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

 

[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