# Find a basis for the range and the kernel of T:R3->R2 defined byT(x1, x2, x3) = (x1+x2, x1+x2+x3)Find a basis for the range and the kernel of T:`RR^(3)``|->` `RR^(2)` defined byT(`x_(1)`,...

Find a basis for the range and the kernel of T:R3->R2 defined by

T(x1, x2, x3) = (x1+x2, x1+x2+x3)

Find a basis for the range and the kernel of T:`RR^(3)``|->` `RR^(2)` defined by

T(`x_(1)`, `x_(2)`, `x_(3)`) = (` ``x_(1)+x_(2)`, ` ``x_(1)+x_(2)+x_(3)`)

*print*Print*list*Cite

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)[[1],[1]]+x_3[[0],[1]],` and **since `[[1],[1]]` and `[[0],[1]]` 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

`x_1+x_2=0`

`x_1+x_2+x_3=0,`

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

`[[x_1],[-x_1],[0]]=x_1[[1],[-1],[0]],`

**so a basis for the null space is the single vector (which we expected by what was said above)**

`[[1],[-1],[0]].`

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.