Question

let T be multiplication by the matrix A= (top: 1,0 -2; bottom: -2,1,2)
a) find a basis for the range of T
b) find a basis for the kernel of T
c) the rank and nullity of T
d) the rank and nullity of A
Please, explain me.

Answer 1

For the basis of the range of \(T\), just look at \(T\begin{bmatrix}x\\y\\z\end{bmatrix}\):$$T\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1&0&-2\\-2&1&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}x+0y-2z\\-2x+1y+2z\end{bmatrix}=\begin{bmatrix}1\\-2\end{bmatrix}x+\begin{bmatrix}0\\1\end{bmatrix}y+\begin{bmatrix}-2\\2\end{bmatrix}z$$which means every vector in our range is a linear combination of our column vectors, i.e. the range is spanned by our column vectors; in fact, the column space of our matrix is exactly the range of our linear transform.
Since we know \(\begin{bmatrix}1\\-2\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix},\begin{bmatrix}-2\\2\end{bmatrix}\) span our range, we can use them to determine a basis by picking linearly independent vectors from them. Since the collection of all three is dependent, we kick out \(\begin{bmatrix}-2\\2\end{bmatrix}\) and keep \(\begin{bmatrix}1\\-2\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix}\) as our basis since they're linearly independent.

Answer 2

oops, my earlier line got cut off.$$\begin{bmatrix}x+0y-2z\\-2x+1y+2z\end{bmatrix}=\begin{bmatrix}1\\-2\end{bmatrix}x+\begin{bmatrix}0\\1\end{bmatrix}y+\begin{bmatrix}-2\\2\end{bmatrix}z$$

Answer 3

I got that stuff but come up in another way, I take rref A to get (1,-2) and (0,1) . friend, I cannot "kick out " something without any reason or any logic. Anyway, i am there

Answer 4

@Hoa we kick it out because we understand that the other two vectors are linearly independent without it.

Answer 5

surely we understand, but my prof wants the steps to get this, so, the range of T is 2 right?

Answer 6

The rank is the dimension of the range, which is indeed 2.

Answer 7

The nullity is then the dimension of our kernel (null space). Together, by the rank-nullity theorem, they sum to the dimension of our linear transformation itself! :-)
(which is \(3\)).

This means the dimension of our kernel is \(3-2=1\).

Answer 8

oh oh, we mess up there, the rref is \[\left[\begin{matrix}1 & 0&-2 \\ 0 & 1& -2\end{matrix}\right]\] so the basis for the range of T is \[\left(\begin{matrix}1 \\ 0\end{matrix}\right) and \left(\begin{matrix}0 \\ 1\end{matrix}\right)\]

Answer 9

So all we need is one vector in the null space of our matrix and we will have a basis. So, @Hoa, can you find one vector in the null space? :-) hint: reduce

Answer 10

@Hoa both are valid bases for \(\mathbb{R}^2\).

Answer 11

ok, x-2z =0 and
y-2z =0

Answer 12

Consider:$$\begin{bmatrix}1\\-2\end{bmatrix}+2\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}$$
... so they definitely are a fine basis :-) they're linearly independent and span our vector space, which is all that is needed (though yours is also orthonormal and canonical)

Answer 13

and z is free variable and I can express that free by \[\left(\begin{matrix}2 \\ 2\\1\end{matrix}\right)\]

Answer 14

Exactly correct @Hoa :-) which is then a fine basis for our kernel as its dimension is \(1\).

Answer 15

hi , what caniconal mean?

Answer 16

standard

Answer 17

ok, got it

Answer 18

A linear transform and the matrix underlying the transform have the same rank/nullity.

Answer 19

you mean, A and T?

Answer 20

that is a specific example, yes

Answer 21

http://math.stackexchange.com/questions/131950/the-rank-of-a-linear-transformation-matrix

Answer 22

so, in conclusion, with that kind of question, I have to get rref of A , consider the column space to get the pivots and free variable, number of pivots is the range of T , number of free is the ker (T) and base on that stuff, figure out the basis of them, right?

Answer 23

Sure... but in linear algebra there are often many neat approaches to a problem.

Answer 24

Thanks for your help. I appreciate. I rarely get help from that field.

Answer 25

from calculus, many good friends there, but to linear, it is not hard but complicated, and it sounds like they don't want to spend time with that. I don't know why

Answer 26

By the way, I typically try figuring out trivial linear combinations of the other column vectors I suspect of ruining linear independence... e.g. I merely did the following in my head:$$-2\begin{bmatrix}2\\-2\end{bmatrix}+2\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}-4+2\\4+2\end{bmatrix}=\begin{bmatrix}-2\\2\end{bmatrix}$$

Answer 27

nope, you make mistake there, right the first one -2*2 + 2 *0 = -4 , not -2

Answer 28

ooops