How to determine if a set spans a vector space

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$lbrace (2, -6), (-1,4), (-3, 9)rbrace$

I row reduced this to get the rows $(2,-1,-3); (0,1,0)$. I want to check if this spans $mathbbR^2$. I don't see exactly what to call the free variables. If I use $c_1, c_2, c_3$ for each column respectively then I get $c_2= b$ and $2c_1 - c_2 -3c_3 = a$. $c_3$ is always allowed to be anything. What decides if I have two free variables?

linear-algebra

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edited Jul 29 at 22:17

高田航

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asked Jul 29 at 21:56

Kenisha Stills

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up vote
1
down vote

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$lbrace (2, -6), (-1,4), (-3, 9)rbrace$

I row reduced this to get the rows $(2,-1,-3); (0,1,0)$. I want to check if this spans $mathbbR^2$. I don't see exactly what to call the free variables. If I use $c_1, c_2, c_3$ for each column respectively then I get $c_2= b$ and $2c_1 - c_2 -3c_3 = a$. $c_3$ is always allowed to be anything. What decides if I have two free variables?

linear-algebra

share|cite|improve this question

edited Jul 29 at 22:17

高田航

1,116318

asked Jul 29 at 21:56

Kenisha Stills

111

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

$lbrace (2, -6), (-1,4), (-3, 9)rbrace$

I row reduced this to get the rows $(2,-1,-3); (0,1,0)$. I want to check if this spans $mathbbR^2$. I don't see exactly what to call the free variables. If I use $c_1, c_2, c_3$ for each column respectively then I get $c_2= b$ and $2c_1 - c_2 -3c_3 = a$. $c_3$ is always allowed to be anything. What decides if I have two free variables?

linear-algebra

share|cite|improve this question

edited Jul 29 at 22:17

高田航

1,116318

asked Jul 29 at 21:56

Kenisha Stills

111

$lbrace (2, -6), (-1,4), (-3, 9)rbrace$

I row reduced this to get the rows $(2,-1,-3); (0,1,0)$. I want to check if this spans $mathbbR^2$. I don't see exactly what to call the free variables. If I use $c_1, c_2, c_3$ for each column respectively then I get $c_2= b$ and $2c_1 - c_2 -3c_3 = a$. $c_3$ is always allowed to be anything. What decides if I have two free variables?

linear-algebra

share|cite|improve this question

edited Jul 29 at 22:17

高田航

1,116318

asked Jul 29 at 21:56

Kenisha Stills

111

share|cite|improve this question

share|cite|improve this question

edited Jul 29 at 22:17

高田航

1,116318

edited Jul 29 at 22:17

高田航

1,116318

edited Jul 29 at 22:17

高田航

1,116318

1,116318

asked Jul 29 at 21:56

Kenisha Stills

111

asked Jul 29 at 21:56

Kenisha Stills

111

asked Jul 29 at 21:56

Kenisha Stills

111

111

add a comment | 

add a comment | 

2 Answers
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oldest

votes

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0
down vote

We know that the set $S=lbrace langle 2,-6rangle,langle -1,4rangle,langle -3,9ranglerbrace$ spans $mathbbR^2$ if there are exactly two linearly independent vectors in $S$. Observe that $$-frac32cdotlangle2,-6rangle=langle-3,9rangle$$
so these two vectors are not linearly independent. To test if $langle -1,4rangle$ is independent of the other two vectors, evaluate
$$textRREFbeginpmatrix
-1 & 4 \
2 & -6
endpmatrix=beginpmatrix
1 & 0\
0 & 1
endpmatrix=I_2$$
Because the matrix formed by the two vectors is row equivalent to the $2times 2$ identity matrix, they are linearly independent. Therefore, $S$ spans $mathbbR^2$.

share|cite|improve this answer

answered Jul 29 at 22:13

高田航

1,116318

add a comment | 

up vote
0
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The trick to putting the vectors in a matrix is that row reduction preserves linear independence of the columns. So to determine which of the original columns are independent, just check which columns in the reduced matrix are. Thus, for instance, since columns $1$ and $2$ in the reduced matrix are independent, so are $1$ and $2$ from the original vectors.

