Finding basis, rank, and nullity of T_A for a 3×4 matrix

[stunner5000pt]

Fpr the matrix find a basis for hte kernel and image of [itex} T_{A} [/itex] and find the rank and nullity of $T_{A}$
T is a linear transformation

$$\left(\begin{array}{cccc} 2&1&-1&3 \\ 1&0&3&1 \\ 1&1&-4&2 \end{array} \right)$$

the kernel of T simply means to find the null space of A right?
so when i row reduce i get
$$\left(\begin{array}{cccc} 1&0&3&1 \\ 0&1&-7&0 \\ 0&0&0&0 \end{array} \right)$$

so do i simply find a 3x1 line matrix X such taht AX = 0
The image means something to do iwth the solution... but there is no augmented form given here... is there??

the basis of A will be the rank T right? Is base A = 2?? So the rank of T = 2?
the dimensio of the kernel is T is nullity of T... but i need to find the basis for the kernel first

 

[0rthodontist]

The _rank_ of A will be the rank of T. Otherwise you are correct, what you need to do is find all the vectors x such that Ax = 0, from which you can extract a basis.

 

[stunner5000pt]

ok s oteh rank of kernel of T is 2

what about he nullity, though?? SO basically it asks how many vectors are forming the basis of th kernel?

 

[0rthodontist]

No, the _dimension_ of the kernel of T is 2. That _is_ the nullity.

 

[stunner5000pt]

ok ok
so the number of independent rows in A: is the rank of A and that is 2
that is the tank of Ta yes?

also the rank of the kernel is the number of linearly independent vecotrs that can be formed from AX = 0 right

p.s. are you an orthodontist?