Question

Let U,V be subspaces of R^n.
Show that U∩V={0} if and only if S∪T is a linearly independent set of vectors for every linearly independent set S={u1,u2,...uk} ⊆ U and every linearly independent set T = {v1,v2,...vl} ⊆ V

Answer 1

Let $U\cap V=\{0\}.$

Lets $S=u_1,u_2,...,u_n\sub U\space and\space T=v_1,v_1,...,v_n\sub V$

are two linearly independent set of vectors.

Claim$\coloneqq S\cup T$ is linearly independent.

Suppose not, then there exist a vector from $S\cup T$, say $u_r\not=0.$

Then $u_r=a_1u_1+...+a_lu_l+b_1v_1+...+b_lv_l$ for some

$a_i,b_i\space in\space F\space and\space u_i\in U,v_i\in T$

$\implies u_r-a_1u_1-...-a_lu_l=b_1v_1+...+b_lv_l$

Hence, this gives a contraiction of the given condition

i.e.,$U\cap V=\{0\}.$

Conversely,

Suppose $S$ and $T$ are given set of linearly independent sub of $U$ and $V$ respectively.

Claim$\coloneqq U\cap V=\{0\}.$

Suppose not, then there is a non-zero vector in this intersection, which contradicts the

assumption.