Mixing
How to quickly tell if a set is linearly independent or not?
$begingroup$
Before proving if a set is a basis for $R^n$, I have to determine if the set is linearly independent or not.
We aren't allowed to use matrices, and I want to save time during a quiz.
What are some things I should watch out for?
linear-algebra
asked Jan 21 at 4:13
mingming
3906
$endgroup$
add a comment |
$begingroup$
Before proving if a set is a basis for $R^n$, I have to determine if the set is linearly independent or not.
We aren't allowed to use matrices, and I want to save time during a quiz.
What are some things I should watch out for?
linear-algebra
asked Jan 21 at 4:13
mingming
3906
$endgroup$
1
$begingroup$
I had a typo in my previous comment (and it's been more than 5 minutes so I'm fixing it here): If you have $n+1$ (or more) vectors, they must be linearly dependent. If you can see that one is a multiple of another vector (or a sum of two or more vectors), then the set is linearly dependent. This is not always easy to see, however.
$endgroup$
– JavaMan
Jan 21 at 4:23
$begingroup$
If the set contains the zero vector its dependent
$endgroup$
– Ben
Jan 21 at 4:25
$begingroup$
If there are $n$ vectors but the $i$th coordinate is always zero then its dependent (more generally: if there are $n-k$ vectors and $k+1$ many coordinates which are always 0)
$endgroup$
– Ben
Jan 21 at 4:27
$begingroup$
Note that the quick way to check this is to set up the linear independence equation in matrix form, then ask a computer to compute the coefficient matrix's determinant. (It's good to remember that problems often have better solutions than what a class forces you to use.)
$endgroup$
– rwbogl
Jan 21 at 5:14
$begingroup$
Thanks for the help guys!
$endgroup$
– ming
Jan 21 at 5:53
add a comment |
$begingroup$
Before proving if a set is a basis for $R^n$, I have to determine if the set is linearly independent or not.
We aren't allowed to use matrices, and I want to save time during a quiz.
What are some things I should watch out for?
linear-algebra
asked Jan 21 at 4:13
mingming
3906
$endgroup$
Before proving if a set is a basis for $R^n$, I have to determine if the set is linearly independent or not.
We aren't allowed to use matrices, and I want to save time during a quiz.
What are some things I should watch out for?
linear-algebra
linear-algebra
asked Jan 21 at 4:13
mingming
3906
asked Jan 21 at 4:13
mingming
3906
asked Jan 21 at 4:13
mingming
3906
asked Jan 21 at 4:13
mingming
3906
asked Jan 21 at 4:13
mingming
3906
3906
1
$begingroup$
I had a typo in my previous comment (and it's been more than 5 minutes so I'm fixing it here): If you have $n+1$ (or more) vectors, they must be linearly dependent. If you can see that one is a multiple of another vector (or a sum of two or more vectors), then the set is linearly dependent. This is not always easy to see, however.
$endgroup$
– JavaMan
Jan 21 at 4:23
$begingroup$
If the set contains the zero vector its dependent
$endgroup$
– Ben
Jan 21 at 4:25
$begingroup$
If there are $n$ vectors but the $i$th coordinate is always zero then its dependent (more generally: if there are $n-k$ vectors and $k+1$ many coordinates which are always 0)
$endgroup$
– Ben
Jan 21 at 4:27
$begingroup$
Note that the quick way to check this is to set up the linear independence equation in matrix form, then ask a computer to compute the coefficient matrix's determinant. (It's good to remember that problems often have better solutions than what a class forces you to use.)
$endgroup$
– rwbogl
Jan 21 at 5:14
$begingroup$
Thanks for the help guys!
$endgroup$
– ming
Jan 21 at 5:53
add a comment |
1
$begingroup$
I had a typo in my previous comment (and it's been more than 5 minutes so I'm fixing it here): If you have $n+1$ (or more) vectors, they must be linearly dependent. If you can see that one is a multiple of another vector (or a sum of two or more vectors), then the set is linearly dependent. This is not always easy to see, however.
$endgroup$
– JavaMan
Jan 21 at 4:23
$begingroup$
If the set contains the zero vector its dependent
$endgroup$
– Ben
Jan 21 at 4:25
$begingroup$
If there are $n$ vectors but the $i$th coordinate is always zero then its dependent (more generally: if there are $n-k$ vectors and $k+1$ many coordinates which are always 0)
$endgroup$
– Ben
Jan 21 at 4:27
$begingroup$
Note that the quick way to check this is to set up the linear independence equation in matrix form, then ask a computer to compute the coefficient matrix's determinant. (It's good to remember that problems often have better solutions than what a class forces you to use.)