But, it should be noted that since you only need $2$ vectors to span $mathbb R^2$, and since checking if two vectors are independent just means checking if they are multiples of each other, this trick is barely necessary here...

Addressing your work at the end, use $c_1,c_2,c_3$ for the coordinates (clearly what you meant). Then $c_2=b$, if you choose a specific $(a,b)$ in the image. Then note that $2c_1-b-3c_3=a$. So choose $c_3$ freely and then $c_1$ is determined. Or vice versa (choose any $c_1$ and then $c_3$ is determined). So, the solution space for a given $(a,b)$ is only $1$-dimensional (but that there's a solution at all is what we're interested in). Incidentally, $1$ is also the nullity.

share|cite|improve this answer

edited Jul 30 at 18:52

answered Jul 29 at 22:29

Chris Custer

5,2722622

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm not sure if I'm saying the same thing, but it looks to me that there's only one free variable. My teacher keeps saying that there's two though.
  – Kenisha Stills
  Jul 31 at 13:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The dimension of the solution space is $1$. The dimension of the image is $2$. You're confusing the two...
  – Chris Custer
  Jul 31 at 17:30
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

We know that the set $S=lbrace langle 2,-6rangle,langle -1,4rangle,langle -3,9ranglerbrace$ spans $mathbbR^2$ if there are exactly two linearly independent vectors in $S$. Observe that $$-frac32cdotlangle2,-6rangle=langle-3,9rangle$$
so these two vectors are not linearly independent. To test if $langle -1,4rangle$ is independent of the other two vectors, evaluate
$$textRREFbeginpmatrix
-1 & 4 \
2 & -6
endpmatrix=beginpmatrix
1 & 0\
0 & 1
endpmatrix=I_2$$
Because the matrix formed by the two vectors is row equivalent to the $2times 2$ identity matrix, they are linearly independent. Therefore, $S$ spans $mathbbR^2$.

share|cite|improve this answer

answered Jul 29 at 22:13

高田航

1,116318

add a comment | 

up vote
0
down vote

We know that the set $S=lbrace langle 2,-6rangle,langle -1,4rangle,langle -3,9ranglerbrace$ spans $mathbbR^2$ if there are exactly two linearly independent vectors in $S$. Observe that $$-frac32cdotlangle2,-6rangle=langle-3,9rangle$$
so these two vectors are not linearly independent. To test if $langle -1,4rangle$ is independent of the other two vectors, evaluate
$$textRREFbeginpmatrix
-1 & 4 \
2 & -6
endpmatrix=beginpmatrix
1 & 0\
0 & 1
endpmatrix=I_2$$
Because the matrix formed by the two vectors is row equivalent to the $2times 2$ identity matrix, they are linearly independent. Therefore, $S$ spans $mathbbR^2$.

share|cite|improve this answer

answered Jul 29 at 22:13

高田航

1,116318

add a comment | 

up vote
0
down vote

up vote
0
down vote

We know that the set $S=lbrace langle 2,-6rangle,langle -1,4rangle,langle -3,9ranglerbrace$ spans $mathbbR^2$ if there are exactly two linearly independent vectors in $S$. Observe that $$-frac32cdotlangle2,-6rangle=langle-3,9rangle$$
so these two vectors are not linearly independent. To test if $langle -1,4rangle$ is independent of the other two vectors, evaluate
$$textRREFbeginpmatrix
-1 & 4 \
2 & -6
endpmatrix=beginpmatrix
1 & 0\
0 & 1
endpmatrix=I_2$$
Because the matrix formed by the two vectors is row equivalent to the $2times 2$ identity matrix, they are linearly independent. Therefore, $S$ spans $mathbbR^2$.

