Find a basis for W which is subset of V

  • Thread starter Thread starter songoku
  • Start date Start date
  • Tags Tags
    Basis
songoku
Messages
2,475
Reaction score
389
Homework Statement
Please see below
Relevant Equations
Span
Linear Independent
1681474137895.png


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
 
Last edited by a moderator:
Physics news on Phys.org
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).
 
Thank you very much pasmith
 
songoku said:
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.
 
Ah ok, thank you very much Steve4Physics
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top