Answer to Question #283063 in Linear Algebra for Rizwana

Question

Suppose U and V are subspace of R^n with U intersection V={0}. if {u1,.....uk} is a basis for U and {v1,.....vL} is a basis for V, prove that {u1.....uk,v1.......vL} is a basis for u+v.

Answer 1

Let us prove that every vector in $U+V$ can be written as a linear combination of  $\{u_1,…,u_k,v_1,…,v_l\}$.

$U+V=\{u+v:\ u\in U,\ v\in V\}$

Since $\{u_1,…,u_k\}$ is a basis of $U$, we can write $u=\alpha_1 u_1+…+\alpha_ku_k$ .

Since $\{v_1,…,v_l\}$ is a basis of $V$, we can write $v=\beta_1 v_1+…+\beta_lv_l$ .

So, $u+v=\alpha_1u_1+…+\alpha _ku_k+\beta _1 v_1+…+\beta _lv_l$ .

Now we need to prove that $\{u_1,…,u_k,v_1,…,v_l\}$ is linearly independent.

Suppose we have $\alpha_1u_1+…+\alpha _ku_k+\beta _1 v_1+…+\beta _lv_l=0$ .

It can be written as $\sum_{i=1}^k\alpha _iu_i=-\sum_{j=1}^l\beta_jv_j$ .

We can see the left hand side is in $U$ and the right hand side is in $V$ . It means that both are in $U\cap V$. Since $U\cap V=\{0\}$ , it follows that $\sum_{i=1}^k\alpha _iu_i=\sum_{j=1}^l\beta_jv_j=0$ .

Using the fact that $\{u_1,…,u_k\}$ is a basis of $U$ and $\{v_1,…,v_l\}$ is a basis of $V$ , we get that $\alpha_1=…=\alpha_k=0$ and $\beta_1=…=\beta_l=0$ .

So, $\{u_1,…,u_k,v_1,…,v_l\}$ is linearly independent.

Now we can conclude that $\{u_1,…,u_k,v_1,…,v_l\}$ is a basis for $U+V$ .