# linear transformation

• Jan 11th 2007, 06:36 AM
0123
linear transformation
talking of linear functions: the theorem says a function f: Rn->Rm is linear if and only if there exists the matrix A such that f(x)=Ax, the A is unique and the euclidean basis in Rn and Rm are fixed. Ok, I understood the proof and the remark that the linear application f is associated with matrix A, whose columns are the images of the vectors of the euclidean basis of Rn according to f. But then, how do we arrive to say that rank of A is equal to the dimension of the image space Im(f) ?:confused:
I would be extremely grateful if you could show me the logic behind this.. infinitely many times thanks
• Jan 11th 2007, 07:14 AM
ThePerfectHacker
Quote:

Originally Posted by 0123
But then, how do we arrive to say that rank of A is equal to the dimension of the image space Im(f) ?

Say, for $\displaystyle m\times n$ matrix,
$\displaystyle f_{A}:\mathbb{R}^n\to \mathbb{R}^m$
Now any element in $\displaystyle \mathbb{R}^n$ can be expressed as,
$\displaystyle c_1\bold{e}_1+...+c_n\bold{e_n}$
Then, the image is the set of all linear combinations,
$\displaystyle S=\{ f_A(c_1\bold{e}_1+...+c_n\bold{e_n}) \}$
It is a linear transformation,
$\displaystyle S=\{ c_1f_A(\bold{e}_1)+...+c_nf_A(\bold{e}_n) \}$
But note that,
$\displaystyle f_A(\bold{e}_1),...,f_A(\bold{e}_n)$
Correspond to the coloum vectors of the matrix $\displaystyle A$.
Thus, $\displaystyle S$ is a space spanned by the linear combinations of the coloum vectors. That means it has a dimension which is called the rank of $\displaystyle A$. Alternatievly, it is the dimension of the set mentioned above, which is the dimension of the image space.
• Jan 11th 2007, 07:27 AM
0123
Umm..let's check if I have undestood the entire thing rightly:
the image is the space geneated by the columns of A. The rank is the max number of linearly indipendent vectors, so the rank is the dimension of the image. Then, since the image is generated by the columns of the matrix, if these are vectors linearly independent then they are a basis of the image. Otherwise let's eliminate the dependent ones and the lefts will be the basis.
Am I wrong or have I understood your explaination?
• Jan 11th 2007, 08:43 AM
ThePerfectHacker
Quote:

Originally Posted by 0123
Umm..let's check if I have undestood the entire thing rightly:
the image is the space geneated by the columns of A. The rank is the max number of linearly indipendent vectors, so the rank is the dimension of the image. Then, since the image is generated by the columns of the matrix, if these are vectors linearly independent then they are a basis of the image. Otherwise let's eliminate the dependent ones and the lefts will be the basis.
Am I wrong or have I understood your explaination?

Sounds good to me.

What I have shown is that the image of the function is the space spanned by all linear combinations of the column vectors. Further, the colomun space by defintion is space spanned by linear combinations of the colomn vectors. Thus, they are really the same thing. Saying the "Rank" means the dimension of basis for the latter case. And saying "Dimension" means the dimension of basis for former case. Which is the same thing.