Questions about linear independent and spanning set and basi

In summary, the definitions of spanning set, linear independence, and basis of a vector space were provided. A set is a spanning set if it is a set of vectors in the vector space that can be used to create all other vectors in the space through linear combinations. A set of vectors is linearly independent if there is no non-trivial linear combination that results in the zero vector. A basis of a vector space is a set of linearly independent vectors that span the entire space, and the number of vectors in the basis is equal to the dimension of the space. Based on these definitions, a set of 6 vectors in R5 cannot be a basis for R5 because it is not linearly independent. Additionally, a set of
  • #1
DavidDai
7
0
Member warned about posting with no effort shown

Homework Statement



Definition of spanning set:
Let be vectors in the vector space . The set of all linear combination of the vectors is a subspace ( say ) of . The subspace is called the space spanned by the vectors The set is called a spanning set of .

Definition of linear independence:
Suppose that is a vector space. The set of vectors from is linearly dependent if there is a relation oflinear dependence on that is not trivial. In the case where the only relation of linear dependence on is the trivial one, then is a linearly independent set of vectors.

Definition of basis of vector space:
1. It spans the space.
2. Its vectors are independent.
3. The number of vectors in the basis is equal to the dimension of the space.
True or false? Given reason

1. A set of 5 Vectors in R5 must be a basis for R5
2. A set of 6 Vectors in R5 cannot be a basis for R5
3. A set of 7 vectors in R5 must be a spanning set for R5
4. A set of 6 polynomials in R5 must be a basis for R5
5. A set of 6 polynomials in R5 may be a basis for R5Anyone help me to explain these question.. I want to know the reason cz it's confused me a lot and i always get mess about these kind of question
appericated it!
 
Last edited:
Physics news on Phys.org
  • #2
DavidDai said:

Homework Statement


True or false? Given reason

Homework Equations



1. A set of 5 Vectors in R5 must be a basis for R5
2. A set of 6 Vectors in R5 cannot be a basis for R5
3. A set of 7 vectors in R5 must be a spanning set for R5
4. A set of 6 polynomials in R5 must be a basis for R5
5. A set of 6 polynomials in R5 may be a basis for R5

The Attempt at a Solution


Anyone help me to explain these question.. I want to know the reason cz it's confused me a lot and i always get mess about these kind of question
appericated it!
I assume that what you wrote under "Relevant equations" is actually the "The problem statement", and that R5 stand for ##\mathbb R^5##.

For starters, maybe you could define the used concepts: basis and spanning set.
 
  • Like
Likes SammyS
  • #3
Samy_A said:
I assume that what you wrote under "Relevant equations" is actually the "The problem statement", and that R5 stand for ##\mathbb R^5##.

For starters, maybe you could define the used concepts: basis and spanning set.
Thanks. I just edited the form of the question and now it should be ok..btw, Can u help me to this question?
 
  • #4
DavidDai said:
Thanks. I just edited the form of the question and now it should be ok..btw, Can u help me to this question?
Thanks.

The forum rules expect you to show some effort in solving the exercise, before other forum members jump into help.
That's why I suggested you at least start with defining what a basis is, and what a spanning set is.
 
  • #5
Samy_A said:
Thanks.

The forum rules expect you to show some effort in solving the exercise, before other forum members jump into help.
That's why I suggested you at least start with defining what a basis is, and what a spanning set is.
Thanks for reminding. This is my first time to ask question here so sorry about that.
 
  • #6
DavidDai said:

Homework Statement



Definition of spanning set:
Let be vectors in the vector space . The set of all linear combination of the vectors is a subspace ( say ) of . The subspace is called the space spanned by the vectors The set is called a spanning set of .

Definition of linear independence:
Suppose that is a vector space. The set of vectors from is linearly dependent if there is a relation oflinear dependence on that is not trivial. In the case where the only relation of linear dependence on is the trivial one, then is a linearly independent set of vectors.

