1. Using the Dimension Theorem...

Let $L:R^{3}\rightarrow R^{2}$ be the linear mapping defined by $L\left(x,y,z\right)=\left(x-y,y-z\right)$.
Find the standard 2x3 matrix representation of L and compute L(1,1,1).
Show that the kernel of L is a subspace of R^{3}. What is the dimension of the kernel?
State the Dimension Theorem. Hence or otherwise find the image of R^{3} under the transformation L.

So I have $L = \left( \begin{array}{ccc}1 & -1 & 0 \\ 0 & 1 & -1 \end{array} \right)$. L(1,1,1)=(0,0).
The kernel of L has basis (1,1,1) and has dimension 1, and I've proven it is a subspace of R^{3}.

My problem is with the final piece of the question.
The Dimension Theorem states that given a vector space, any two bases have the same cardinality...

How am I supposed to use that to find the image of R^{3} under L?

Many thanks...

2. Originally Posted by Unenlightened
Let $L:R^{3}\rightarrow R^{2}$ be the linear mapping defined by $L\left(x,y,z\right)=\left(x-y,y-z\right)$.
Find the standard 2x3 matrix representation of L and compute L(1,1,1).
Show that the kernel of L is a subspace of R^{3}. What is the dimension of the kernel?
State the Dimension Theorem. Hence or otherwise find the image of R^{3} under the transformation L.

So I have $L = \left( \begin{array}{ccc}1 & -1 & 0 \\ 0 & 1 & -1 \end{array} \right)$. L(1,1,1)=(0,0).
The kernel of L has basis (1,1,1) and has dimension 1, and I've proven it is a subspace of R^{3}.

My problem is with the final piece of the question.
The Dimension Theorem states that given a vector space, any two bases have the same cardinality...

How am I supposed to use that to find the image of R^{3} under L?

Many thanks...
1. Convince yourself that Img(L) is a sub-space of R^2.
2. Now note (1,0) and (0,1) belong to Img(L). This Img(L) contains the sub-space spanned by (1,0) and (0,1)
3. (1) and (2) above give Img(L) = R^2

Other wise use rank-nullity theorem to get dim(Img(L)) = 2. As dimension of R^2 = 2 and Img(L) is a sub-space of R^2. Img(L)=R^2 is an immediate consequence.