Understanding nullspace (kernel) of a matrix

[NewtonianAlch]

Homework Statement

Find the kernel of the matrix:

[PLAIN]http://img256.imageshack.us/img256/9015/53369959.jpg

The Attempt at a Solution

So I row-reduce it and get:

[PLAIN]http://img812.imageshack.us/img812/1391/97980793.jpg

The system of equations the row-reduced form equals 0.

So I set x_{3} and x_{4} as the free variables and solve for x_{2}: x_{2} = -x_{3} - x_{4}

Substitute that into the top equation to get x_{1} -4(x_{3}+x_{4}) +2x_{3} +7x_{4} = 0

Solve for x_{1}: x_{1} = 2x_{3} - 3x_{4}

From this we get:

Vector(2x_{3}-3x_{4}, -x_{3}-x_{4}, x_{3}, x_{4})

So ker(A) = x = x_{3}(2,-1,1,0) + x_{4}(-3,-1,0,1)

What does this mean essentially? I know how to solve it, but I don't really understand what I'm doing or what this is useful for. As far as I understand, the kernel is a subspace of a linear map, so what does this translate to in practical terms?

 

[Deveno]

well a linear transformation maps one vector space into another. the kernel is a measure of how many vectors get annihilated by the linear transformation (we lose entire dimensions at a time).

in this case, we lose 2 dimensions, meaning there is only 2 left for the image space. the matrix A condenses R^4 down to a two dimensional plane in R^3, by sending an entire 2-dimensional plane to the origin.

it's easier to understand what this means with some really basic examples (we will assume the standard bases are used):

suppose T:R^3-->R^2 is the map T(x,y,z) = (x,y). we just dump the 3rd dimension. this has the matrix:

[1 0 0]
[0 1 0], and it's pretty clear that the nullspace of T is {(0,0,z) : z in R} (the z-axis).

or suppose L:R^3-->R^3 is the map L(x,y,z) = (x,x,x). this has the matrix

[1 0 0]
[1 0 0]
[1 0 0], and has nullspace {(0,y,z) : y,z in R}. here our image space is just the line x(1,1,1), anything in the yz-plane maps to the origin.

the nullspace tells you how much information you are losing because of a linear transformation T. if a nullspace is just {0}, you know that T is preserving a faithful copy of the original space (although it may "stretch" or "rotate" vectors, we can still trade coordinates one-for-one). a matrix for such a T just sends a basis straight over to some other basis.

in terms of solving systems of linear equations, finding x for which Ax = 0, and finding nullspace(A), are the same problem.