Suppose $U_1, ... , U_m$ are finite-dimensional subspaces of $V$ and $\beta_1,...,\beta_m$ are their bases, respectively. Prove that: $U_1+...+U_m$ is a direct sum if and only if $\beta=(\beta_1,...,\beta_m)$ is a basis of $U_1+...+U_m$ .
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If the sum is not direct between say $U_1$ and $U_2$ then, by definition of direct sum, we have that $U_1\cap U_2\neq\{0\}$.
Choose a non-zero vector of $U_1\cap U_2$, say $v$ defined by the basis $\beta_1$ of $U_1$ as
$$v:=\sum_{k=1}^{m_1}c_ku_k,\quad c_k\in\Bbb F,u_k\in\beta_1\tag{1}$$
and defined under the basis $\beta_2$ of $U_2$
$$v:=\sum_{k=1}^{m_2}h_ks_k,\quad h_k\in\Bbb F,s_k\in\beta_2\tag{2}$$
From $(1)$ and $(2)$ we have that
$$v-v=0=\sum_{k=1}^{m_1}c_ku_k+\sum_{k=1}^{m_2}h_ks_k\in\operatorname{span}(\beta_1\cup\beta_2)$$
Because $v\neq 0$ then there is some $c_k$ or $h_k$ that is non-zero, hence $\beta_1$ and $\beta_2$ are not linearly independent by definition.
By the other side if $\beta_1\cup\beta_2$ are linearly independent then we have that
$$U_1=\operatorname{span}(\beta_1),\quad U_2=\operatorname{span}(\beta_2)$$
So $U_1\cap U_2=\{0\}$, then $U_1+U_2$ is a direct sum, by definition.
The sum is not a direct sum if and only if there is a nontrivial element in the intersection of a pair of U's if and only if that non-trivial element is a linear combinations of different sets of basis elements (some betas from each U that it belongs to) if and only if the betas are linearly dependent if and only if the betas are not a basis.