Question

consider T(x）= A (x)

[1 2 0 3 6 1 2 4 1 1 2 3 -1 2 9 2 1 5 10 11 0 (a) Find a basis for the nullspace (kernel) of T. (b) Find a basis for the range of T. (c) What are the values of the rank and nullity?

Answer 1

$A=\left[ \begin{array}{ccccc} 1 & 2 & 0 & -1 & 1 \\\\ 3 & 6 & 1 & 2 & 5 \\\\ 2 & 4 & 1 & 9 & 10 \\\\ 1 & 2 & 3 & 2 & 0 \end{array} \right] \longrightarrow_{R_2=R_2-3R_1} \left[ \begin{array}{ccccc} 1 & 2 & 0 & -1 & 1 \\\\ 0 & 0 & 1 & 5 & 2 \\\\ 2 & 4 & 1 & 9 & 10 \\\\ 1 & 2 & 3 & 2 & 0 \end{array} \right] \\ \\\longrightarrow_{R_3=R_3-2R_1} \left[ \begin{array}{ccccc} 1 & 2 & 0 & -1 & 1 \\\\ 0 & 0 & 1 & 5 & 2 \\\\ 0 & 0 & 1 & 11 & 8 \\\\ 1 & 2 & 3 & 2 & 0 \end{array} \right] \longrightarrow_{R_4=R_4-R_1} \left[ \begin{array}{ccccc} 1 & 2 & 0 & -1 & 1 \\\\ 0 & 0 & 1 & 5 & 2 \\\\ 0 & 0 & 1 & 11 & 8 \\\\ 0 & 0 & 3 & 3 & -1 \end{array} \right]$

$\\\longrightarrow_{R_3=R_3-R_2} \left[ \begin{array}{ccccc} 1 & 2 & 0 & -1 & 1 \\\\ 0 & 0 & 1 & 5 & 2 \\\\ 0 & 0 & 0 & 6 & 6 \\\\ 0 & 0 & 3 & 3 & -1 \end{array} \right] \longrightarrow_{R_4=R_4-3R_2}\left[ \begin{array}{ccccc} 1 & 2 & 0 & -1 & 1 \\\\ 0 & 0 & 1 & 5 & 2 \\\\ 0 & 0 & 0 & 6 & 6 \\\\ 0 & 0 & 0 & -12 & -7 \end{array} \right]\\ \longrightarrow_{R_4=R_4+2R_3} \left[ \begin{array}{ccccc} 1 & 2 & 0 & -1 & 1 \\\\ 0 & 0 & 1 & 5 & 2 \\\\ 0 & 0 & 0 & 6 & 6 \\\\ 0 & 0 & 0 & 0 & 5 \end{array} \right]$

$\longrightarrow_{R_5=\frac{1}{5}R_5,~R_4=\frac{1}{6}R_4} \left[ \begin{array}{ccccc} 1 & 2 & 0 & -1 & 1 \\\\ 0 & 0 & 1 & 5 & 2 \\\\ 0 & 0 & 0 & 1 & 1 \\\\ 0 & 0 & 0 & 0 & 1 \end{array} \right]=B$

a) B is the row reduced form of A, thus for null space A, we consider BX=O, then

$\left[ \begin{array}{ccccc} 1 & 2 & 0 & -1 & 1 \\\\ 0 & 0 & 1 & 5 & 2 \\\\ 0 & 0 & 0 & 1 & 1 \\\\ 0 & 0 & 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{c} x_{1} \\\\ x_{2} \\\\ x_{3} \\\\ x_{4}\\\\ x_{5} \end{array} \right] =\left[ \begin{array}{c} 0 \\\\ 0 \\\\ 0 \\\\ 0\\\\ 0 \end{array} \right] \\ \\=>x_1+2x_2-x_4+x_5=0,~x_3+5x_4+2x_5=0,~x_4+x_5=0,~x_5=0.\\ =>x_5=0=>x_4=0=>x_3=0=> x_1+2x_2=0. \\(x_1,x_2,x_3,x_4,x_5)=(-2x_2,x_2,0,0,0)=(-2,1,0,0,0)x_2.$

Thus basis for null space is {(-2,1,0,0,0)}.

b) Not that 1st, 3rd,4th and 5th columns of B contains leading 1. Thus the corresponding columns of the original matrix B wil give the basis elements for column space, that is range.

Thus basis for range is {(1,3,2,1),(0,1,1,3),(-1,2,9,2),(1,5,10,0).

c) rank = 4, and nullity =1.