share|cite|improve this answer

answered Jul 29 at 22:13

高田航

1,116318

We know that the set $S=lbrace langle 2,-6rangle,langle -1,4rangle,langle -3,9ranglerbrace$ spans $mathbbR^2$ if there are exactly two linearly independent vectors in $S$. Observe that $$-frac32cdotlangle2,-6rangle=langle-3,9rangle$$
so these two vectors are not linearly independent. To test if $langle -1,4rangle$ is independent of the other two vectors, evaluate
$$textRREFbeginpmatrix
-1 & 4 \
2 & -6
endpmatrix=beginpmatrix
1 & 0\
0 & 1
endpmatrix=I_2$$
Because the matrix formed by the two vectors is row equivalent to the $2times 2$ identity matrix, they are linearly independent. Therefore, $S$ spans $mathbbR^2$.

share|cite|improve this answer

answered Jul 29 at 22:13

高田航

1,116318

share|cite|improve this answer

share|cite|improve this answer

answered Jul 29 at 22:13

高田航

1,116318

answered Jul 29 at 22:13

高田航

1,116318

answered Jul 29 at 22:13

高田航

1,116318

1,116318

add a comment | 

add a comment | 

up vote
0
down vote

The trick to putting the vectors in a matrix is that row reduction preserves linear independence of the columns. So to determine which of the original columns are independent, just check which columns in the reduced matrix are. Thus, for instance, since columns $1$ and $2$ in the reduced matrix are independent, so are $1$ and $2$ from the original vectors.

But, it should be noted that since you only need $2$ vectors to span $mathbb R^2$, and since checking if two vectors are independent just means checking if they are multiples of each other, this trick is barely necessary here...

Addressing your work at the end, use $c_1,c_2,c_3$ for the coordinates (clearly what you meant). Then $c_2=b$, if you choose a specific $(a,b)$ in the image. Then note that $2c_1-b-3c_3=a$. So choose $c_3$ freely and then $c_1$ is determined. Or vice versa (choose any $c_1$ and then $c_3$ is determined). So, the solution space for a given $(a,b)$ is only $1$-dimensional (but that there's a solution at all is what we're interested in). Incidentally, $1$ is also the nullity.

share|cite|improve this answer

edited Jul 30 at 18:52

answered Jul 29 at 22:29

Chris Custer

5,2722622

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm not sure if I'm saying the same thing, but it looks to me that there's only one free variable. My teacher keeps saying that there's two though.
  – Kenisha Stills
  Jul 31 at 13:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The dimension of the solution space is $1$. The dimension of the image is $2$. You're confusing the two...
  – Chris Custer
  Jul 31 at 17:30
  

  
  
  
  

add a comment | 

up vote
0
down vote

The trick to putting the vectors in a matrix is that row reduction preserves linear independence of the columns. So to determine which of the original columns are independent, just check which columns in the reduced matrix are. Thus, for instance, since columns $1$ and $2$ in the reduced matrix are independent, so are $1$ and $2$ from the original vectors.

But, it should be noted that since you only need $2$ vectors to span $mathbb R^2$, and since checking if two vectors are independent just means checking if they are multiples of each other, this trick is barely necessary here...

Addressing your work at the end, use $c_1,c_2,c_3$ for the coordinates (clearly what you meant). Then $c_2=b$, if you choose a specific $(a,b)$ in the image. Then note that $2c_1-b-3c_3=a$. So choose $c_3$ freely and then $c_1$ is determined. Or vice versa (choose any $c_1$ and then $c_3$ is determined). So, the solution space for a given $(a,b)$ is only $1$-dimensional (but that there's a solution at all is what we're interested in). Incidentally, $1$ is also the nullity.

share|cite|improve this answer

edited Jul 30 at 18:52

answered Jul 29 at 22:29

Chris Custer

5,2722622

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm not sure if I'm saying the same thing, but it looks to me that there's only one free variable. My teacher keeps saying that there's two though.
  – Kenisha Stills
  Jul 31 at 13:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The dimension of the solution space is $1$. The dimension of the image is $2$. You're confusing the two...
  – Chris Custer
  Jul 31 at 17:30
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

The trick to putting the vectors in a matrix is that row reduction preserves linear independence of the columns. So to determine which of the original columns are independent, just check which columns in the reduced matrix are. Thus, for instance, since columns $1$ and $2$ in the reduced matrix are independent, so are $1$ and $2$ from the original vectors.

