# Basis for range and kernel of T and prove rank nullity theorem

$$T \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} = \begin{bmatrix} x_1 +3x_2\\ x_2 \\ \end{bmatrix}$$

By considering $ker(T)$ first $$ker(T)=\{(x_1,x_2) |T(x_1,x_2)=0\} \\ x_1+3x_2=0 \implies x_1=-3x_2 \\ ker(T)=(-3x_2,x_2)=x_2(-3,1) \\ \therefore basis(ker(T))=\{(-3,1)\}$$

and by considering $$im(T)=\{(x_1+3x_2),(x_2)\} = \{x_1(1,0)+x_2(3,1)\} \\ \begin{pmatrix} 1 & 0\\ 3 & 1 \\ \end{pmatrix}$$

can be reduced to $$\begin{pmatrix} 1 & 0\\ 0 & 1\\ \end{pmatrix}$$ So$$basis(im(T))=\{(1,0),(0,1)\}$$

Is the above correct?

But when considering the rank nullity theorem :- $$dim(im(T)) +dim(ker(T)) = dim(v)=3$$ but it's not is it?

$ker(T) =(0,0)$
As $T(x_1,x_2)=0 \Rightarrow x_1+3x_2=0$ and $x_2=0$
• So $dim(ker(T))=0$ or $1$ ? Feb 5, 2016 at 11:01