Row, column, null spaces.

  • Thread starter dcramps
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  • #1
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Homework Statement


Consider the matrix A:

1 4 5 0 9
3 -2 1 0 -1
-1 0 -1 0 -1
2 3 5 1 8

(Sorry I don't know how to do TeX matrices on this site)

Find a basis for the row, column, and null space.


Homework Equations


The Attempt at a Solution


I reduced to row echelon form, which got me:
1 0 1 0 1
0 1 1 0 2
0 0 0 1 0
0 0 0 0 0

Row space:
I took all non-zero rows to be the vectors for the row space

Column space:
I found the columns from the ref version that were linearly independent.
v1 = [ 1 0 0 0 ]
v2 = [ 0 1 0 0 ]
v3 = [ 1 1 0 0 ]
v4 = [ 0 0 1 0 ]
v5 = [ 1 2 0 0 ]

v1+2*v2 = v5, so those 3 are dependent.
v1+v2 = v3, so those 3 are dependent.

Only v4 is independent since no combinations of the others are equal to it, and no combination of it is equal to any of the others, so I just take v4 as the column space? A bit confused here but that is my understanding.

Null space:
in row echelon

x1 = -x3 -x5
x2 = -x3 -2x5
x4 = 0

So x3 and x5 are the 'free variables' but I'm not sure where to go from here.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
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Homework Statement


Consider the matrix A:

1 4 5 0 9
3 -2 1 0 -1
-1 0 -1 0 -1
2 3 5 1 8

(Sorry I don't know how to do TeX matrices on this site)

Find a basis for the row, column, and null space.


Homework Equations


The Attempt at a Solution


I reduced to row echelon form, which got me:
1 0 1 0 1
0 1 1 0 2
0 0 0 1 0
0 0 0 0 0

Row space:
I took all non-zero rows to be the vectors for the row space

Column space:
I found the columns from the ref version that were linearly independent.
v1 = [ 1 0 0 0 ]
v2 = [ 0 1 0 0 ]
v3 = [ 1 1 0 0 ]
v4 = [ 0 0 1 0 ]
v5 = [ 1 2 0 0 ]

v1+2*v2 = v5, so those 3 are dependent.
v1+v2 = v3, so those 3 are dependent.

Only v4 is independent since no combinations of the others are equal to it, and no combination of it is equal to any of the others, so I just take v4 as the column space? A bit confused here but that is my understanding..
No, of course, not. Saying that v1, v2, and v3 are dependent only means that you don't need all three of them, not that you don't need any. Also it makes no sense to say that a single vector, like v4, is "independent". Sets of vectors are "independent" or "dependent". The fact that v3 and v5 can be written in terms of v1 and v2 means that you don't need v3 and v5. The set {v1, v2, v4} is the basis.


Null space:
in row echelon

x1 = -x3 -x5
x2 = -x3 -2x5
x4 = 0

So x3 and x5 are the 'free variables' but I'm not sure where to go from here.
So [x1, x2, x3, x4, x5]= [-x3- x5, -x3- 2x5, x3, 0, x5]= [-x3,-x3, x3, 0, 0]+ [-x5, -2x5, 0, 0, x5]
= x3[-1, -1, 1, 0, 0]+ x5[-1, -2, 0, 0, 1].

Those are your basis vectors.
 
  • #3
43
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No, of course, not. Saying that v1, v2, and v3 are dependent only means that you don't need all three of them, not that you don't need any. Also it makes no sense to say that a single vector, like v4, is "independent". Sets of vectors are "independent" or "dependent". The fact that v3 and v5 can be written in terms of v1 and v2 means that you don't need v3 and v5. The set {v1, v2, v4} is the basis.
Ah, right. That makes much sense. Thank you :)

So [x1, x2, x3, x4, x5]= [-x3- x5, -x3- 2x5, x3, 0, x5]= [-x3,-x3, x3, 0, 0]+ [-x5, -2x5, 0, 0, x5]
= x3[-1, -1, 1, 0, 0]+ x5[-1, -2, 0, 0, 1].

Those are your basis vectors.

So you're just calculating [-x3- x5, -x3- 2x5, x3, 0, x5], then splitting it up into terms of x3 and x5, then factoring out?
 
  • #4
vela
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So you're just calculating [-x3- x5, -x3- 2x5, x3, 0, x5], then splitting it up into terms of x3 and x5, then factoring out?
Yeah. He's just expressing the solution to the null space equations in vector form. Some people find it confusing because x3 and x5 play two roles here: one as a coordinate and the other as an arbitrary constant. As you noted, x3 and x5 are your free variables, so they can take on any value. Let's say x3=a and x5=b. Once these two are chosen, the rest of the coordinates are determined by the equations you found, so you get:

[tex]\begin{align*}
x_1 &= -x_3-x_5 = -a-b \\
x_2 &= -x_3-2x_5 = -a-2b \\
x_3 &= a \\
x_4 &= 0 \\
x_5 &= b
\end{align*}[/tex]

In vector notation, you can write this as

[tex]\begin{pmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{pmatrix} = \begin{pmatrix}-a-b\\-a-2b\\a\\0\\b\end{pmatrix}=a\begin{pmatrix}-1\\-1\\1\\0\\0\end{pmatrix}+b\begin{pmatrix}-1\\-2\\0\\0\\1\end{pmatrix}[/tex]

The reason to write it like that is now you can explicitly see that the null space is spanned by the two vectors (-1,-1,1,0,0) and (-1,-2,0,0,1).
 

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