_j$ is a subspace of $\mathbf{F}^n$ of those vectors whose coordinates are all $0$, except possibly in the $j^\text{th}$ slot (thus, for example, $U_2 = \{ (0, x, 0, \dots, 0) \in \mathbf{F}^n | x \in \mathbf{F} \}$. Then $$\mathbf{F}^n = U_1 \oplus \dots \oplus U_n$$, as you should verify.

The definition of direct sum is given as

"The sum $U_1+\dots+U_n$ is called a direct sum if each element of $U_1 + \dots + U_m$ can be written in only one way as a sum $u_1+ \dots + u_m$, where each $u_j$ is in $U_j$."

Proof: Let $v$ be an arbitrary vector.

$(\leftarrow)$ Suppose that $v \in U_1 \oplus \dots \oplus U_n$. Then we know that $v$ can be written as $v = u_1 + u_2 + \dots + u_n$ for $u_i \in U_i$. Since each $u_i \in \mathbf{F}^n$ and $\mathbf{F}^n$ is a vector space, $v = u_1 + u_2 + \dots + u_n \in \mathbf{F}^n$ by closure under addition.

$(\rightarrow)$ Now suppose that $v \in \mathbf{F} ^ n$. Then we can write $v$ as $v = (v_1, v_2, \dots, v_n)$. For $1 \leq i \leq n$, let $u_i \in U_i$ be a vector with $v_i$ as its $i^\text{th}$ coordinate, and zero everywhere else. It follows that $v = u_1 + u_2 + \dots + u_n$, so $v \in U_1 + U_2 + \dots + U_n$. Now let $w_1 \in U_1, w_2 \in U_2, \dots w_n \in U_n$ be arbitrary such that $v = w_1 + \dots + w_n$ ...

How do I finish this part of the proof formally (it makes sense intuitively)? I was thinking of showing that $u_1 + u_2 + \dots + u_n = v = w_1 + \dots + w_n$ and $u_1 + u_2 + \dots + u_n - (w_1 + \dots + w_n) = 0$, so $(u_1 - w_1) + \dots + (u_n - w_n) = 0$, but I get stuck here. Any hints would be appreciated!

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  • $\begingroup$ @Matematleta That seems to be stated later on in the book. How do I finish it without that knowledge? $\endgroup$
    – Iyeeke
    Sep 11, 2020 at 23:52
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    $\begingroup$ @Matematleta, having $U_{i} \cap U_{j} = \{0 \}$ if $i \neq j$ is not enough when dealing with $3$ or more subspaces. To quote an example from Axler, consider $$ U_{1} = \{ (x,y, 0) \in \mathbb{R}^{3} \mid x, y \in \mathbb{R} \}, $$ $$ U_{2} = \{ (0,0, z) \in \mathbb{R}^{3} \mid z \in \mathbb{R} \}, $$ $$ U_{3} = \{ (0, y, y) \in \mathbb{R}^{3} \mid y \in \mathbb{R} \}. $$ They satisfy the previous condition but their sum is not a direct one. $\endgroup$
    – Kevin Aquino
    Sep 12, 2020 at 0:37
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    $\begingroup$ @Matematleta The definition is given as "The sum $U_1 + \dots + U_n$ is called a direct sum if each element of $U_1 + \dots + U_m$ can be written in only one way as a sum $u_1 + \dots + u_m$, where each $u_j$ is in $U_j$." $\endgroup$
    – Iyeeke
    Sep 12, 2020 at 1:17
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    $\begingroup$ @Kevin López Aquino Yes, correct sorry I meant to write and should have written $U_i\cap\sum_{j\neq i}U_i = \{0\}$ for $1\le i\le n$ $\endgroup$
    – Matematleta
    Sep 12, 2020 at 4:07

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Following your idea from

$$u_{1}+\cdots+u_{n}=w_{1}+\cdots+w_{n}$$

$$u_{1}-w_{1}=w_{2}+\cdots+w_{n}-u_{2}-\cdots-u_{n}$$

The left side is in $U_{1}$, while the right side is in $U_{2}\oplus\cdots \oplus U_{n}$, and the intersections between this two subspaces is $\{ 0 \}$ because they are in direct sum. Then as $u_{1}-w_{1}$ belong to both of them, it must be zero; this implies $u_{1}=w_{1}$. If you applies the same pocedure to the rest of the coordinates you willl conclude all the $u_{i}$ are equall to the $w_{i}$.

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