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I learned that the following two statements are equivalent $$V = W_1 \bigoplus W_2$$ and that $\gamma_1\cup\gamma_2$ is an ordered basis for $V$ where $\gamma_i$ is an ordered basis for $W_i$ ($1\leq i\leq k$). However, consider $V = \mathbb{R}^2$, $W_1 = span({e_1, e_2})$ and $W_2 = span({e_1})$. Then we have that $\gamma_1\cup\gamma_2 = {e_1, e_2}$, which is a basis for $V$. But very clearly we have that $V$ is not a direct sum of $W_1$ and $W_2$. What am I misunderstanding?

Gary Song
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  • What is your definition of $V=W_1\oplus W_2$? – wormram Jun 10 '23 at 01:20
  • Regardless, the correct condition is as follows. Let $V$ be a finite dimensional vector space. Let $W_1,W_2$ be subspaces of $V$ with bases ${b_1,\ldots,b_m}$ and ${c_1,\ldots,c_n}$ respectively. Then $V=W_1\oplus W_2$ if and only if the vectors $b_1,\ldots,b_m$ and $c_1,\ldots,c_n$ form a basis for $V$. Do you see how this is different than what you wrote? What you wrote becomes true if you replace union with multiset union. – wormram Jun 10 '23 at 01:31
  • If $W_1, W_2$ are subspaces of $V$ then you cannot always get a (internal) direct sum $V = W_1 \oplus W_2$. One of the conditions required is $W_1 \cap W_2 = {0}$ which your e.g. doesn't satisfy. – balddraz Jun 10 '23 at 01:31
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    The correct equivalence is: $V=W_1 \oplus W_2$ iff $\gamma_1 \cup \gamma_2$ is a basis for $V$ and* $\gamma_1 \cap \gamma_2 = \varnothing$.* – azif00 Jun 10 '23 at 03:24
  • I'm not entirely sure what the example you gave is trying to show. You may want to look at the third bullet point in the answer written by Surb here: https://math.stackexchange.com/a/1163321/1053712 . Remember for direct sum it isn't enough to sum the subspaces, you also have the condition that their intersection is only {0}. – neon tangerine Jun 10 '23 at 04:11

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In order for the two spaces $W_1\oplus W_2=V$, it must be the case that $\gamma_1\cap\gamma_2=\{\vec 0\}$, i.e., the spaces spanned by the basis vectors of both spaces are all linearly independent from each other.

If this is not the case, then we cannot use the direct sum (by definition), instead, we could use the following:

$$W_1+W_2=\{ae_1+be_2+ce_1:a,b,c\in\mathbb{R}\}=\{(a+c)e_1+be_2:a,b,c\in\mathbb{R}\}$$

But $(a+c)$ is just some constant (call it $d$), so:

$$W_1+W_2=\{de_1+be_2:b,d\in\mathbb{R}\}=W_1$$

But this is just one of the subspaces we started with.

The direct sum is used to denote that the two subspaces satisfy $W_1\cap W_2=\{\vec 0\}$. Note that when this is the case $W_1\oplus W_2$ yields a new subspace spanned by $\gamma_1\cup\gamma_2$. For example, if $W_1=\{0\}\times\mathbb{R}$ and $W_2=\mathbb{R}\times\{0\}$, then $W_1\cap W_2=\{\vec 0\}$ and $W_1\oplus W_2=V=\mathbb{R}^2$.

This link gives a good description regarding the uses of the direct sum.

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