$endgroup$
– rwbogl
Jan 21 at 5:14
$begingroup$
Thanks for the help guys!
$endgroup$
– ming
Jan 21 at 5:53
1
1
$begingroup$
I had a typo in my previous comment (and it's been more than 5 minutes so I'm fixing it here): If you have $n+1$ (or more) vectors, they must be linearly dependent. If you can see that one is a multiple of another vector (or a sum of two or more vectors), then the set is linearly dependent. This is not always easy to see, however.
$endgroup$
– JavaMan
Jan 21 at 4:23
$begingroup$
I had a typo in my previous comment (and it's been more than 5 minutes so I'm fixing it here): If you have $n+1$ (or more) vectors, they must be linearly dependent. If you can see that one is a multiple of another vector (or a sum of two or more vectors), then the set is linearly dependent. This is not always easy to see, however.
$endgroup$
– JavaMan
Jan 21 at 4:23
$begingroup$
If the set contains the zero vector its dependent
$endgroup$
– Ben
Jan 21 at 4:25
$begingroup$
If the set contains the zero vector its dependent
$endgroup$
– Ben
Jan 21 at 4:25
$begingroup$
If there are $n$ vectors but the $i$th coordinate is always zero then its dependent (more generally: if there are $n-k$ vectors and $k+1$ many coordinates which are always 0)
$endgroup$
– Ben
Jan 21 at 4:27
$begingroup$
If there are $n$ vectors but the $i$th coordinate is always zero then its dependent (more generally: if there are $n-k$ vectors and $k+1$ many coordinates which are always 0)
$endgroup$
– Ben
Jan 21 at 4:27
$begingroup$
Note that the quick way to check this is to set up the linear independence equation in matrix form, then ask a computer to compute the coefficient matrix's determinant. (It's good to remember that problems often have better solutions than what a class forces you to use.)
$endgroup$
– rwbogl
Jan 21 at 5:14
$begingroup$
Note that the quick way to check this is to set up the linear independence equation in matrix form, then ask a computer to compute the coefficient matrix's determinant. (It's good to remember that problems often have better solutions than what a class forces you to use.)
$endgroup$
– rwbogl
Jan 21 at 5:14
$begingroup$
Thanks for the help guys!
$endgroup$
– ming
Jan 21 at 5:53
$begingroup$
Thanks for the help guys!
$endgroup$
– ming
Jan 21 at 5:53
add a comment |
2 Answers
2
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oldest
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$begingroup$
- The set contains the zero vector
- You can write an element as a linear combination of the others
- If the dimension of your vector space is $n$, then any set that contains more $n$ will be linearly dependent.
The first and third points are easy to check, while the second will just require some quick arithmetic.
answered Jan 21 at 4:25
MetricMetric
1,23649
$endgroup$
add a comment |
$begingroup$
The two basic observations, as others have pointed out, are that if there are more vectors than the dimension of the vector space, or if the zero vector is in the set, you surely have a dependent set. Additionally if there are identical vectors in the set, the set is dependent too (obviously). Other than this, there is no easy and quick way to check whether a set of vectors is dependant without using matrices (if you are allowed, just use row reduction to find the rank of the matrix whose column vectors are in the set, which conveniently tells you the number of independent vectors as well).
answered Jan 21 at 5:08
YiFanYiFan
4,4711627
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
- The set contains the zero vector
- You can write an element as a linear combination of the others
- If the dimension of your vector space is $n$, then any set that contains more $n$ will be linearly dependent.
The first and third points are easy to check, while the second will just require some quick arithmetic.
answered Jan 21 at 4:25
MetricMetric
1,23649
$endgroup$
add a comment |
$begingroup$
- The set contains the zero vector
- You can write an element as a linear combination of the others
- If the dimension of your vector space is $n$, then any set that contains more $n$ will be linearly dependent.
The first and third points are easy to check, while the second will just require some quick arithmetic.
answered Jan 21 at 4:25
MetricMetric
1,23649
$endgroup$
add a comment |
$begingroup$
- The set contains the zero vector
- You can write an element as a linear combination of the others
- If the dimension of your vector space is $n$, then any set that contains more $n$ will be linearly dependent.
