Rank and Nullity of a linear transformation

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June 28th 2009, 01:51 PM
meg0529

Rank and Nullity of a linear transformation

Okay so i have been breaking my head over this for a couple of hours but don't really know if I'm even doing the right thing.

Okay so we are looking at a transformation defined in the following way:

T $\left(\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{2 1}&a_{22}&a_{23}\end{array}\right)$ = $\left(\begin{array}{cc}2a_{11}-a_{12}&a_{13}+2a_{12}\\0&0\end{array}\right)$

They want to know if this is a Linear transformation, which as per my calculations it is. And then they want to know the nullity and the rank of the transformation. So since the dimension of vector space "V" is 2*3=6 the nullity and rank need to add up to this. Now I'm a little confused as to how to proceed with finding the null space and range, It seems that the null space needs the top row to be zeros and the bottom row can take on any values, but then how many matrices does that make? if it was just the 0 matrix then the null space would have dimension 0 but there are obviously other vectors, so my first question is what does the null space look like?

Then for the range, the thing that I think we are supposed to do is, take the basis for all 2X3 matrices and then transform each element, so this gave me 6 matrices, $\begin{array}{cc}1&0\\0&0\end{array} , \begin{array}{cc}-1&1\\0&0\end{array} , \begin{array}{cc}0&1\\0&0\end{array}$ with the remaining three matrices being the zero matrix.

So then since only the first 3 are linearly independent the rank is 3? which means the nullity must be three? But I don't know what the null space looks like so I don't know how to verify this. I don't even know if I'm doing this stuff right(Crying)!

Thank you so much for your help!
June 28th 2009, 06:35 PM
pop

First of all, is there a typo in your transformation? It says $a_{a2}$ . Like you said, if we find the nullity, we can use the dimension theorem to find the rank of T and we are finished the question.

Recall that the null space of T (from X -> Y) is the set of x element of X such that T(x) = 0. It is obvious that the zero element of Y in this case is the zero 2x2 matrix. So essentially, we need to find the solutions to two equations: $2a_{11} -a_{a2} = 0$ and $a_{13}+2a_{12} = 0$ . Unfortunately I can't do this for you right now since I think there was a typo. But when you solve for this, it will yield a set of solutions. The dimension of this set is the nullity. Then apply the dimension theorem and you are done.
June 28th 2009, 07:54 PM
meg0529

so sorry it was $a_{12}$
June 29th 2009, 09:10 AM
HallsofIvy

Quote:

Originally Posted by meg0529

Okay so i have been breaking my head over this for a couple of hours but don't really know if I'm even doing the right thing.

Okay so we are looking at a transformation defined in the following way:

T $\left(\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{2 1}&a_{22}&a_{23}\end{array}\right)$ = $\left(\begin{array}{cc}2a_{11}-a_{12}&a_{13}+2a_{12}\\0&0\end{array}\right)$

They want to know if this is a Linear transformation, which as per my calculations it is. And then they want to know the nullity and the rank of the transformation. So since the dimension of vector space "V" is 2*3=6 the nullity and rank need to add up to this. Now I'm a little confused as to how to proceed with finding the null space and range, It seems that the null space needs the top row to be zeros and the bottom row can take on any values, but then how many matrices does that make? if it was just the 0 matrix then the null space would have dimension 0 but there are obviously other vectors, so my first question is what does the null space look like?

Then for the range, the thing that I think we are supposed to do is, take the basis for all 2X3 matrices and then transform each element, so this gave me 6 matrices, $\begin{array}{cc}1&0\\0&0\end{array} , \begin{array}{cc}-1&1\\0&0\end{array} , \begin{array}{cc}0&1\\0&0\end{array}$ with the remaining three matrices being the zero matrix.

So then since only the first 3 are linearly independent the rank is 3? which means the nullity must be three? But I don't know what the null space looks like so I don't know how to verify this. I don't even know if I'm doing this stuff right(Crying)!

Thank you so much for your help!

In order to be in the null space, you must have
$\left(\begin{array}{cc}2a_{11}-a_{12}&a_{13}+2a_{12}\\0&0\end{array}\right)= \left(\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}\right)$
so you must have $2a_{11}- a_{12}= 0$ and $a_{13}+ 2a_{12}= 0$ . You can solve those for $a_{12}= 2a_{11}$ and $a_{12}= -a_{13}$ . That means that given 4 of the values of the a's we can calculate the other two so the null space has dimension 4 and the nullity is 4.