A question about rank/image of a linear transformation

**arbolis** · February 10th 2009, 05:04 PM

Let $V$ and $W$ be K-vectorial spaces and $T:V\to W$ a linear transformation.
From my notes, we call the rank of $T$ as $R(T)=T(V)=\{ w\in W : \exists v \in V \text{such that } T(v)=w \}$ .
And in the exercises I have there is no mention of the rank of a linear transformation. However I've to deal with the image of a linear transformation in the exercises and I've no mention of it in my notes. So my question is "does the rank and the image of a linear transformation is in fact the same thing?".

**math2009** · February 10th 2009, 05:17 PM

Both rank and image are different concepts.

$rank(A) \in R$ ,it's a scalar. $im(A)$ is vector.
If $R(T)=T(V)$ , it mean $W\subseteq R$ .After linear transformation, dimension may be decreased.

**Jameson** · February 10th 2009, 05:21 PM

I believe that the Rank is the dimension of the image of T.

**arbolis** · February 10th 2009, 05:33 PM

Ok. I'm confused now. Reading my notes (the professor's ones that is), it reads "proposition : Let $V$ and $W$ be K- vector spaces, $T:V\to W$ a linear transformation. Hence $R(T)$ is a subspace of $W$ and $\ker (T)$ is a subspace of $V$ ."
Also according to his definition, the rank is a vector. Unless "rango" (in Spanish) doesn't mean "rank" in this specific case, but I really doubt it.
I forgot to precise but in my first post $v$ and $w$ are vectors, not scalars.

**NonCommAlg** · February 10th 2009, 05:40 PM

Originally Posted by arbolis

Ok. I'm confused now. Reading my notes (the professor's ones that is), it reads "proposition : Let $V$ and $W$ be K- vector spaces, $T:V\to W$ a linear transformation. Hence $R(T)$ is a subspace of $W$ and $\ker (T)$ is a subspace of $V$ ."
Also according to his definition, the rank is a vector. Unless "rango" (in Spanish) doesn't mean "rank" in this specific case, but I really doubt it.
I forgot to precise but in my first post $v$ and $w$ are vectors, not scalars.

there's nothing confusing here: R(T) and ker(T) are subspaces and so they have dimensions. as Jameson said, the rank of T is the dimension of R(T). also the nullity of T is the dimension of ker(T).

**arbolis** · February 10th 2009, 05:44 PM

as Jameson said, the rank of T is the dimension of R(T).

Ok so the rank is a scalar, not a vector.

**arbolis** · February 11th 2009, 02:45 PM

Nevermind. I met my professor today by casualty and he told me that image and "rango" (range maybe, not rank!, sorry) are the same thing for him. So it explains his definition of "rango" of a linear transformation.
Thank you anyway.

**math2009** · February 11th 2009, 04:54 PM

range is for functions, it's similar to image of algebra.
I guest "rango " is range.

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