The first and third points are easy to check, while the second will just require some quick arithmetic.
answered Jan 21 at 4:25
MetricMetric
1,23649
$endgroup$
- The set contains the zero vector
- You can write an element as a linear combination of the others
- If the dimension of your vector space is $n$, then any set that contains more $n$ will be linearly dependent.
The first and third points are easy to check, while the second will just require some quick arithmetic.
answered Jan 21 at 4:25
MetricMetric
1,23649
answered Jan 21 at 4:25
MetricMetric
1,23649
answered Jan 21 at 4:25
MetricMetric
1,23649
answered Jan 21 at 4:25
MetricMetric
1,23649
1,23649
add a comment |
add a comment |
$begingroup$
The two basic observations, as others have pointed out, are that if there are more vectors than the dimension of the vector space, or if the zero vector is in the set, you surely have a dependent set. Additionally if there are identical vectors in the set, the set is dependent too (obviously). Other than this, there is no easy and quick way to check whether a set of vectors is dependant without using matrices (if you are allowed, just use row reduction to find the rank of the matrix whose column vectors are in the set, which conveniently tells you the number of independent vectors as well).
answered Jan 21 at 5:08
YiFanYiFan
4,4711627
$endgroup$
add a comment |
$begingroup$
The two basic observations, as others have pointed out, are that if there are more vectors than the dimension of the vector space, or if the zero vector is in the set, you surely have a dependent set. Additionally if there are identical vectors in the set, the set is dependent too (obviously). Other than this, there is no easy and quick way to check whether a set of vectors is dependant without using matrices (if you are allowed, just use row reduction to find the rank of the matrix whose column vectors are in the set, which conveniently tells you the number of independent vectors as well).
answered Jan 21 at 5:08
YiFanYiFan
4,4711627
$endgroup$
add a comment |
$begingroup$
The two basic observations, as others have pointed out, are that if there are more vectors than the dimension of the vector space, or if the zero vector is in the set, you surely have a dependent set. Additionally if there are identical vectors in the set, the set is dependent too (obviously). Other than this, there is no easy and quick way to check whether a set of vectors is dependant without using matrices (if you are allowed, just use row reduction to find the rank of the matrix whose column vectors are in the set, which conveniently tells you the number of independent vectors as well).
answered Jan 21 at 5:08
YiFanYiFan
4,4711627
$endgroup$
The two basic observations, as others have pointed out, are that if there are more vectors than the dimension of the vector space, or if the zero vector is in the set, you surely have a dependent set. Additionally if there are identical vectors in the set, the set is dependent too (obviously). Other than this, there is no easy and quick way to check whether a set of vectors is dependant without using matrices (if you are allowed, just use row reduction to find the rank of the matrix whose column vectors are in the set, which conveniently tells you the number of independent vectors as well).
answered Jan 21 at 5:08
YiFanYiFan
4,4711627
answered Jan 21 at 5:08
YiFanYiFan
4,4711627
answered Jan 21 at 5:08
YiFanYiFan
4,4711627
answered Jan 21 at 5:08
YiFanYiFan
4,4711627
4,4711627
add a comment |
add a comment |
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$begingroup$
I had a typo in my previous comment (and it's been more than 5 minutes so I'm fixing it here): If you have $n+1$ (or more) vectors, they must be linearly dependent. If you can see that one is a multiple of another vector (or a sum of two or more vectors), then the set is linearly dependent. This is not always easy to see, however.
$endgroup$
– JavaMan
Jan 21 at 4:23
$begingroup$
If the set contains the zero vector its dependent
$endgroup$
– Ben
Jan 21 at 4:25
$begingroup$
If there are $n$ vectors but the $i$th coordinate is always zero then its dependent (more generally: if there are $n-k$ vectors and $k+1$ many coordinates which are always 0)
$endgroup$
– Ben
Jan 21 at 4:27
$begingroup$
Note that the quick way to check this is to set up the linear independence equation in matrix form, then ask a computer to compute the coefficient matrix's determinant. (It's good to remember that problems often have better solutions than what a class forces you to use.)
$endgroup$
– rwbogl
Jan 21 at 5:14
$begingroup$
Thanks for the help guys!
$endgroup$
– ming
Jan 21 at 5:53