Find a basis for the null space of the given linear transfor

Find a basis for the null space of the given linear transformations given by T(x, y, z) = 3x - 2y + 2 for all x, y, z F. defined by T(x, y. z, w) = (t + y 4- 2z, x + y + 2. - x + y + 2u;) for ail x, 1/, 2, w 6 F. I_3: P_3 = F^2 given by T(p) = (p(l),p(-l)) Find a basis for the range of T_2 which is given in question 1

Solution

1.

a

T(x,y,z)=0

3x-2y+z=0

z=2y-3x

So, (x,y,z)=(x,y,2y-3x)=x(1,0,-3)+y(0,1,2)

Basis is : {(1,0,-3),(0,1,2)}

b)

T(x,y,z,w)=0

So

x+y+2z=0

x=-y-2z

-x+y+2w=0

x=y+2w=-y-2z

2w=-2y-2z

w=-y-z

So,(x,y,z,w)=(-y-2z,y,z,-y-z)=y(-1,1,0,-1)+z(-2,0,1,-1)

c)

Let, p=ax^3+bx^2+cx+d

p(1)=0

p(-1)=0

a+b+c+d=0

-a+b-c+d=0

This gives

b=-d,a=-c

So,

p(x)=ax^3+bx^2-ax-b=a(x^3-x)+b(x^2-1)

2.

By rank nulliyt theorem

nullity T + dim range of T=4

We saw dimension of nullity of T =2

So dimension of range of T is 2

So we need only find two vectors in range of T

T(0,0,0,1)=(0,0,2)

T(1,0,0,0)=(1,1,-1)