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Basis for nullspace - Free variables and basis for $N(A)$

Find a basis for the range and kernel of $T$.To find Basis and kernel of matrix AFind the basis for kernel (nullspace) of matrix (eigenspaces)Method for finding basis of an imageFinding basis for column space of matrixfind basis for kernel and columnsFor the following system, give a succinct description of the set of solutions.Determining whether equation belongs to span

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1

$begingroup$

I'm trying to understand the solution for a textbook question where I am asked to find a basis for $N(A)$. I took a screenshot of the page, and I circled the portion I don't quite understand. I don't understand where the values for $s$ and $t$ came from? Like how did they get $s = 1$, and $t = -2$ for the first part in that tuple? I tried finding a correlation between these values and the RREF matrix but I just couldn't see the connection. Any help or explanation would be appreciated.

linear-algebra matrices

share|cite|improve this question

edited Apr 16 at 8:11

Asaf Karagila♦

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asked Apr 16 at 5:42

GilmoreGirlingGilmoreGirling

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know about free variables?
  $endgroup$
  – Tojrah
  Apr 16 at 5:45
  

  
  
  
  

add a comment
 | 

1

$begingroup$

I'm trying to understand the solution for a textbook question where I am asked to find a basis for $N(A)$. I took a screenshot of the page, and I circled the portion I don't quite understand. I don't understand where the values for $s$ and $t$ came from? Like how did they get $s = 1$, and $t = -2$ for the first part in that tuple? I tried finding a correlation between these values and the RREF matrix but I just couldn't see the connection. Any help or explanation would be appreciated.

linear-algebra matrices

share|cite|improve this question

edited Apr 16 at 8:11

Asaf Karagila♦

316k3535 gold badges457457 silver badges795795 bronze badges

asked Apr 16 at 5:42

GilmoreGirlingGilmoreGirling

11577 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know about free variables?
  $endgroup$
  – Tojrah
  Apr 16 at 5:45
  

  
  
  
  

add a comment
 | 

$begingroup$

I'm trying to understand the solution for a textbook question where I am asked to find a basis for $N(A)$. I took a screenshot of the page, and I circled the portion I don't quite understand. I don't understand where the values for $s$ and $t$ came from? Like how did they get $s = 1$, and $t = -2$ for the first part in that tuple? I tried finding a correlation between these values and the RREF matrix but I just couldn't see the connection. Any help or explanation would be appreciated.

linear-algebra matrices

share|cite|improve this question

edited Apr 16 at 8:11

Asaf Karagila♦

316k3535 gold badges457457 silver badges795795 bronze badges

asked Apr 16 at 5:42

GilmoreGirlingGilmoreGirling

11577 bronze badges

$endgroup$

share|cite|improve this question

edited Apr 16 at 8:11

Asaf Karagila♦

316k3535 gold badges457457 silver badges795795 bronze badges

asked Apr 16 at 5:42

GilmoreGirlingGilmoreGirling

11577 bronze badges

share|cite|improve this question

edited Apr 16 at 8:11

Asaf Karagila♦

316k3535 gold badges457457 silver badges795795 bronze badges

asked Apr 16 at 5:42

GilmoreGirlingGilmoreGirling

11577 bronze badges

edited Apr 16 at 8:11

Asaf Karagila♦

316k3535 gold badges457457 silver badges795795 bronze badges

edited Apr 16 at 8:11

Asaf Karagila♦

316k3535 gold badges457457 silver badges795795 bronze badges

asked Apr 16 at 5:42

GilmoreGirlingGilmoreGirling

11577 bronze badges

asked Apr 16 at 5:42

GilmoreGirlingGilmoreGirling

11577 bronze badges

2 Answers
2

active

oldest

votes

3

$begingroup$

We already transform $A$ into $$beginpmatrix 1&-1&0&2\0&0&1&-1endpmatrix$$
Since first and third column contains the pivot, so $x_1$ and $x_3$ are pivot variables. That is, $x_2$ and $x_4$ are free.
The task now is to solve $$x_1-x_2+2x_4=0\x_3-x_4=0$$
Set $x_2=1$ and $x_4=0$ to see $$beginpmatrix 1\1\0\0 endpmatrix$$ is a special solution. Similarly set $x_2=0$ and $x_4=1$ to see $$beginpmatrix -2\0\1\1 endpmatrix$$ is a special solution. Now the combination $$sbeginpmatrix 1\1\0\0 endpmatrix+tbeginpmatrix -2\0\1\1 endpmatrix$$ gives all solution s to $Ax=0$

share|cite|improve this answer

edited Apr 16 at 6:10

answered Apr 16 at 5:56

Chinnapparaj RChinnapparaj R

9,20133 gold badges1010 silver badges3333 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

Columns 1 and 3 have the pivots. So the other two columns (2 and 4) correspond to the free variables.

