Question

Please put the solution in the form of a formal proof, Thank You.

Let T: R3-R2 be the linear map given by a 2c (a) Find a basis of the range space. (Be sure to justify that it spans and is linearly independent.) (b) Find a basis of the null space. (Be sure to justify that it spans and is linearly independent.) (c) Use parts (a) and (b) to verify the rank-nullity theorem.

Answer 1

$T:\mathbb{R}^3\rightarrow \mathbb{R}^2\\ T(a,b,c)=(a+2c,c)$

First will find matrix of T,

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

$T(0,1,0)=(0,0)=0(1,0)+0(0,1)--(2)$

$T(0,0,1)=(2,1)=2(1,0)+1(0,1)--(3)$

By(1)(2)(3)

$[T]=\begin{bmatrix} 1 & 0&2 \\ 0 & 0& 1 \end{bmatrix}$

$Rref[T]=\begin{bmatrix} 1 & 0&0 \\ 0 & 0& 1 \end{bmatrix}$

Rank of [T] is 2 .

And basis for Range(T) is

Null(T) is the set of solution space of $[T]x=0$

which is equivalent to

$Rref[T]x=0\\ \begin{bmatrix} 1 & 0&0 \\ 0 & 0& 1 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}=0\\ x_1=0,x_3=0\\ x=\begin{bmatrix} 0\\ 1\\ 0\end{bmatrix}$

dim(Null(T))=1

$Null(T)=Span\{\begin{bmatrix} 0\\ 1\\ 0\end{bmatrix}\}$

Basis for Nll(T) is

010

(c)

Rank nullity theorem,

Rank(T) + Nullity(T) = Rank R" 2+1=3