Question

Let U,V be subspaces of Rn. 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}.

Let $S={u_1,u_2,....….,u_n}\subset U and \space T={v_1,v_2,.....,.....v_n}\space \subset V$

are two linearly independent sets of vectors.

Claim:$=S\cup T$ is linearly independent.

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

Then $u_r=$ $a_1u_1+..........+a_lu_l+b_1v_1+.............+b_lv_l$ for some $a_i, b_i \space in \space F and \space u_i \in U,\space v_i \in T$

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

Hence, this gives a contradiction of the given condition

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

Conversely,

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

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

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