Then call $x_4=t$. Then the last equation says $$x_3 - x_4 =0 leftrightarrow x_3 = x_4= t$$

Call the other free variable $x_2=s$.
Then the first equation becomes, after substiting what we know so far:

$$x_1 - s + 2t = 0$$ from which

$$x_1 = s -2t$$ follows.

share|cite|improve this answer

answered Apr 16 at 5:55

Henno BrandsmaHenno Brandsma

133k44 gold badges5353 silver badges139139 bronze badges

$endgroup$

add a comment
 | 

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3

$begingroup$

We already transform $A$ into $$beginpmatrix 1&-1&0&2\0&0&1&-1endpmatrix$$
Since first and third column contains the pivot, so $x_1$ and $x_3$ are pivot variables. That is, $x_2$ and $x_4$ are free.
The task now is to solve $$x_1-x_2+2x_4=0\x_3-x_4=0$$
Set $x_2=1$ and $x_4=0$ to see $$beginpmatrix 1\1\0\0 endpmatrix$$ is a special solution. Similarly set $x_2=0$ and $x_4=1$ to see $$beginpmatrix -2\0\1\1 endpmatrix$$ is a special solution. Now the combination $$sbeginpmatrix 1\1\0\0 endpmatrix+tbeginpmatrix -2\0\1\1 endpmatrix$$ gives all solution s to $Ax=0$

share|cite|improve this answer

edited Apr 16 at 6:10

answered Apr 16 at 5:56

Chinnapparaj RChinnapparaj R

9,20133 gold badges1010 silver badges3333 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

We already transform $A$ into $$beginpmatrix 1&-1&0&2\0&0&1&-1endpmatrix$$
Since first and third column contains the pivot, so $x_1$ and $x_3$ are pivot variables. That is, $x_2$ and $x_4$ are free.
The task now is to solve $$x_1-x_2+2x_4=0\x_3-x_4=0$$
Set $x_2=1$ and $x_4=0$ to see $$beginpmatrix 1\1\0\0 endpmatrix$$ is a special solution. Similarly set $x_2=0$ and $x_4=1$ to see $$beginpmatrix -2\0\1\1 endpmatrix$$ is a special solution. Now the combination $$sbeginpmatrix 1\1\0\0 endpmatrix+tbeginpmatrix -2\0\1\1 endpmatrix$$ gives all solution s to $Ax=0$

share|cite|improve this answer

edited Apr 16 at 6:10

answered Apr 16 at 5:56

Chinnapparaj RChinnapparaj R

9,20133 gold badges1010 silver badges3333 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

We already transform $A$ into $$beginpmatrix 1&-1&0&2\0&0&1&-1endpmatrix$$
Since first and third column contains the pivot, so $x_1$ and $x_3$ are pivot variables. That is, $x_2$ and $x_4$ are free.
The task now is to solve $$x_1-x_2+2x_4=0\x_3-x_4=0$$
Set $x_2=1$ and $x_4=0$ to see $$beginpmatrix 1\1\0\0 endpmatrix$$ is a special solution. Similarly set $x_2=0$ and $x_4=1$ to see $$beginpmatrix -2\0\1\1 endpmatrix$$ is a special solution. Now the combination $$sbeginpmatrix 1\1\0\0 endpmatrix+tbeginpmatrix -2\0\1\1 endpmatrix$$ gives all solution s to $Ax=0$

share|cite|improve this answer

edited Apr 16 at 6:10

answered Apr 16 at 5:56

Chinnapparaj RChinnapparaj R

9,20133 gold badges1010 silver badges3333 bronze badges

$endgroup$

share|cite|improve this answer

edited Apr 16 at 6:10

answered Apr 16 at 5:56

Chinnapparaj RChinnapparaj R

9,20133 gold badges1010 silver badges3333 bronze badges

answered Apr 16 at 5:56

Chinnapparaj RChinnapparaj R

9,20133 gold badges1010 silver badges3333 bronze badges

answered Apr 16 at 5:56

Chinnapparaj RChinnapparaj R

9,20133 gold badges1010 silver badges3333 bronze badges

4

$begingroup$

Columns 1 and 3 have the pivots. So the other two columns (2 and 4) correspond to the free variables.

Then call $x_4=t$. Then the last equation says $$x_3 - x_4 =0 leftrightarrow x_3 = x_4= t$$

Call the other free variable $x_2=s$.
Then the first equation becomes, after substiting what we know so far:

$$x_1 - s + 2t = 0$$ from which

$$x_1 = s -2t$$ follows.

share|cite|improve this answer

answered Apr 16 at 5:55

Henno BrandsmaHenno Brandsma

133k44 gold badges5353 silver badges139139 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

Columns 1 and 3 have the pivots. So the other two columns (2 and 4) correspond to the free variables.

Then call $x_4=t$. Then the last equation says $$x_3 - x_4 =0 leftrightarrow x_3 = x_4= t$$

Call the other free variable $x_2=s$.
Then the first equation becomes, after substiting what we know so far:

$$x_1 - s + 2t = 0$$ from which

$$x_1 = s -2t$$ follows.

share|cite|improve this answer

answered Apr 16 at 5:55

Henno BrandsmaHenno Brandsma

133k44 gold badges5353 silver badges139139 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Columns 1 and 3 have the pivots. So the other two columns (2 and 4) correspond to the free variables.

Then call $x_4=t$. Then the last equation says $$x_3 - x_4 =0 leftrightarrow x_3 = x_4= t$$

Call the other free variable $x_2=s$.
Then the first equation becomes, after substiting what we know so far:

$$x_1 - s + 2t = 0$$ from which

$$x_1 = s -2t$$ follows.

share|cite|improve this answer

answered Apr 16 at 5:55

Henno BrandsmaHenno Brandsma

133k44 gold badges5353 silver badges139139 bronze badges

$endgroup$

share|cite|improve this answer

answered Apr 16 at 5:55

Henno BrandsmaHenno Brandsma

133k44 gold badges5353 silver badges139139 bronze badges

answered Apr 16 at 5:55

Henno BrandsmaHenno Brandsma

133k44 gold badges5353 silver badges139139 bronze badges

answered Apr 16 at 5:55

Henno BrandsmaHenno Brandsma

133k44 gold badges5353 silver badges139139 bronze badges