Let A V1 V2 V3 V4 with the columns are V1 131T V2 2 6 2T

Let A = [V_1 V_2 V_3 V_4] with the columns are V_1 = (1,3,1)^T, V_2 = (2, 6, 2)^T, V_3 = (-2, -5, 0)^T, V_4 = (1, 4, 3)^T. Find bases and dimensions of Null(A), Col(A) and Row(A) and rank (A). Is x = (1, 0, 1, 1)^T in the Null(A)? What is a basis for the row space of A^T? (No calculation is needed).

Solution

a)

We denote: transpose by \' ie :x^T=x\'

We can see:

v_2=2v_1

v_4-3v_1=(-2,-5,0)=v_3

So we see that:v_2 and v_3 both belong to span of:v1,v4 which are linearly indepenent.

So bases of col(A)={v1,v4}

dim col(A)=2

dim row (A)=dim col(A)=2

Rows of A are:

First row: r_1=(1,2,-2,1)

Second row:r_2=(3,6,-5,4)

Third row:r_3=(1,2,0,3)

First and third row and clearly linearly independent as third entry of third row is zero but third entry of first row is non zero. And dim row(A)=2

Hence basis of row(A)={r_1,r_3}

rank(A)=dim row(A)=2

By rank nullity theorem, for matrix of size:mxn

rank(A)+dim null(A)=n

Hence, dim null(A)=2

We look for x so that:Ax=0 to find bases for null(A)

Augmented matrix for :Ax=0 is A itself

Performing row operations:r_2=r_-2r_1,r_3=r_3-r_1

This gives the matrix with the following rows:

First row: r_1=(1,2,-2,1)

Second row:r_2=(0,0,1,1)

Third row:r_3=(0,0,2,2)

Let, x=(x1,x2,x3,x4)\'

Hence, x3+x4=0 ie x3=-x4

x1+2x2-2x3+x4=0

x1+2x2+2x4+x4=0

x1=-2x2-3x4

Hence, x=(-2x2-3x4,x2,-x4,x4)\'=x2(-2,1,0,0)\'+x4(-3,0,-1,1)\'

Hence basis for: null(A)={(-2,1,0,0)\',(-3,0,-1,1)\'}={X,Y}

(b)

Let, x=(1,0,1,1)\' be in null (A) so there exist :a,b so that:

x=aX+bY=(-2a-3b,a,-b,b)=(1,0,1,1)

So there are no solutions looking at last two coordinates as we get:b=1,b=-1

c)

rows of A\' are columns of A

Hence, basis of row space of A\' is basis of col(A)