Solving Linear Space Basis: Find q(t) & q'(t) Dimension

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Homework Help Overview

The discussion revolves around finding a basis for a linear space defined by the condition q'(1) = q(2), where q is a subset of Q_2, the space of polynomials of degree less than or equal to 2. Participants are exploring the implications of this condition on the coefficients of the polynomial.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the formulation of q(t) and its derivative q'(t), with attempts to express the relationship between coefficients a, b, and c based on the given condition. Some express uncertainty about their approach and seek clarification on how to derive a basis from their findings.

Discussion Status

There is ongoing exploration of the relationships between the coefficients of the polynomial and the conditions imposed by the problem. Some participants have offered guidance on how to simplify the expressions and consider the implications of the equations derived from the condition. However, there is no explicit consensus on the final basis or dimension yet.

Contextual Notes

Participants note the potential complexity of the problem and the need to ensure that the proposed functions span the subspace and are linearly independent. There is also mention of constraints related to homework rules and the need for clarity in the derivation process.

tangibleLime
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Homework Statement



Find a basis of the linear space and thus determine the dimension.

q(t) : q'(1)=q(2)

Where q is a subset of Q_2: all polynomials of degree less than or equal to 2. I had to change it from p to q so the forum didn't make it a smiley...

The Attempt at a Solution



I'm not sure what to do here. I created an arbitrary form of q(t) and q'(t) and found a basis for each of these.

q(t) = a+bt+ct^{2}
\ss =(1,t,t^2)

q'(t) = 1+2ct
\ss = (1,t)

Now I'm lost, and I'm pretty certain I was never on the right track. Any help would be extremely appreciated.

Thanks.
 
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tangibleLime said:

Homework Statement



Find a basis of the linear space and thus determine the dimension.

q(t) : q'(1)=q(2)

Where q is a subset of Q_2: all polynomials of degree less than or equal to 2. I had to change it from p to q so the forum didn't make it a smiley...

The Attempt at a Solution



I'm not sure what to do here. I created an arbitrary form of q(t) and q'(t) and found a basis for each of these.

q(t) = a+bt+ct^{2}
\ss =(1,t,t^2)

q'(t) = 1+2ct
\ss = (1,t)

You aren't asked for the basis of the derivative subspace so that last equation is irrelevant. Also, you haven't written down what q'(1) = q(2) tells you about a, b, and c. Do that so you can see what the subspace looks like.
 
My apologies, meant to edit that part in but got sidetracked.

Here's my work. I take p'(1) = p(2) and solve for a, b and c. I use these new values in a new generic 2-degree polynomial, but what I come up with just looks like a horrible mess. I also noted the answer (from the back of the book) in the top left... can't figure out how to get those basis vectors from this new polynomial (though I'm certain I'm way off target.)

[PLAIN]http://mikederoche.com/temp/linalg1.jpg
 
Last edited by a moderator:
You aren't so far off target. From your a + b + 2c = 0, you can let two of the variables be arbitrary and solve for the third one in terms of them. So you only have two of them in your generic polynomial. Then group your polynomial on the two constants (factor them out).
 
Thanks for the help! I got the correct answer, here is my work for any other lost travelers who find this post via Google:

(Starting with a+b+2c = 0)

[PLAIN]http://mikederoche.com/temp/linalg2.jpg
 
Last edited by a moderator:
OK. But remember that to show you have a basis you need to be sure you have proved:

1. Your two functions span the subspace
2. They are linearly independent.
 

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