find the kernel of the linear transformation :-

Question

1-T:R³→R³,T(x,y,z)=(0,0,0)

2-T:R³→R³,T(x,y,z)=(x,0,z)

3-T:R³→R³,T(x,y,z)=(z,y,x)

4-T:R²→R²,T(x,y)=(x-y,y-x)

Answer 1

1-T:R³→ R³: T(x,y,z) =(0,0,0); To get matrix A of this linear transformation, we have T(1,0,0) = T(0,1,0) = T(0,0,1) = (0,0,0); This means A = [(0,0,0) (0,0,0) (0,0,0)} and because any vector from R³ is solution

of equation Ax = 0. Kernel is R³

2-T:.R³→ R³:T(x,y,z) =(x,0,z); then T(1,0,0) = (1,0,0); T(0,1, 0) = (0,0,0) and T(0,0,1) = (0,0,1);

Then matrix A = [(1,0,0) (0,0,0) (0,0,1)];

Equation Ax = 0, x = 0, y = t (any real number), z = 0.

R³→ R³:Kernel is subset of R³: t(0, 1, 0), axis y

3-T:R³→ R³:T(x,y,z) = (z,y,x); to get matrix A: T(1,0,0) = (0,0,1), T(0,1,0) = (0,1,0), T(0,0,1) - (1,0,0);

A = [(0,0,1) (0,1,0) (1,0,0)];

Equation Ax = 0: z = 0, y = 0, x = 0 only 0 solution

Kernel of T only zero vector (0,0,0)

4-T:R²→R²: T(x,y) = (x - y,y - x).

To get matrix A of this linear transformation: T(1,0) = (1, -1); T(0,1) = (-1, 1)

Matrix A = [(1,-1) (-1,1)]. Equation Ax = 0 and x - y = 0, - x + y = 0. Solution is x = y.

So kernel of T is span of vector (1,1): K(T) = t(1,1) where t is any real number