Easy calculation of basis of the null space

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td21
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Homework Statement


find the basis of the nullspace of this matrix [tex]\begin{pmatrix} 1&1&1&-1 \\ 0&0&1&3 \end{pmatrix}[/tex]


Homework Equations





The Attempt at a Solution


i forget it.
i first substitue 0 and 1 for last row but what about the first row? Substiute 0 and 1 again and this will give 4 basis? but i think there is only 2 basis.
 

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  • #2
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Homework Statement


find the basis of the nullspace of this matrix [tex]\begin{pmatrix} 1&1&1&-1 \\ 0&0&1&3 \end{pmatrix}[/tex]


Homework Equations





The Attempt at a Solution


i forget it.
i first substitue 0 and 1 for last row but what about the first row? Substiute 0 and 1 again and this will give 4 basis? but i think there is only 2 basis.

I don't understand what you mean by "Substitute 0 and 1 for the last row". What you need to do is to put this matrix into Row Reduced Echelon Form. This is equivalent to solving the equation:

(1,0)x_1 + (1,0)x_2 + (1,1)x_3 + (-1,3)x_4 = (0,0)

Then, you will have some expressions that relate the values of x_1,x_2,x_3,x_4. Use this to construct a basis for the null space.
 
  • #3
HallsofIvy
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Equivalently, if you call the vector <x, y, z, w>, you get x+ y+ z- w= 0 and z+ 3w= 0. From the latter, z= -3w. Putting that into the first equation, x+ y+ z- (-3w)= x+ y+ 4w= 0 so x= -y- 4w.

That is, any such vector can be written <x, y, z, w>= <-y- 4w, y, w, -3w> and you can get your basis vectors from that.
 
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  • #4
td21
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I don't understand what you mean by "Substitute 0 and 1 for the last row". What you need to do is to put this matrix into Row Reduced Echelon Form. This is equivalent to solving the equation:

(1,0)x_1 + (1,0)x_2 + (1,1)x_3 + (-1,3)x_4 = (0,0)

Then, you will have some expressions that relate the values of x_1,x_2,x_3,x_4. Use this to construct a basis for the null space.
Well, sub 0 and 1 to free variable technique is done as follow, but failed...
1)From the last row, sub last element as 0, obtain 1 vector: (ab00)
2)From the last row, sub last element as 1, obtain 1 vector: (cd-31)



1*)from1), sub b as o, obtain (0000)[discard]
1**)from1), sub b as 1, obtain (-1100)[1st basis]

2*)from2), sub d as o, obtain (40-31)[2nd basis]
2**)from2), sub d as 1, obatin (31-31)[3rd basis???but it is the 1st basis + 2nd basis????]
 

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