$V$ and $W$ are vector spaces. $T:V\rightarrow W$ is a linear transformation

$V = \mathbb R^3 = (x,y,z)$ , dimension = 3.

$W = \mathbb R^2 = (x,y)$, dimension = 2.

$T(x,y,z) = (x,y)$

The basis of the nullity is ${ (0,0,0) }$ (right?). Then its dimension is 0.

The rank nullity theorem says:

$rank(T) + null(T) = dim (V)$.


$rank(T) + 0 = 3$

$rank(T) = 3.$

Well, I know there is some thing wrong here. How can rank(T) = 3 if dim(W) = 2 ?

Thank you for helping!

  • $\begingroup$ Is $(0,0,0)$ ever the basis for anything? No, it is not. The empty space has dimension zero, not one. In any event, the nullity is not dimension zero here. What is $T(0,0,1)$? $\endgroup$ May 14 '14 at 0:09

The map you are describing is $T:\Bbb R^3\to\Bbb R^2$ defined by $T(x,y,z)=(x,y)$. A vector $(x,y,z)$ is in the kernel of $T$ if and only if $x=y=0$. Hence every vector in the kernel of $T$ is of the form $$(0,0,z)=z\cdot(0,0,1)$$ That is, $$ \ker T=\operatorname{Span}\{(0,0,1)\} $$ so $\dim\ker T=1$. Clearly $T$ is surjective so $$\operatorname{rank}T=\dim\operatorname{image}T=\dim\Bbb R^2=2$$ Thus $$ \operatorname{rank}T+\dim\ker T=2+1=3 $$ and the Rank-Nullity theorem survives!


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