Linear Algebra Powertool

Dimension

On Line Section 2.2

Worksheet by Russell Blyth

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Outline

This worksheet covers material that corresponds to section 2.2 of the text.

The basic objectives are:

• 1) Use Gauss-Jordan elimination to find a basis for the column space of a matrix
• 2) Use Gauss-Jordan elimination to find a basis for the null space of a matrix
• 3) Note a relationship between the dimensions of these two spaces.

The Dimensions of the Column Space and Null Space of a Matrix

• Consider a matrix A, as given below. How do we find a basis for the column space of the matrix? How do we find a basis for the null space of the matrix?

Let A be the matrix

> A := matrix(5,6,[[-1,2,6,-8,-14,3],[2,4,1,-8,5,-1],[-3,1,4,-9,-10,0],[3,-2,-1,12,-1,4],[5,7,11,-11,-19,9]]);

Compute the rank of A:

> rank(A);

How does the rank of A relate to the dimensions of the vector spaces we are interested in studying? Let's row reduce A next:

> RedMat := gaussjord(A);

We conclude that:

• 1) The rank of the matrix is 3, so the set of 6 column vectors cannot be linearly independent. The biggest subset of linearly independent vectors has size 3.
• 2) The pivots of the reduced echelon form correspond to independent vectors. Thus the first three columns of vectors in A form a linearly independent set.

Let's see how to write the final three column vectors of A as linear combinations of the first three columns:

> B:= delcols(A,4..6);
v4:=col(A,4);v5:=col(A,5);v6:=col(A,6);
linsolve(B,v4);
linsolve(B,v5);
linsolve(B,v6);

Check:

> v1:=col(A,1);v2:=col(A,2);v3:=col(A,3);
evalm(2*v1-3*v2); evalm(2*v2-3*v3); evalm(v1-v2+v3);

Thus the first three columns of A are a basis for the column space (why are they linearly independent?). The remaining three columns of A are each linear combinations of the first three columns. Note that the vectors in the column space are vectors in

What is the dimension of the column space of A?

Now find a basis for the null space of A, that is, for the solution space of AX = 0.

> Z:=vector(5,[0,0,0,0,0]);
nullspace:=linsolve(A,Z);

Get a basis by

> y1:=subs(_t[1]=1,_t[2]=0,_t[3]=0,op(nullspace));
y2:=subs(_t[1]=0,_t[2]=1,_t[3]=0,op(nullspace));
y3:=subs(_t[1]=0,_t[2]=0,_t[3]=1,op(nullspace));

Let's check these:

> evalm(A&*y1); evalm(A&*y2); evalm(A&*y3);

So {y1, y2, y3} is a basis for the null space of A. Note that the vectors in the null space are in .

What is the dimension of the null space?

What is the sum of the rank (the dimension of the column space) and the dimension of the null space?

Maple provides direct commands to find bases for the column space and null space:

> colspan(A,'d'); d;

> kernel(A);

Exercises:

1) Repeat all the above for the matrix:

> E:=matrix(4,6,[[-1,2,0,4,5,-3],[3,-7,2,0,1,4],[2,-5,2,4,6,1],[4,-9,2,-4,-4,7]]);

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