Rank & Kernel of A: Solving Linear Equations

[tomeatworld]

Homework Statement

Let A=[{1,3,2,2},{1,1,0,-2},{0,1,1,2}]
i) Find the rank
ii) Viewing A as a linear map from M_4x1 to M_3x1, find a basis for the kernel of A and verify directly that these basis vectors are indeed linearly independent.

Homework Equations

None

The Attempt at a Solution

i) is easy enough. Reduce rows to get: A=[{1,3,2,2},{0,1,1,2},{0,0,0,0}] so rank is 2.
ii) I'm not exactly sure of the question here. At first, I thought it was just find the kernel of the matrix and I had some trouble with that. Using the reduced matrix:
x₁ + 3x₂ + 2x₃ + 2x₄ = 0
and
x₂ + x₃ + 2x₄ = 0
but how do I solve this for 4 variables with only 2 equations :/ Any help is appreciated!

 

[talolard]

Multiply the vector by the original matrix, not the reduced row form. You get a pretty comfy vector.

 

[tomeatworld]

How so?

 

[talolard]

If your vector is {x,y,z,w} then you should get the equations
w=(x+y)/2
z= -x-2y
So that's actually two vectors that you will get. If you don't know how to turn those equations into vectors i recomened you go here http://tutorial.math.lamar.edu/Classes/LinAlg/LinAlg.aspx and read up a bit.

 

[Mark44]

Homework Statement

Let A=[{1,3,2,2},{1,1,0,-2},{0,1,1,2}]
i) Find the rank
ii) Viewing A as a linear map from M_4x1 to M_3x1, find a basis for the kernel of A and verify directly that these basis vectors are indeed linearly independent.

Homework Equations

None

The Attempt at a Solution

i) is easy enough. Reduce rows to get: A=[{1,3,2,2},{0,1,1,2},{0,0,0,0}] so rank is 2.
ii) I'm not exactly sure of the question here. At first, I thought it was just find the kernel of the matrix and I had some trouble with that. Using the reduced matrix:
x₁ + 3x₂ + 2x₃ + 2x₄ = 0
and
x₂ + x₃ + 2x₄ = 0
but how do I solve this for 4 variables with only 2 equations :/ Any help is appreciated!

Work with your matrix to get it in reduced row echelon form. In this form all entries above or below a leading 1 entry are zero. Since your matrix looks like this:
1 3 2 2
0 1 1 2

it's not in reduced row echelon form.

After getting to this form use the first row to write an equation that has x₁ in terms of x₃ and x₄. Use the second row to write an equation that has x₂ in terms of x₃ and x₄. The last two equations are simply x₃ = x₃ and x₄ = x₄. This will show two vectors that span the nullspace.

 

[tomeatworld]

Ah of course. So you'd get:

1 0 -1 -4
0 1 1 2

so you end with the two vectors: {(1,0,-1,-4),(0,1,1,2)} which are the basis for the kernel. Sound good?

 

[Mark44]

No, that's not it. The rows don't make a basis for the kernel.

The matrix you have represents the equation Ax = 0, where the first two rows of A are as you show.

Solve for x₁ in the first row (equation) and for x₂ in the second row (equation) to get the following.
x₁ = x₃ + x₄
x₂ = -x₃ - 2x₄

If you add equations for x₃ and x₄, you get the following system.

x₁ = x₃ + x₄
x₂ = -x₃ - 2x₄
x₃ = x₃
x₄ = ...... x₄

Every vector in the nullspace is a linear combination of two vectors that are lurking in the system above.