Prove whether or not a transformation is one to one.

[thatguythere]

Homework Statement

Let T: R³ -> M₂₂ by
T\begin{bmatrix}
a \\
b \\
c
\label{T}
\end{bmatrix}
=
\begin{bmatrix}
a-b & b-c \\
a+b & b+c\\
\end{bmatrix}

Is this transformation one-to-one?

Homework Equations

The Attempt at a Solution

I am not really certain how to go about this. A nudge in the right direction?

 

[Dick]

Homework Statement

Let T: R³ -> M₂₂ by
T\begin{bmatrix}
a \\
b \\
c
\label{T}
\end{bmatrix}
=
\begin{bmatrix}
a-b & b-c \\
a+b & b+c\\
\end{bmatrix}

Is this transformation one-to-one?

Homework Equations

The Attempt at a Solution

I am not really certain how to go about this. A nudge in the right direction?

A linear transformation is one-to-one if it's kernel is {0}. Try computing the kernel.

 

[HallsofIvy]

First, use "tex" and "/tex" at the beginning and end of an entire equation, not for individual parts:
$$T\begin{bmatrix}a \\ b \\ c \end{bmatrix}= \begin{bmatrix}a- b & b- c \\ a+ b & b+ c\end{bmatrix}$$.

Now, can we assume that you understand what "one to one" means? T is one to one if Tu= Tv only if u= v. Taking
$$u= \begin{bmatrix}a \\ b \\ c\end{bmatrix}$$
and
$$v=\begin{bmatrix}x \\ y \\ z\end{bmatrix}$$
then Tu= Tv means that
$$\begin{bmatrix}a- b & b- c \\ a+ b & b+ c\end{bmatrix}= \begin{bmatrix}x- y & y- z \\ x+ y & y+ z\end{bmatrix}$$
which is the same as a- b= x- y, b- c= y- z, a+ b= x+ y, and b+ c= y+ z.

Can you prove from those equations that a= x, b= y, and c= z?
(Dick's method is equivalent to this, taking x= y= z= 0.)

 

[thatguythere]

So if T(u) = T(v) then 0 = T(u)-T(v) = T(u-v) so u-v is in kerT and u-v=0 so u=v?
But only if the kerT=0.

 

[Dick]

So if T(u) = T(v) then 0 = T(u)-T(v) = T(u-v) so u-v is in kerT and u-v=0 so u=v?
But only if the kerT=0.

That's it exactly. So just find ker(T).