Question

Is the map T : R^2 ? R^3 given by T(x, y) = (x + y, x, 2y) a linear transformation? Explain

Answer 1

Since we have shown 1) and 2), we have proved that the given is a linear transformation.

I have defined 1) and 2) below and showed both.

Explanation:

We want to show the following:

1)
`T(u)+T(v)=T(u+v)`

and

2)
`T(cu)=cT(u)`

where u and v are in R^2 and c is a real number.

Let
`u=(r,s)` and
`v=(m,n)`.

We are going to find sides have 1) and then see if we have the same thing.

Let's start with the left side:

`T(u)+T(v)`

`T((r,s))+T((m,n))`

`(r+s,r,2s)+(m+n,m,2n)` (Applied T's rule)

`(r+s+m+n,r+m,2s+2n)`.

Let's do the right side now:

`T(u+v)`

`T((r,s)+(m,n))`

`T((r+m),(s+n))`

`(r+m+s+n,r+m,2(s+n))` (Applied T's rule)

The first elements of each match (the r+m+s+n).

The second elements match (the r+m).

The third elements are actually also equal by distributive property. That is 2(s+n)=2s+2n.

So 1) has been show.

Let's look at 2) now.

We will do the same for it as we did for showing 1).

Start with the left hand side:

`T(cu)`

`T(c(r,s))`

`T((cr,cs))`

`(cr+cs,cr,2(cs))` (Applied T's rule)

`(cr+cs,cr,2cs)`

Now the right side:

`cT(u)`

`cT((r,s))`

`c[(r+s,r,2s)]` (Applied T's rule)

`(c(r+s),cr,c(2s))`

`(c(r+s),cr,2cs)`

The first elements are the same by distributive property.

The second elements are the same.

The third elements are also the same.

Since we have shown 1) and 2), we have proved that the given is a linear transformation.