What could be a example of a linear transformation T:R^4 ----> R^3 such that

kernel T = span{(1,0,0,0),(0,0,1,0)} and image T = span{(1,0,0),(1,0,0)}. How do I justify this?

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- Apr 5th 2011, 04:27 PMxycodeExample of a linear transformation T:R^4 --> R^3?
What could be a example of a linear transformation T:R^4 ----> R^3 such that

kernel T = span{(1,0,0,0),(0,0,1,0)} and image T = span{(1,0,0),(1,0,0)}. How do I justify this? - Apr 5th 2011, 06:47 PMtonio

Basic and most important fact about linear transformations: given two vector spaces $\displaystyle V,U$ and

a basis $\displaystyle \{v_i\}_{i\in I}$ of $\displaystyle V$ and a set of vectors $\displaystyle \{u_i\}_{i\in I}$ in $\displaystyle U$, there exists al.t. $\displaystyle T:V\rightarrow U$__unique__

s.t. $\displaystyle Tv_j=u_j\,,\,\,\forall\,\,j\in I$

Well, prove the above (very easy) and try to use it to solve your problem.

Tonio - Apr 5th 2011, 07:04 PMDeveno
ok, we're looking for a 3x4 matrix that sends all vectors of the form (a,0,b,0) to (0,0,0), and has images of the form...um, i'm not sure. i think you made a typo. perhaps you meant to write T = span{(1,0,0),(0,1,0)}. i will provisionally adopt this, but if you meant a different span set, you'll have to adjust accordingly.

we might decide to make life easy on ourselves by setting the first column and the third column of our matrix to 0. that will ensure we get the kernel we want. seting the entire 3rd row to 0 will ensure we get the image we want. so one possibility is A =

[0 1 0 1 ]

[0 1 0 -1]

[0 0 0 0 ]

which sends (x1,x2,x3,x4) to (x2+x4,x2-x4,0). note that A(1,0,0,0) = (0,0,0), A(0,0,1,0) = (0,0,0), so any linear combination of those two vectors gets mapped to 0.

note also that A(1/2,1/2,0) = (1,0,0), and A(1/2,-1/2,0) = (0,1,0). if you just want the linear transformation, then set:

T(x1,x2,x3,x4) = (x2+x4,x2-x4,0).