Confusion on criteria for direct sum

Question

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?

Answer 1

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.