Whats the answer for these karnel and range of a linear transformation questions?

Question

Q1:find the kernel of the linear transformation.

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)

Q2:the linear transformation T is represented by T(v)=Av.Find a basis for (a) the kernel of T and (b) the range of T.

1-A= ⌈ 1 2⌉

⌊ 3 4 ⌋

Q3:the linear transformation T is defined by T(x)=Ax.Find (a) ker(T), (b) nullity(T), (c) range(T), (d) rank(T).

1-A= ⌈ 4 1⌉

0 0

⌊ 2 -3⌋

Answer 1

Q1: the kernel is everything that maps to the 0 vector, which is all of R3. The range is the 0 vector.

1.2 x=z=0 maps to 0, so the kernel is any vector of form (0,y,0) or the y axis. The range is the xz plane.

1.3 the kernel is the 0 vector, the range is R3

1. 4 the kernel is the line x=y, the range is the line y=-x

Q2 I'm not sure what is meant by 1-A. To find the kernel, solve for v in the matrix equation Av=0. A span for the range is (T(e1), T(e2)) where e1 and e2 or the elementary vectors. Now to find the basis, remove a vector if is a linear combination of the others.

Q3 this is similar to the previous question. Rank is the dimension of the space generated by the columns of A. Nullity is the dimension of the null space of the transformation, or the number of columns without leading entries.