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If I have a direct sum $V = V_1 \oplus V_2$ and a subspace $W \subset V$, it it necessarily true that $W = W_1 \oplus W_2$ where $W_1 \subset V_1$ and $W_2 \subset V_2$?

I believe this is true since we should be able take $W_1 :=W \cap V_1$ and $W_2 := W \cap V_2$, but I just want to make sure there isn't a flaw this this argument. Thanks!

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Nope, it is not true. The vector space $K^2$ (where $K$ is the base field) is the direct sum of the span of its first basis vector and the span of its second basis vector, but the span of the sum of the two basis vectors is not a "direct subsum"! – darij grinberg Jan 17 at 18:34
up vote 3 down vote accepted

Note that $\langle(1,1)\rangle\subset\mathbb{R}^{2}=\langle (0,1)\rangle\oplus\langle (1,0)\rangle$ can not be written as a direct sum of subspaces of $\langle(0,1)\rangle$ and $\langle(1,0)\rangle$.

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In your attempt, while the sum $W_1 + W_2$ is direct (that is, $W_1 \cap W_2 = 0$), it might not equal all of $W$.

There is a simple counterexample for $V = \mathbb{R}^2$. $V$ is the direct sum of the $x$ and $y$ axes, but if we set $W$ equal to the line $y = x$, then $W$ intersected with the $x$ and $y$-axes is zero.

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No. Consider the simple case $$V_1, V_2 := \Bbb F, \qquad W := \langle (1, 1) \rangle \subset V := \Bbb F \oplus \Bbb F .$$ In this case, $W \cap V_a = \{ 0 \}$ for $a = 1, 2$, so your construction already fails for this example.

In fact, the only subspaces of $V_a$, $a = 1, 2$, are $0$ and $V_a$ itself, so the only $1$-dimensional subspaces of $V$ that are the direct sum of a subspace of $V_1$ and a subspace of $V_2$ are $V_1$ and $V_2$ themselves, and neither of these is $W$. So, in general, there are not $W_a \subseteq V_a$, $a = 1, 2$, such that a general subspace $W \subset V_1 \oplus V_2$ is equal to $W_1 \oplus W_2$.

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