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Consider case 1:

Suppose we have subspaces $U_1, U_2, U_3$ of $V$.

Let $V = \mathbb{F^{3}}$.

Let $U_1 = \{(x,y,0) \in \mathbb{F^{3}}: x,y \in \mathbb{F}\}$, $U_2 = \{(0,0,z) \in \mathbb{F^{3}}: z \in \mathbb{F}\}$, and $U_3 = \{(0,y,y) \in \mathbb{F^{3}}: y \in \mathbb{F}\}$.

It is clear that $U_1 + U_2 + U_3 = \mathbb{F^{3}}$, because the set of vectors of the form $(x,y,z)$ can be expressed as $(x,y,z) = (x,y,0) + (0,0,z) + (0,0,0) = U_1 + U_2 + U_3$.

Because $(0,0,0) = (x,y_1+y_3,z+y_3)$, there is more than one way to write $0$ as a sum $u_1+\cdots+u_m$ for each $u_j \in U_j$.

Since $0$ cannot be expressed as a unique combination, it is clear that $U_1 \oplus U_2 \oplus U_3 \neq \mathbb{F^{3}}$ .

It makes sense that $\mathbb{F^{3}}$ is not equivalent to the direct sum of the three subspaces, because $U_3 \subseteq U_1 + U_2$. Since $U_3$ is a line through the origin on the $yz-axis$, and $U_1 \oplus U_2 = \mathbb{F^{3}}$, it is clear that $U_1 + U_2 = U_1 + U_2 + U_3$.

*I would like to make sure the above reasoning is accurate.

Now consider another case:

Let $V = \mathbb{F^{2}}$.

Let $U_1 = \{(x_1,0) \in \mathbb{F^2}: x_1 \in \mathbb{F}\}$ and $W = \{(x_2,x_2) \in \mathbb{F^2}: x_2 \in \mathbb{F}\}$.

Geometrically, I interpret $U_1$ as the $x-axis$ and $W$ as the line $y=x$.

Since $(0,0) = (x_1+x_2,x_2)$ is only $0$ for $x_1 = x_2 = 0$, we can deduce that $U_1 \cap W = \{0\}$.

This means $U_1 \oplus W = \mathbb{F^2}$.

*If my understanding of a direct sum in $\mathbb{R^{n}}$ is correct, in order for any collection of (nontrivial) subspaces $U_1, \ldots, U_n$ to be considered a direct sum, it must be that $U_1 \not\subset \cdots \not\subset U_n$ and $U_1 \not\supset \cdots \not\supset U_n$ for $u_1 \in U_1,\ldots, u_n \in U_n$

Is this a correct way to think about a direct sum of subspaces?

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    $\begingroup$ Everything you wrote is correct, except the last bit. First, it is not sufficient to characterize direct sums, and second it is not technically correct as written, because the zero vector alone is a subspace that can be a direct summand. If you insist that the subspaces are nontrivial, then the starred statement is true. $\endgroup$
    – vadim123
    Sep 19 '13 at 23:31
  • $\begingroup$ Thanks for the clarification $\endgroup$
    – St Vincent
    Sep 19 '13 at 23:37
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Let $V = \mathbb{F^{2}}$.

Let $U_1 = \{(x_1,0) \in \mathbb{F^2}: x_1 \in \mathbb{F}\}$ and $W = \{(x_2,x_2) \in \mathbb{F^2}: x_2 \in \mathbb{F}\}$.

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