Question

determine whether the following function is a linear transformation T:M22 ->M23 ;where B is a fixed 2*3 matrix and T(A)=AB

Answer 1

@TuringTest I remember you teaching this to me but i totally forgot everything

Answer 2

let's see
you know the two tests for a linear transformation?

Answer 3

yes...

Answer 4

have you tried to test the one for scalar multiplication? \[T(cA)=cT(A)\]?

Answer 5

im confusd by this part T(A)= AB ...i dont get how can i represent it

Answer 6

well the idea is that the transformation takes a 2x2 matrix and multiplies it by a 2x3 matrix, producing a 2x3 matrix

Answer 7

now if I have a matrix\[A=\left[\begin{matrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{matrix}\right]\]then if I multiply each element in the matrix by a constant c, can I factor it out of the matrix\[\left[\begin{matrix}ca_{11}&ca_{12}\\ca_{21}&ca_{22}\end{matrix}\right]=cA\]????

Answer 8

i understand this one......so what about T(A)=AB dont we satisfy each condition regarding this statement T(A)=AB

Answer 9

well the question is now asking that given the fact that\[T(A)=AB\]\[cT(A)=cAB\]is the same as\[T(cA)\]?
(i.e is it the same if we multiply c after the transformation of A or before?)

Answer 10

U mean i have to prove both sides to be equal??

Answer 11

yes, you must prove\[T(cA)=cT(A)\]

Answer 12

depending on how rigorous you want to be, this proof is not that hard

Answer 13

what is \[cT(A)\]?

Answer 14

this one i got it cT(A)=T(cA) ......what i dont understand is that 'do i multiply those matrices A & B'

Answer 15

no, you just have to show that the constant can be multiplied before or after teh multiplication
given the properties of matrices in general you don't need to resort to actually writing things out, though I did for you above show you that you can multiply every elemnt in a matrix by a constant c, then factor it out of the matrix
that is,\[cA=M\]where \(M\) is the matrix \(A\) with each element multiplied by \(c\)

Answer 16

so...\[T(A)=AB\]is should be obvious that if we multiply by c after we get\[c[T(A)]=c(AB)\]

Answer 17

now for the other, by the property I just stated you can say\[T(cA)=T(M)\]where \(M\) is as I said above just the new matrix you get by multiplying each elemt in \(A\) by the scalar \(c\)

Answer 18

\[T(M)=MB\]by definition of the transformation, and since \(M=cA\) we get\[T(cA)=cAB=cT(A)\]and we prove our statement

Answer 19

that was an overkill proof in my opinion by the way

Answer 20

next test is\[T(M)+T(N)=T(M+N)\]meditate on how you might approach that while I take a shower (think about the properties of matrix multiplication and addition)...
see ya in a bit

Answer 21

firstly i let M=[a11 a12: a21 a22] && N=[b11 b12: b21 b22]
then i wil get T(M+N) = T[(a11+b11) (a12+b12):(a21+b21) (a22+b22)]

T(M+N) IS NOT EQUAL TO AB
so it means that the condition does not hold

Answer 22

that is wrong
essentially the question is asking if matrix multiplication is distributive

Answer 23

\[T(M+N)=(M+N)B\]\[T(M)+T(N)=MB+NB\]so the question becomes\[(M+N)B=MB+NB\]

Answer 24

again, since we are assuming that N and M have the same dimension 2x2 we can add the two matrices and call this a third matrix P, where each element is the sum of the corresponding elemnts in N and M

Answer 25

is B a constant or a matrix this is what confuses me

Answer 26

B is a 2x3 matrix\[T(M+N)=T(P)=PB\]since each element in \(P\) is the sum of the two corresponding ones in \(M\) and \(N\), say the first entry in \(P\) which is\[m_{11}+n_{11}\]gets multiplied by the corresponding entry in \(B\)\[(m_{11}+n_{11})b_{11}\]since multiplication is distributive, we can write this as\[m_{11}b_{11}+n_{11}b_{11}\]and since we already established that we can add and separate matrices by addition we can write the set of all these distributions as two separate matrices\[MB+NB=T(M)+T(N)\]and so we have demonstrated it (I might not want to use the word prove here, but we showed it makes sense at least)

Answer 27

We could have just said that since matrix multiplication is distributive \[T(M+N)=(M+N)B=MB+NB=T(M)+T(N)\]but I wanted to demonstrate that it was indeed distributive so long as the matrices can be multiplied

Answer 28

i got it now but it was very tricky...thanks

Answer 29

then it means that it is linear transformation

Answer 30

yes, and I think the second version I present makes it easy if you already know matrix multiplication is distributive
I think you should review the properties of matrix operations, that would help you here
http://tutorial.math.lamar.edu/Classes/LinAlg/PropsOfMatrixArith.aspx

Answer 31

rules 5 and 9 on this page are really all we used to prove that this is a linear transformation

Answer 32

i got them....