But, it should be noted that since you only need $2$ vectors to span $mathbb R^2$, and since checking if two vectors are independent just means checking if they are multiples of each other, this trick is barely necessary here...

Addressing your work at the end, use $c_1,c_2,c_3$ for the coordinates (clearly what you meant). Then $c_2=b$, if you choose a specific $(a,b)$ in the image. Then note that $2c_1-b-3c_3=a$. So choose $c_3$ freely and then $c_1$ is determined. Or vice versa (choose any $c_1$ and then $c_3$ is determined). So, the solution space for a given $(a,b)$ is only $1$-dimensional (but that there's a solution at all is what we're interested in). Incidentally, $1$ is also the nullity.

share|cite|improve this answer

edited Jul 30 at 18:52

answered Jul 29 at 22:29

Chris Custer

5,2722622

The trick to putting the vectors in a matrix is that row reduction preserves linear independence of the columns. So to determine which of the original columns are independent, just check which columns in the reduced matrix are. Thus, for instance, since columns $1$ and $2$ in the reduced matrix are independent, so are $1$ and $2$ from the original vectors.

But, it should be noted that since you only need $2$ vectors to span $mathbb R^2$, and since checking if two vectors are independent just means checking if they are multiples of each other, this trick is barely necessary here...

Addressing your work at the end, use $c_1,c_2,c_3$ for the coordinates (clearly what you meant). Then $c_2=b$, if you choose a specific $(a,b)$ in the image. Then note that $2c_1-b-3c_3=a$. So choose $c_3$ freely and then $c_1$ is determined. Or vice versa (choose any $c_1$ and then $c_3$ is determined). So, the solution space for a given $(a,b)$ is only $1$-dimensional (but that there's a solution at all is what we're interested in). Incidentally, $1$ is also the nullity.

share|cite|improve this answer

edited Jul 30 at 18:52

answered Jul 29 at 22:29

Chris Custer

5,2722622

share|cite|improve this answer

share|cite|improve this answer

edited Jul 30 at 18:52

edited Jul 30 at 18:52

edited Jul 30 at 18:52

answered Jul 29 at 22:29

Chris Custer

5,2722622

answered Jul 29 at 22:29

Chris Custer

5,2722622

answered Jul 29 at 22:29

Chris Custer

5,2722622

5,2722622

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm not sure if I'm saying the same thing, but it looks to me that there's only one free variable. My teacher keeps saying that there's two though.
  – Kenisha Stills
  Jul 31 at 13:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The dimension of the solution space is $1$. The dimension of the image is $2$. You're confusing the two...
  – Chris Custer
  Jul 31 at 17:30
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm not sure if I'm saying the same thing, but it looks to me that there's only one free variable. My teacher keeps saying that there's two though.
  – Kenisha Stills
  Jul 31 at 13:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The dimension of the solution space is $1$. The dimension of the image is $2$. You're confusing the two...
  – Chris Custer
  Jul 31 at 17:30
  

  
  
  
  

I'm not sure if I'm saying the same thing, but it looks to me that there's only one free variable. My teacher keeps saying that there's two though.
– Kenisha Stills
Jul 31 at 13:22

I'm not sure if I'm saying the same thing, but it looks to me that there's only one free variable. My teacher keeps saying that there's two though.
– Kenisha Stills
Jul 31 at 13:22

The dimension of the solution space is $1$. The dimension of the image is $2$. You're confusing the two...
– Chris Custer
Jul 31 at 17:30

The dimension of the solution space is $1$. The dimension of the image is $2$. You're confusing the two...
– Chris Custer
Jul 31 at 17:30

add a comment | 

 

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