Let A denote the following matrix 1 1 2 1 2 1 1 0 1 2 3 1 0

Let A denote the following matrix: (1 -1 2 1 2 1 -1 0 1 2 -3 -1 0 3 -5 -2 4 -1 3 2) Find a basis for the null space of A, a basis for the row space of A and a basis for the column space of A. Also determine the rank of A.

Solution

Ans-

Since P3P3 has dimension 4, the set {p1,p2,p3,p4}{p1,p2,p3,p4} will span P3P3 if and only if they are independent. So, test if the vectors in {p1,p2,p3,p4}{p1,p2,p3,p4} are independent:

Assume

c1p1+c2p2+c3p3+c4p4=0.c1p1+c2p2+c3p3+c4p4=0.

Then

c1(x32x2+x+1)+c2(x2x+2)+c3(2x3+3x+4)+c4(3x2+2x+1)=0.c1(x32x2+x+1)+c2(x2x+2)+c3(2x3+3x+4)+c4(3x2+2x+1)=0.

Collecting like terms, the above can be written as

(c1+2c3)x3+(2c1+c2+3c4)x2+(c1c2+3c3+2c4)x+(c1+2c2+4c3+c4)=0.(c1+2c3)x3+(2c1+c2+3c4)x2+(c1c2+3c3+2c4)x+(c1+2c2+4c3+c4)=0.

A polynomial is the zero polynomial if and only if all its coefficients are 0; so, the above is equivalent to the following system of equations:

c1+2c32c1+c2+3c4c1c2+3c3+2c4c1+2c2+4c3+c4=0=0=0=0(1)(1)c1+2c3=02c1+c2+3c4=0c1c2+3c3+2c4=0c1+2c2+4c3+c4=0

The coefficient matrix of the above system is

A=1211011220340321A=[1020210311321241]

An echelon form of AA is

1000010024300357.[1020014300350007].

This implies the system above has only the trivial solution: c1=c2=c3=c4=0c1=c2=c3=c4=0; thus{p1,p2,p3,p4}{p1,p2,p3,p4} is an independent set and so spans P3P3.

Alternatively you could show that the equation you wrote always has a solution. To do this, write the corresponding system of equations. You\'ll wind up with the system as in (1)(1), but with the right hand side replace by the coefficients of yy.