non-square matrix determinants to see if they form basis or span a set
$begingroup$
Just a question. Let's say that you are given four vectors: ${v_1=[1,0,0] v_2=[0,2,0], v_3=[0,0,3] v_4=[1,1,1].}$. And with the help of the determinant of a matrix containing these vectors in column form, I have to show that they span $R^3$ but do not form a basis for $R^3.$
Obviously, there is another way to attack this problem. By having it in 3x4 matrix form just as mentioned above and reducing it to reduced row echelon form gives you that the v4 vector does not have a pivot, then therefore it can be thrown out of the set. Now you are left with three vectors in your set, and therefore they DO span R3 and also form a basis since they are linearly independent unlike before when you actually had 4 vectors and v4 was dependent on the previous vectors. Is this way of thinking right? I know that I have my book here, but it is useless.
In order for a set of vectors to span (Lets say R3) and form a basis for R3, they must be linearly independent and also the amount of vectors must be equivalent to the n-dimension. Is this right thinking?
That is my first question, and my second question is whether the general theorem of nxn matrices that say that if det is nonzero, then the vector set span Rn and are linearly independent. Well, basically if this theorem also can be applied to mxn matrices as far as the determinant goes when determining spanning and linearly independence? .. If not, how do I do i show that my vectors do span r3 but dont form a basis in r3 with determinant when having an mxn matrix which my vectors create in column form?...
linear-algebra vector-spaces determinant
edited Dec 25 '18 at 0:25
Namaste
1
asked Dec 24 '18 at 21:57
mjalmjal
91
$endgroup$
add a comment |
$begingroup$
Just a question. Let's say that you are given four vectors: ${v_1=[1,0,0] v_2=[0,2,0], v_3=[0,0,3] v_4=[1,1,1].}$. And with the help of the determinant of a matrix containing these vectors in column form, I have to show that they span $R^3$ but do not form a basis for $R^3.$
Obviously, there is another way to attack this problem. By having it in 3x4 matrix form just as mentioned above and reducing it to reduced row echelon form gives you that the v4 vector does not have a pivot, then therefore it can be thrown out of the set. Now you are left with three vectors in your set, and therefore they DO span R3 and also form a basis since they are linearly independent unlike before when you actually had 4 vectors and v4 was dependent on the previous vectors. Is this way of thinking right? I know that I have my book here, but it is useless.
In order for a set of vectors to span (Lets say R3) and form a basis for R3, they must be linearly independent and also the amount of vectors must be equivalent to the n-dimension. Is this right thinking?
That is my first question, and my second question is whether the general theorem of nxn matrices that say that if det is nonzero, then the vector set span Rn and are linearly independent. Well, basically if this theorem also can be applied to mxn matrices as far as the determinant goes when determining spanning and linearly independence? .. If not, how do I do i show that my vectors do span r3 but dont form a basis in r3 with determinant when having an mxn matrix which my vectors create in column form?...
linear-algebra vector-spaces determinant
edited Dec 25 '18 at 0:25
Namaste
1
asked Dec 24 '18 at 21:57
mjalmjal
91
$endgroup$
$begingroup$
How are you defining a determinant for non-square matrices?
$endgroup$
– Math1000
Dec 24 '18 at 21:59
add a comment |
$begingroup$
Just a question. Let's say that you are given four vectors: ${v_1=[1,0,0] v_2=[0,2,0], v_3=[0,0,3] v_4=[1,1,1].}$. And with the help of the determinant of a matrix containing these vectors in column form, I have to show that they span $R^3$ but do not form a basis for $R^3.$
Obviously, there is another way to attack this problem. By having it in 3x4 matrix form just as mentioned above and reducing it to reduced row echelon form gives you that the v4 vector does not have a pivot, then therefore it can be thrown out of the set. Now you are left with three vectors in your set, and therefore they DO span R3 and also form a basis since they are linearly independent unlike before when you actually had 4 vectors and v4 was dependent on the previous vectors. Is this way of thinking right? I know that I have my book here, but it is useless.
In order for a set of vectors to span (Lets say R3) and form a basis for R3, they must be linearly independent and also the amount of vectors must be equivalent to the n-dimension. Is this right thinking?
That is my first question, and my second question is whether the general theorem of nxn matrices that say that if det is nonzero, then the vector set span Rn and are linearly independent. Well, basically if this theorem also can be applied to mxn matrices as far as the determinant goes when determining spanning and linearly independence? .. If not, how do I do i show that my vectors do span r3 but dont form a basis in r3 with determinant when having an mxn matrix which my vectors create in column form?...
linear-algebra vector-spaces determinant
edited Dec 25 '18 at 0:25
Namaste
1
asked Dec 24 '18 at 21:57
mjalmjal
91
$endgroup$
Just a question. Let's say that you are given four vectors: ${v_1=[1,0,0] v_2=[0,2,0], v_3=[0,0,3] v_4=[1,1,1].}$. And with the help of the determinant of a matrix containing these vectors in column form, I have to show that they span $R^3$ but do not form a basis for $R^3.$
Obviously, there is another way to attack this problem. By having it in 3x4 matrix form just as mentioned above and reducing it to reduced row echelon form gives you that the v4 vector does not have a pivot, then therefore it can be thrown out of the set. Now you are left with three vectors in your set, and therefore they DO span R3 and also form a basis since they are linearly independent unlike before when you actually had 4 vectors and v4 was dependent on the previous vectors. Is this way of thinking right? I know that I have my book here, but it is useless.
