Can I show my solution? Linear Algebra

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Homework Help Overview

The discussion revolves around extending linearly independent subsets to form bases in vector spaces, specifically in the context of linear algebra. The original poster presents two scenarios involving vector spaces R^4 and polynomial space P(sub3), questioning the correctness of their proposed bases.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to verify the linear independence of given vectors and seeks to understand the criteria for forming a basis. They question the validity of their bases compared to a textbook answer and explore the implications of non-trivial solutions in linear combinations.

Discussion Status

Participants are engaged in clarifying the conditions under which a set of vectors can be considered a basis. Some guidance has been offered regarding the need for non-trivial solutions in linear combinations, and there is an ongoing exploration of the implications of different bases.

Contextual Notes

There is mention of a requirement to include certain natural basis vectors, which may influence the construction of the proposed bases. The discussion also reflects on the existence of multiple valid bases for a given set of linearly independent vectors.

franz32
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Here's the question.

A theorem says that any linearly independent subset S of vector space V can be extended to a basis for V.

1. S = {(1, 1, 0, 0) ; (1, 0, 1, 0)}, V = R^4.

In this part, the two vectors are linearly independent by verification. Then making a basis T with dimension of 4. I arrived at an answer of

T = { (1, 1, 0, 0) ; (1, 0, 1, 0); (1, 0, 0, 0); (0, 0, 0,1)}
Am I correct?

2. S = { t^3 - t + 1; t^3 + 2} , V = P(sub3)

Again, these are linearly independent by verification.
I used the natural basis { t^3, t^2, t, 1}.

And then, since the dimension is 4, here's my answer.

T = {t^3 -t +1; t^3 + 2; t^3; t^2}

But my book has a different answer...
instead of t^3 and t^2, it was t and 1 respectively. Are both
of these correct?
 
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1 You should be able to check your answer yourself. Any set of 4 vectors is a basis for R4 if and only if they are linearly independent. Are (1, 1, 0, 0) ; (1, 0, 1, 0); (1, 0, 0, 0); (0, 0, 0,1) linearly independent? If a(1, 1, 0, 0)+ b(1, 0, 1, 0)+ c(1, 0, 0, 0)+ d(0, 0, 0,1)= (0, 0, 0, 0) what are a, b, c, and d?


2 It's alright that your book has a different answer. Given a set of independent vectors smaller than a basis, there exist an infinite number of bases containing that set.

Again, if a(t^3 -t +1)+ b(t^3 + 2)+ c(t^3)+ d(t^2)= 0 (for all t) what are a, b, c, and d?
 
Hello again sir.

Sir,

so you are implying that if the scalars a, b, c, d are all zeros, then my answer for the basis is correct?
 
No, it needs to be a non-trivial solution.
 
...

So, to be nontrivial is to have at least a scalar not equal to 0 right?

I hope you don't mind helping me here... in my questions.

1. S = {(1, 1, 0, 0) ; (1, 0, 1, 0)}, V = R^4.

In this part, the two vectors are linearly independent by verification. Then making a basis T with dimension of 4. I arrived at an answer of

T = { (1, 1, 0, 0) ; (1, 0, 1, 0); (1, 0, 0, 0); (0, 0, 0,1)}
According to our professor, I must include some of the natural basis from left to right. But how am I going to show that my basis T has a nontrivial solution? ANd as well as in 2.
 

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