Question

Determine whether is in the range of the linear operator T. T: R^3 ----> R^3; T(x,y,z) = (x-y, x+y+z,x+2z); w=(1, 2, -1)

Answer 1

whether w is*

Answer 2

you want to create the matrix that represents that transformation, row reduce it, and find its Column Space. Then check to see if w is in the Column Space.

Answer 3

do u mean the matrix? if so i got [1 -1 0; 1 1 1; 1 0 1]?

Answer 4

for the first part

Answer 5

since the last equation was x + 2z, the last row should be 1 0 2. Other than that, you got it. So row reduce that matrix, see which columns are pivot columns, and the pivot columns will make up your column space.

Answer 6

I got an identity when I rref?

Answer 7

then you are done ;)

Answer 8

since you got the identity, that means all three columns are linearly independent, and they span all of R^3. So any vector in R^3 is in the range of that transformation. Including w.

Answer 9

but how do you know for sure w is part of it?

Answer 10

okay so what if it's not an identity?

Answer 11

you would need to check to see if w was in the span of the column vectors

Answer 12

check out
\[T\left(\frac{7}{3},\frac{4}{3},\frac{-5}{3}\right)\]

Answer 13

oh kk I get it, but what do u mean check out.. that transformation?

Answer 14

RREF this matrix

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

Answer 15

I got [1 0 0 7/3; 0 1 0 4/3; 0 0 1 -5/3]

Answer 16

notice the 7/3,4/3,-5/3

this tells us that

\[T\left(\frac{7}{3},\frac{4}{3},\frac{-5}{3}\right)=(1,2,-1)\]

and thus w is in the range

Answer 17

Sorry I still don't see it, how do you know just from the 7/3,4/3,-5/3 that it's in the range of (1, 2 ,-1)?