$UT$ is not Invertible Let $T$ be a linear transformation from $\mathbb{R}$$^3$ into $\mathbb{R}$$^2$, and let $U$ be a linear transformation from $\mathbb{R}$$^2$ to $\mathbb{R}$$^3$. Prove that the transformation $UT$ is not invertible.
Is there any generalisation of this?
 A: Remember the rank-nullity theorem:http://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem:
If you have a linear map $L :V\rightarrow W$ , then the dimension of $V$ equals the sum of the rank of $L$ ( the dimension of the image)+ the nullity , equivalently the dimension of the nullspace. In our case, we have a linear map $T: \mathbb R^2 \rightarrow \mathbb R^3 $ , so that $dimV=2$. Then $$ 2=Dim(ImT)+Dim(NullT) $$
From this it follows that $Dim(ImT) \leq 2$.
Now, we have a composition of maps $$T: \mathbb R^3 \rightarrow \mathbb R^2 U: \mathbb R^2 \rightarrow \mathbb R^3 $$
Even if the image of $T$ has rank $2$, the image $UT$ will be the image of a 2-D space so that the composed map $UT$ cannot have a three-dimensional image in $\mathbb R^3$, which is what you need for an injection into $\mathbb R^3$.
A: 
Generalization:$V$ and $W$ are finite-dimensional vector spaces over $F.$ Let $T$ be a linear transformation from $V$ to $W,$ and let $U$ be a linear transformation from $W$ to $V$ with $\dim V >\dim W.$ Then $UT$ is not invertible.

We see that $UT$ is a linear transformation from $V$ to $V.$ Let $\dim V=m.$ If $\{\alpha_i \}_{i=1}^{i=m}$ is a basis for $V,$ then $UT$ is completely determined by its action on these basis vectors.
Consider $\{T(\alpha_i)\}_{i=1}^{i=m}.$ This set is clearly linearly dependent. Thus there exists $c_i$'s $\in F,$ not all zero such that $\sum_{i=1}^{i=m}c_iT(\alpha_i) =0.$ Applying $U$ on both sides we get $\sum_{i=1}^{i=m}c_iUT(\alpha_i) =0.$
The set $\{UT(\alpha_i)\}_{i=1}^{i=m}$ is not a basis for $V$ and thus $UT$ is not invertible.
