Answer to Question #283063 in Linear Algebra for Rizwana

Question #283063

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.

1
Expert's answer
2021-12-28T12:44:14-0500

Let us prove that every vector in U+VU+V can be written as a linear combination of  {u1,,uk,v1,,vl}\{u_1,…,u_k,v_1,…,v_l\}.


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

Since {u1,,uk}\{u_1,…,u_k\} is a basis of UU, we can write u=α1u1++αkuku=\alpha_1 u_1+…+\alpha_ku_k .

Since {v1,,vl}\{v_1,…,v_l\} is a basis of VV, we can write v=β1v1++βlvlv=\beta_1 v_1+…+\beta_lv_l .

So, u+v=α1u1++αkuk+β1v1++βlvlu+v=\alpha_1u_1+…+\alpha _ku_k+\beta _1 v_1+…+\beta _lv_l .


Now we need to prove that {u1,,uk,v1,,vl}\{u_1,…,u_k,v_1,…,v_l\} is linearly independent.

Suppose we have α1u1++αkuk+β1v1++βlvl=0\alpha_1u_1+…+\alpha _ku_k+\beta _1 v_1+…+\beta _lv_l=0 .

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


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


Using the fact that {u1,,uk}\{u_1,…,u_k\} is a basis of UU and {v1,,vl}\{v_1,…,v_l\} is a basis of VV , we get that α1==αk=0\alpha_1=…=\alpha_k=0 and β1==βl=0\beta_1=…=\beta_l=0 .

So, {u1,,uk,v1,,vl}\{u_1,…,u_k,v_1,…,v_l\} is linearly independent.


Now we can conclude that {u1,,uk,v1,,vl}\{u_1,…,u_k,v_1,…,v_l\} is a basis for U+VU+V .



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment