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Find a basis for the range and the kernel of T:R3->R2 defined byT(x1, x2, x3) =...
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High School Teacher
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I'll start with the range and use column vectors to more easily distinguish between scalars and vectors. Vectors in the range can be written as
`[[x_1+x_2],[x_1+x_2+x_3]]=(x_1+x_2)[,]+x_3[,],` and since `[,]` and `[,]` are independent, they form a basis for the range. Since the two dimensional range is contained in `RR^2,` which is also two dimensional, the range must be all of `RR^2.`
The dimension of the kernel must be dim(`RR^3)-`dim(range T)= `3-2=1` by the Rank-Nullity Theorem (this isn't necessary to solve the problem, but it's helpful information to know in advance). If
`[[x_1],[x_2],[x_3]]` is in the null space of `T,` then we get the system of equations
which is easily solved to get `x_3=0, x_2=-x_1,` where `x_1` is a free variable. Thus the kernel consists of vectors of the form
so a basis for the null space is the single vector (which we expected by what was said above)
One final relevant point is that if you represent `T` by the matrix
`[[1,1,0],[1,1,1]],` then the basis we found for the range is just the last two columns. This is no coincidence.
Posted by degeneratecircle on January 30, 2013 at 6:35 AM (Answer #1)
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