Saturday, January 22, 2011

Nullspace of a vector

This post is a note to myself about linear algebra.

Let's consider two vectors in R2:


a = ( 3, 4) = [  3 ]
[ 4 ]

b = (-3, 1) = [ -3 ]
[ 1 ]


We can see from a plot that these vectors a and b are obviously not pointing in the same direction.



Also we notice that a is not a multiple (or a linear combination) of b (or vice versa). There is no constant such that when the elements of a are multiplied by the constant we get b.

Now consider a point P in R2, say


P = (2,-4)


We can construct a linear combination of a and b that reaches this P or any other point.

How? Simply place one of the vectors at the origin and move along it (perhaps in a negative direction), and place the second vector at P and move along it, and find where the two lines meet.



Call the coordinates of the point where the lines cross x and y. There are actually two possibilities depending on which vector we choose for each role. Suppose we move in the reverse direction from the origin along a and from P along b. From the point-slope equation we know that:


(0 - y) / ( 0 - x) =  4/3  (from a)

(Py - y) / (Px - x) = -1/3 (from b)
(-4 - y) / ( 2 - x) = -1/3


We can solve these two equations pretty easily. From the first


y = 4/3 x


and from the second:


(-12 - 3y) = -(2 - x)
x = -10 - 3y
= -10 - 3(4/3) x

5x = -10
x = -2
y = -8/3


The other solution is symmetrical (we're dealing with a parallelogram). We would need to go +2 across and +8/3 up from P to reach x, y = (4, -4/3).

Suppose we want to know the actual multipliers for a and b, i.e. the fraction of the length of a and b that we travel along each vector.

From the figure, we can estimate that the values will be about -0.7 and -1.25.



Call these multipliers u and v. In matrix language


M = [ 3 -3 ]
[ 4 1 ]

[ 3 -3 ] [ u ] = [ 2 ]
[ 4 1 ] [ v ] [ -4 ]


One way we can solve the system (call it the Algebra 2 way) is as follows:


3u - 3v = 2
4u + v = -4


Solve the second equation for v:


v = -4u - 4


Plug into the first equation:


3u - 3(-4u - 4) = 2
15u = -10
u = -2/3
v = 8/3 - 4 = -4/3


Finally, consider P not just as a point but as a vector c:


a = ( 3, 4)
b = (-3, 1)
c = ( 2,-4)




Since we could have chosen P to be any point, c could be any vector in the x,y-plane and it would be constructed as a linear combination of a and b:


u a + v b


We say that c is in the column space of the matrix whose columns are the vectors a and b:


M = [ 3 -3 ]
[ 4 1 ]


because we can get to c by taking a linear combination of the columns of M.

And of course, having a + b = c, we can also add -c to get:


u a + v b - c = 0


If we think of another matrix as the column vectors a, b and c lined up:


[ 3 -3  2 ] [ u ] = [ 0 ]
[ 4 1 -4 ] [ v ] = [ 0 ]
[ w ] = [ 0 ]


The combination u = -2/3, v = -4/3, w = -1 solves this equation and brings us back to zero:


-2/3 [ 3 ] -4/3 [ -3 ] - [  2 ] = [ 0 ]
[ 4 ] [ 1 ] [ -4 ] [ 0 ]

[ -2 ] + [ 4 ] + [ -2 ] = [ 0 ]
[ -8/3 ] [ -4/3 ] [ 4 ] [ 0 ]


That solution (a vector in R3):


x = [ u ] = [ -2/3 ]
[ v ] [ -4/3 ]
[ w ] [ -1 ]


is said to be in the nullspace of


A = [ 3 -3  2 ]
[ 4 1 -4 ]


because Ax = 0.

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