# Linear Transformation (Rank and Nullity)

• Apr 8th 2011, 02:54 PM
vedicmath
Linear Transformation (Rank and Nullity)
Hello,

I need some help in this problem....

A = [1 -2 -1, 1 -3 0, 3 -8 -1] and let linear transformation T: R^3--> R^3 be defined by T(x) = Ax, where x = [ x1 x2 x3].

a) Determine rank(T) and nullity (T)

b) Determine bases for the range R(T) and the null space N(T).

(If N(T) = {[0 0 0]} state that N(T) has no basis)...

I know we have to use the Gauss Jordan to figure that matrix out but I am unable to reduce it properly, perhaps there is another way of doing it??

thanks!
• Apr 8th 2011, 03:45 PM
HallsofIvy
I can see no difficult in reducing it- it seems easier than most of this kind.
$\begin{bmatrix}1 & -2 & 1 \\ 1 & -3 & 0 \\ 3 & -8 & -1\end{bmatrix}$

Start by subtracting the first row from the second row and subtracting three times the first row from the third row.
$\begin{bmatrix}1 & -2 & 1 \\ 0 & -1 & -1 \\ 0 & -2 & -2\end{bmatrix}$
Can you finish that?
• Apr 10th 2011, 02:31 PM
vedicmath
thanks for the response Sir. When I started simplifying using Gauss Jordan I ended up getting---
x = 2(y) + z
z = y...If I consider 1st column as x, 2nd as y and 3rd column as z and then set three of these equations equal to 0...
So the N(A), null space of A = Span {z [3 1 1] }

but then this is what I get when I simplify A, then how do I implement it on T(x) = Ax...and then find the rank and null space of T?...
• Apr 10th 2011, 03:11 PM
Deveno
Halls, your first row is incorrect. a13 should be -1.

vedicmath, how many non-zero rows did you end up with in rref(A)? that is the rank of A, so any two columns of A would be a basis (it is customary to choose the rows of A corresponding to the pivot columns of rref(A)).

what is the nullity of A, by the rank-nullity theorem? you have a spanning set, does it have the same number of elements as the dimension of the null space (that is one criterion for a basis)?