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

\[\left[\begin{matrix}1 & 0 & -2 \\ -2 & 1 & 2\end{matrix}\right]\]

Answer 2

Have you row reduced the matrix yet? All of the answers to your questions can be found by first row reducing the matrix.

Answer 3

its rref is \[\left[\begin{matrix}1 & 0 & -2\\ 0 & 1 & -2\end{matrix}\right]\]

Answer 4

Alright. Lets knock out the easy ones first. How many pivot variables does that matrix have? and how many free variables are there? (I'm not sure what your class is calling them, so if you dont know the definitions of some terms I say, let me know).

Answer 5

2 pivots, and 0 free variable

Answer 6

You are correct about the pivots, but there is actually one free variable. It goes by the columns. So two columns have leading ones in them (those are the pivots), and one column doesnt have a leading one (the last one).

Answer 7

oh, yea, my bad, 1 free

Answer 8

but my logic is 2 equations, 3 variables, ---> 1 free

Answer 9

The rank of A = number of pivots
The nullity of A = number free variables.

The rank and nullity of T are the same as the rank and nullity of A (since A is the matrix representation of T).

Answer 10

so the rank of T is 2
nullity of A is 1?

Answer 11

That is correct.

Answer 12

Hi friend, I confused. It sounds like T =A?

Answer 13

I don't think so...different meanings, but then again I have yet to reach that section

Answer 14

@UsukiDoll LOL,

Answer 15

T is a linear transformation
A is simply a matrix

Answer 16

T(x)= Ax, so whether you are talking about a transformation (T), or matrix multiplication (A), they mean the same thing. They are linked.

Answer 17

while they go hand-in-hand, A is just one of the many possible matrix representations of T.

Answer 18

oh man... x*( like I said I'm not on that sectyion

Answer 19

well now I know something in advance ;)

Answer 20

This is Linear Algebra stuff right?

Answer 21

who needs matrices anyway

Answer 22

yes, this is Linear Algebra.

Answer 23

ok I'm on the same page except I'm struggling with a proof and need a refresher course on subspaces...the ones with the u+v and cu properties.

Answer 24

@joemath314159 so, in simple way, I don't get. more explanation please :)

Answer 25

what don't you get Hoa?

Answer 26

everything =)

Answer 27

do not underestimate Hoa, Usukidoll :)

Answer 28

whether A =T or not?

Answer 29

oh. that is comparing apples and oranges, Hoa. they are related, but not the same thing
A is a matrix
T is a transformation which COULD be represented by A.

Answer 30

got it, continue the stuff, I want a and b parts

Answer 31

umm I think he went

Answer 32

probably need @joemath314159 for that.
i don't know how to write matrices

Answer 33

wahhh I need my proof correction to be checked too. ;__:

Answer 34

we need someone really good with matrices~ i'm not

Answer 35

so, use my way, (top..., bottom...)

Answer 36

this better be a connection error. I'm gonna be so sad if it's not

Answer 37

or draw the matrix.

Answer 38

@TaylorNeedsHelp What, write it in latex? Use begin matrix and seperate the parts with & and use \\ for new lines.

Answer 39

drawing is easier

Answer 40

what's latex?

Answer 41

does anyone know proof writing

Answer 42

program for writing math formulas

Answer 43

ok, you can go to Latex Practicing by clicking Find More Subjects and practice your skill there

Answer 44

hehehe.... good luck!!!

Answer 45

in time.

Answer 46

@TaylorNeedsHelp LaTeX is a formatting language tha got adapted into the MathML thing used to show equations here.

Answer 47

O___O

Answer 48

that's how the people are writing those fancy equations ?

Answer 49

or you can use those post to practice right here, I am ok.

Answer 50

yeah

Answer 51

hey, UsukiDoll, are you good with matrices? :)

Answer 52

if you know someone who can help me correct a proof...

Answer 53

gawd I don't even know if my corrections is in the right track or went to spaceville

Answer 54

tit for tat, UsukiDoll? Show me your proof

Answer 55

I'm on I think...basis and dimensions for the matrices

Answer 56

The Kernel is the solution set, or zeros of it.

Answer 57

but I don't know subspaces and subsets too good.

Answer 58

that's why I'm here lol

Answer 59

you're in luck, UsukiDoll, I'm ok with vector spaces

Answer 60

ok do you know how to prove whether or not the tests for u+v and cu work?

Answer 61

or do I just plug stuff into the equations/axioms?

Answer 62

post your question.

Answer 63

not that i have to tell you that

Answer 64

can't I'm stuck with a proof question

Answer 65

So the basis of the kernel is a basis that is related to the parameteric equation that solves for the null space of A.

Answer 66

unless you want me to post here.

Answer 67

get Hoa's permission

Answer 68

Hoa can I hijack your thread? >:)

Answer 69

do threads usually get this long?

Answer 70

The range, well, that is the output of the transformation. Hmm. Not sure how to get the range of the results.

Answer 71

my own thread is lasting for three days and counting

Answer 72

what thread? link

Answer 73

Or basis of the results... I mean.

Answer 74

http://openstudy.com/study#/updates/51651f5ce4b066fca660e3c8

Answer 75

like one of the questions would be Which are the following are subsets?
The set of all nxn symmetric matrices
The set of all nxn diagonal matrices
The set of all nxn nonsingular matrices
do I just use the axiom formulas to determine if there is such a thing or?

Answer 76

subsets need a superset

Answer 77

what the!

Answer 78

You cannot define subsets if they are not subsets of anything... right, @Hoa ?

Answer 79

what is a superset?

Answer 80

all you go to far and go around.
since it fromR^3--> R^2 so, the rank(t) =2 and the basis of R^2 is (0,1) and (1,0) hehehe...It sounds right?

Answer 81

what the heck?

Answer 82

what? I said heck not a swear word

Answer 83

to me, all of UsukiDoll ask for about the subsets are under the "Square Matrices"

Answer 84

Set

Answer 85

hihi.... sorry for breaking the sentence down.

Answer 86

I am not good at matrices, that is why i am here to make question, the good thing is I understand what I learn hihihi.... all you make question, If I know, I will answer hihi..

Answer 87

My prof has been out 3 days this semester, itherwise I would be far enough a long to perhaos have more help. This stuff is coming up soon in our class. Have not looked ahead much because doing test prep.

Answer 88

no question? friends? if so, I gonna close the question

Answer 89

k, I think your your basis for the kernel of T is going to be \[\left(\begin{matrix}2 \\ 2\\1\end{matrix}\right)\] and the basis for the range of T is the two pivot columns of the original matrix \[ \left(\begin{matrix}1 \\ -2\end{matrix}\right) and \left(\begin{matrix}0 \\ 1\end{matrix}\right)\]

Answer 90

If you want to see a reference video, this is what I was looking at:
http://www.youtube.com/watch?v=Vmr6EAQQazA

Answer 91

@Hoa Some good reading here: http://tinyurl.com/cxr5rgo