Linear Transformations and Rank/Nullity Dimensions Question

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January 15th 2012, 01:21 PM
maxgunn555

Linear Transformations and Rank/Nullity Dimensions Question

Just a general question; how do you find the rank and nullity of a linear transformation if you weren't to convert it into matrix form or is it necessary that you must in order to find rank/nullity?

Also say you have found the rank of a matrix (through the column space method) and say you have an answer [c1, c2, 2c1] (except written vertically as a vector of course) how do you know what number the dimension is of the image which is what the rank is supposed to show you.

Lastly how do you find the null space and thus kernel dimension (nullity) of the linear transformation in matrix form?

Thank you.
January 15th 2012, 09:39 PM
FernandoRevilla

Re: Linear Transformations and Rank/Nullity Dimensions Question

Quote:

Originally Posted by maxgunn555

Just a general question; how do you find the rank and nullity of a linear transformation if you weren't to convert it into matrix form or is it necessary that you must in order to find rank/nullity?

It is interesting, but not necessary. Choose for example the linear map $T:\mathbb{R}_2[x]\to \mathbb{R}_2[x]$ , $T[p(x)]=p'(x)$ then you can easily find that $\ker T=\{k: k\in \mathbb{R}\}$ and $\textrm{Im}T=\mathbb{R}_1[x]$ without converting $T$ into matrix form.

Quote:

Also say you have found the rank of a matrix (through the column space method) and say you have an answer [c1, c2, 2c1] (except written vertically as a vector of course) how do you know what number the dimension is of the image which is what the rank is supposed to show you.

I don't understand the exact meaning of your question.

Quote:

Lastly how do you find the null space and thus kernel dimension (nullity) of the linear transformation in matrix form?

$\ker T\equiv AX=0$ , $\dim (\ker T)=\dim V-\textrm{rank}(A)$
January 16th 2012, 09:45 AM
maxgunn555

Re: Linear Transformations and Rank/Nullity Dimensions Question

Thank you very much professor.

But about the question. The wikipedia method of finding the rank seems to leave general answers as a vector but it doesn't show you what the rank would be. i guess however many vectors there are equals the dimension of the kernel.
January 16th 2012, 10:39 AM
FernandoRevilla

Re: Linear Transformations and Rank/Nullity Dimensions Question

Quote:

Originally Posted by maxgunn555

The wikipedia method of finding the rank seems to leave general answers as a vector but it doesn't show you what the rank would be.

If $A$ is an $m\times n$ matrix and its row echelon form is

$E=\begin{bmatrix} a_{11} & & \ldots & \\ 0 &a_{22} & \ldots & \\ 0 & 0 & \ldots & a_{rr}&\ldots\\ 0 & 0 & \ldots & 0&\ldots & 0 \\ \\{0} & {0} &\ldots & 0&\ldots & 0\end{bmatrix}\qquad (a_{ii}\neq 0\quad\forall i=1,\ldots r)$

then, $\boxed{\dim (\ker A)}=n-\textrm{rank}(A)=n-\textrm{rank}(E)=\boxed{n-r}$ .