In order for a set of vectors to span (Lets say R3) and form a basis for R3, they must be linearly independent and also the amount of vectors must be equivalent to the n-dimension. Is this right thinking?
That is my first question, and my second question is whether the general theorem of nxn matrices that say that if det is nonzero, then the vector set span Rn and are linearly independent. Well, basically if this theorem also can be applied to mxn matrices as far as the determinant goes when determining spanning and linearly independence? .. If not, how do I do i show that my vectors do span r3 but dont form a basis in r3 with determinant when having an mxn matrix which my vectors create in column form?...
linear-algebra vector-spaces determinant
linear-algebra vector-spaces determinant
edited Dec 25 '18 at 0:25
Namaste
1
asked Dec 24 '18 at 21:57
mjalmjal
91
edited Dec 25 '18 at 0:25
Namaste
1
asked Dec 24 '18 at 21:57
mjalmjal
91
edited Dec 25 '18 at 0:25
Namaste
1
edited Dec 25 '18 at 0:25
Namaste
1
edited Dec 25 '18 at 0:25
Namaste
1
1
asked Dec 24 '18 at 21:57
mjalmjal
91
asked Dec 24 '18 at 21:57
mjalmjal
91
asked Dec 24 '18 at 21:57
mjalmjal
91
91
$begingroup$
How are you defining a determinant for non-square matrices?
$endgroup$
– Math1000
Dec 24 '18 at 21:59
add a comment |
$begingroup$
How are you defining a determinant for non-square matrices?
$endgroup$
– Math1000
Dec 24 '18 at 21:59
$begingroup$
How are you defining a determinant for non-square matrices?
$endgroup$
– Math1000
Dec 24 '18 at 21:59
$begingroup$
How are you defining a determinant for non-square matrices?
$endgroup$
– Math1000
Dec 24 '18 at 21:59
add a comment |
1 Answer
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oldest
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$begingroup$
I think you understand this material well.
The natural way to determine whether these vectors span is, as you say, to calculate the reduced row echelon form. They can't form a basis since there are four of them in a three dimensional space.
You can use determinants to show that they span, but it's not pretty. You look at the four possible $3 times 3$ matrices formed by omitting each of the four vectors in turn. When you find one with a nonzero determinant you have a basis, so a spanning set.
answered Dec 25 '18 at 0:20
Ethan BolkerEthan Bolker
45.7k553120
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add a comment |
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1 Answer
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$begingroup$
I think you understand this material well.
The natural way to determine whether these vectors span is, as you say, to calculate the reduced row echelon form. They can't form a basis since there are four of them in a three dimensional space.
You can use determinants to show that they span, but it's not pretty. You look at the four possible $3 times 3$ matrices formed by omitting each of the four vectors in turn. When you find one with a nonzero determinant you have a basis, so a spanning set.
answered Dec 25 '18 at 0:20
Ethan BolkerEthan Bolker
45.7k553120
$endgroup$
add a comment |
$begingroup$
I think you understand this material well.
The natural way to determine whether these vectors span is, as you say, to calculate the reduced row echelon form. They can't form a basis since there are four of them in a three dimensional space.
You can use determinants to show that they span, but it's not pretty. You look at the four possible $3 times 3$ matrices formed by omitting each of the four vectors in turn. When you find one with a nonzero determinant you have a basis, so a spanning set.
answered Dec 25 '18 at 0:20
Ethan BolkerEthan Bolker
45.7k553120
$endgroup$
add a comment |
$begingroup$
I think you understand this material well.
The natural way to determine whether these vectors span is, as you say, to calculate the reduced row echelon form. They can't form a basis since there are four of them in a three dimensional space.
You can use determinants to show that they span, but it's not pretty. You look at the four possible $3 times 3$ matrices formed by omitting each of the four vectors in turn. When you find one with a nonzero determinant you have a basis, so a spanning set.
answered Dec 25 '18 at 0:20
Ethan BolkerEthan Bolker
45.7k553120
$endgroup$
I think you understand this material well.
The natural way to determine whether these vectors span is, as you say, to calculate the reduced row echelon form. They can't form a basis since there are four of them in a three dimensional space.
You can use determinants to show that they span, but it's not pretty. You look at the four possible $3 times 3$ matrices formed by omitting each of the four vectors in turn. When you find one with a nonzero determinant you have a basis, so a spanning set.
answered Dec 25 '18 at 0:20
Ethan BolkerEthan Bolker
45.7k553120
answered Dec 25 '18 at 0:20
Ethan BolkerEthan Bolker
45.7k553120
answered Dec 25 '18 at 0:20
Ethan BolkerEthan Bolker
45.7k553120
answered Dec 25 '18 at 0:20
Ethan BolkerEthan Bolker
45.7k553120
45.7k553120
add a comment |
add a comment |
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$begingroup$
How are you defining a determinant for non-square matrices?
$endgroup$
– Math1000
Dec 24 '18 at 21:59