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

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?

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$ker(T) =(0,0)$

As $T(x_1,x_2)=0 \Rightarrow x_1+3x_2=0$ and $x_2=0$

And so the rank nullity theorem still holds.

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  • $\begingroup$ So $dim(ker(T))=0 $ or $1$ ? $\endgroup$
    – S.Dan
    Feb 5, 2016 at 11:01
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    $\begingroup$ dim(ker(T)) =0 . As by convention dimension of zero vector is zero. $\endgroup$ Feb 5, 2016 at 11:02

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