# Rank-nullity theorem

• Jan 15th 2011, 09:38 AM
Wottie
Rank-nullity theorem
This may be a stupid question, but when I was studying the rank-nullity theorem, I was wondering about the next problem:

The rank-nullity theorem states, for a linear map T: V --> W:
"dim(Im T) + dim(Ker T) = dim V"
But what happens if you, for example, have a linear map from R² to R³. Then the dimension of the image is 3 and the dimension of V is 2. But that means that the dimension of the kernel has to be -1, and that's not possible, is it?

I hope someone can help me out with this.
• Jan 15th 2011, 10:00 AM
FernandoRevilla
Quote:

Originally Posted by Wottie
But what happens if you, for example, have a linear map from R² to R³. Then the dimension of the image is 3 and the dimension of V is 2.

Impossible. If $\displaystyle B=\left\{{e_1,e_2}\right\}$ is a basis of $\displaystyle \mathbb{R}^2$ , then

$\displaystyle S=\left\{{f(e_1),f(e_2)}\right\}$

is a generator system of $\displaystyle \textrm{Im}f$ so,

$\displaystyle \dim \textrm{Im}f\leq 2$ .

Fernando Revilla
• Jan 15th 2011, 06:17 PM
HallsofIvy
Quote:

Originally Posted by Wottie
This may be a stupid question, but when I was studying the rank-nullity theorem, I was wondering about the next problem:

The rank-nullity theorem states, for a linear map T: V --> W:
"dim(Im T) + dim(Ker T) = dim V"
But what happens if you, for example, have a linear map from R² to R³. Then the dimension of the image is 3

No, that's impossible. Are you clear on what the "image" of a linear map is? For T:V-> W, the image of T is a subspace of W but not necessarily all of W.

Quote:

and the dimension of V is 2. But that means that the dimension of the kernel has to be -1, and that's not possible, is it?
No, it is not possible- and so you have proved that the dimension of the image of T cannot be larger than the dimension of V (but can be less).

Quote:

I hope someone can help me out with this.