Definition of basis of vector space:
1. It spans the space.
2. Its vectors are independent.
3. The number of vectors in the basis is equal to the dimension of the space.
(I added the bold blue)
Using that third point, can you answer some of the questions?
 
  • #7
Samy_A said:
(I added the bold blue)
Using that third point, can you answer some of the questions?
I can answer question 2 the reason is any set of 6 vectors in R5 is linearly dependent, even though some sets of 6 vectors in R5 span R5. To be a basis it must be a linearly independent spanning set, so if it's linearly dependent, it cannot be a basis.
But actually I know the answer for all of these questions but i don't the reason apart from question2
 
  • #8
DavidDai said:
I can answer question 2 the reason is any set of 6 vectors in R5 is linearly dependent, even though some sets of 6 vectors in R5 span R5. To be a basis it must be a linearly independent spanning set, so if it's linearly dependent, it cannot be a basis.
But actually I know the answer for all of these questions but i don't the reason apart from question2
Ok. What about questions 4 and 5?
 
  • #9
Samy_A said:
Ok. What about questions 4 and 5?
No i don't know how to explain question 1 3 4 5.
 
  • #10
DavidDai said:
No i don't know how to explain question 1 3 4 5.
Let's take 4 as an example:
"4. A set of 6 polynomials in R5 must be a basis for R5"

Can a set of 6 polynomials be a basis for ##\mathbb R^5##? The answer can be found using the definitions you posted.
 
  • #11
Samy_A said:
Let's take 4 as an example:
"4. A set of 6 polynomials in R5 must be a basis for R5"

Can a set of 6 polynomials be a basis for ##\mathbb R^5##? The answer can be found using the definitions you posted.
Thanks for helping. Here is too late now so I have to sleep. I am going to do these question tomorrow.
Appericated for helping!
 
  • #12
Samy_A said:
The forum rules expect you to show some effort in solving the exercise, before other forum members jump into help.
Noted...
 
  • #13
Hi guys. I hv done these question already. Thanks for helping!
 

1. What is the definition of a linearly independent set?

A linearly independent set is a set of vectors where no vector can be written as a linear combination of the other vectors in the set. In other words, none of the vectors in the set are redundant or can be expressed as a combination of the others.

2. How do you determine if a set of vectors is linearly independent?

To determine if a set of vectors is linearly independent, you can use the following steps:

  1. Write the vectors as columns in a matrix.
  2. Use row reduction to put the matrix in reduced row echelon form.
  3. If there is a pivot position in every column, the vectors are linearly independent. If not, they are linearly dependent.

3. What is the difference between a spanning set and a basis?

A spanning set is a set of vectors that can be used to create any vector in a given vector space through linear combinations. A basis, on the other hand, is a linearly independent spanning set. It is the smallest possible set of vectors that can span a vector space.

4. Can a set of vectors have more than one basis?

Yes, a vector space can have multiple bases. This is because there can be different combinations of linearly independent vectors that can span the same vector space. However, all of these bases will have the same number of vectors.

5. How do you find the basis of a vector space?

To find the basis of a vector space, you can use the following steps:

  1. Write the vectors as columns in a matrix.
  2. Use row reduction to put the matrix in reduced row echelon form.
  3. The columns with the pivot positions will form a basis for the vector space.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
3
Views
9K
  • Precalculus Mathematics Homework Help
Replies
3
Views
1K
  • Precalculus Mathematics Homework Help
Replies
5
Views
1K
  • Linear and Abstract Algebra
Replies
6
Views
755
  • Precalculus Mathematics Homework Help
Replies
6
Views
1K
  • Linear and Abstract Algebra
Replies
8
Views
769
  • Precalculus Mathematics Homework Help
Replies
14
Views
5K
  • Precalculus Mathematics Homework Help
Replies
17
Views
2K
  • Calculus and Beyond Homework Help
Replies
14
Views
531
  • Precalculus Mathematics Homework Help
Replies
3
Views
1K
Back
Top