Find a basis for W which is subset of V

[songoku]

Homework Statement

Please see below

Relevant Equations

Span
Linear Independent

I think I can prove W is a subspace of V. I want to ask about basis of W.

Let $$V = a_1+a_2 \sin t+a_3 \cos t +a_4 \sin (2t)+a_5 \cos (2t)$$
$$W = p(t) = q"(t) + q(t)$$
$$=-a_2 \sin t-a_3 \cos t-4a_4 \sin (2t)-4a_5 \cos(2t)+a_1+a_2 \sin t+a_3 \cos t +a_4 \sin (2t)+a_5 \cos (2t)$$
$$=a_1-3a_4 \sin (2t) -3a_5 \cos (2t)$$
Since all elements in W are linearly independent, the basis for W is {1, sin (2t), cos (2t)}

Am I correct? Thanks

 

[pasmith]

Yes. If [itex]B[/itex] is a basis for [itex]V[/itex] and [itex]L : V \to \dots[/itex] is a linear map, then [itex]L(B)[/itex] spans [itex]L(V)[/itex]. If the non-zero elements of [itex]L(B)[/itex] are linearly independent then they will be a basis for [itex]L(V)[/itex].

 

[songoku]

Thank you very much pasmith

 

[Steve4Physics]

Homework Statement: Please see below
Relevant Equations: Span
Linear Independent

View attachment 324875

I think I can prove W is a subspace of V. I want to ask about basis of W.

Let $$V = a_1+a_2 \sin t+a_3 \cos t +a_4 \sin (2t)+a_5 \cos (2t)$$
$$W = p(t) = q"(t) + q(t)$$
$$=-a_2 \sin t-a_3 \cos t-4a_4 \sin (2t)-4a_5 \cos(2t)+a_1+a_2 \sin t+a_3 \cos t +a_4 \sin (2t)+a_5 \cos (2t)$$
$$=a_1-3a_4 \sin (2t) -3a_5 \cos (2t)$$
Since all elements in W are linearly independent, the basis for W is {1, sin (2t), cos (2t)}

Am I correct? Thanks

Your underlying method is correct but perhaps your proof could be improved.

Your equation
##V = a_1+a_2 \sin t+a_3 \cos t +a_4 \sin (2t)+a_5 \cos (2t)##
looks like you have the space ##V## on the left side and a single vector on the right side. You can’t equate these two different things.

A similar comment applies to ##W = p(t) = q"(t) + q(t)##.

A better way to start might be to say:
Since ##q(t) \in V## we can express ##q(t)## most generally as:
##q(t) = a_1+a_2 \sin t+a_3 \cos t +a_4 \sin (2t)+a_5 \cos (2t)##
And take it from there.

 

[songoku]

Ah ok, thank you very much Steve4Physics