Union of two bases in different subspaces question Not sure about this one.
Prove or disprove,
If $U$ and $W$ are linear subspaces of a linear space $V$ with the bases $\mathcal{B}_1$ and $\mathcal{B}_2$ ($\mathcal{B}_1$ for $U$, $\mathcal{B}_2$ for $W$) then $\mathcal{B}_1\cup\mathcal{B}_1$ is a base of $U +W$.
I've tried a lot of examples and it seems like it must be true, but i'm afraid i'm missing something.
Thanks in advance !
 A: This is false since it would imply $dim(U+W)=\dim U+\dim W$ which is true only is the sum is direct. In particular, take $W=U$, but with a different basis $\mathcal B_2$. Would you say that, if $U$ has tow distinct bases with $r$ elements, it has a basis with $2r$ elements?
A: It may be true with additional conditions. I mean, suppose $U\subseteq W$. Then $B_1\cup B_2$ is definitely not linearly independent. For example, you can take $W=\mathrm{span}((0,1,0),(1,0,0))$, $V=\mathbb{R}^3$, $U=\mathrm{span}((0,1,0))$. A basis of $U$ is $\{(0,1,0)\}$, a basis of $W$ is $\{(0,2,0),(1,0,0)\}$, but $B_1\cup B_2=\{(0,1,0),(0,2,0),(1,0,0)\}$ is not a basis since $(0,1,0)-\frac12(0,2,0)+0(1,0,0)=0$ which implies the three vectors are linearly dependent. Surely $B_1\cup B_2$ is a system of generators for $U+W$ which in this case is $W$.
A: The property is true if and only if $U\cap W=\{0\}$.
First of all you can show that $\mathcal{B}_1\cup\mathcal{B}_2$ is a spanning set for $U+W$ (just write down the definitions).
Suppose now that $U\cap W=\{0\}$ and that
$$
\alpha_1u_1+\dots+\alpha_mu_m+\beta_1w_1+\dots+\beta_nw_n=0
$$
where $\mathcal{B}_1=\{u_1,\dots,u_m\}$ and $\mathcal{B}_2=\{w_1,\dots,w_n\}$. Then, setting
$$
v=\alpha_1u_1+\dots+\alpha_mu_m=-(\beta_1w_1+\dots+\beta_nw_n)
$$
we have that $v\in U\cap W$ and so $v=0$. From the linear independence of $\mathcal{B}_1$ and of $\mathcal{B}_2$ we conclude
$$
\alpha_1=0,\dots,\alpha_m=0,\beta_1=0,\dots,\beta_n=0.
$$
Thus $\mathcal{B}_1\cup\mathcal{B}_2$ is linearly independent.
Conversely, suppose $\mathcal{B}_1\cup\mathcal{B}_2$ is a basis of $U+W$, that is, it is linearly independent (because we know it's a spanning set). Let $v\in U\cap W$; then
$$
v=\alpha_1u_1+\dots+\alpha_mu_m=\beta_1w_1+\dots+\beta_nw_n
$$
for suitable scalars. Now
$$
\alpha_1u_1+\dots+\alpha_mu_m+(-\beta_1)w_1+\dots+(-\beta_n)w_n=0
$$
and, by the linear independence of $\mathcal{B}_1\cup\mathcal{B}_2$, we conclude that $\alpha_1=0,\dots,\alpha_m=0$, so $v=0$.
(The proof could be extended also to infinite dimensional spaces.)
