Let $S:R^3 \rightarrow R^2$ and $T :R^2 \rightarrow R^3$ be linear transformations. Is it possible for $S\circ T$ to be the identity linear transformation? Is it possible for $T \circ S$ to be the identity linear transformation? In each case, if it is possible, give an example, and if it is not possible, explain why it is not.
I was able to identify that it will be possible for $S\circ T$ to be the identity linear transformation. I said that for $S\circ T$ to make sense the codomain of $T$ and domain of $S$ must match and in this case they do. Therefore the resulting composite transformation $S\circ T$ goes from the domain of $T$ to the codomain of $S$ and hence is $S\circ T: R^2 \rightarrow R^2$ so the result is standard matrix and we can use the identity matrices of a $2\times 3 $ matrix multiplied by $3\times 2$ matrix as an example which works out as the identity matrix of $2\times 2$ matrix. However I am unsure why $T\circ S$ would not be the identity transformation and unsure how the image and rank would come into the solution
Any help would be much appreciated
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$\begingroup$$S\circ T$ could be the identity. Let $S:\mathbb R^3\to \mathbb R^2$ be given by $S(x,y,z)=(x,y)$. And let $T:\mathbb R^2\to\mathbb R^3$ be given by $T(x,y)=(x,y,0)$. Then $S\circ T=I_{\mathbb R^2}$.
On the other hand, we can't get $T\circ S$ to be the identity of $\mathbb R^3$, because the rank-nullity theorem tells us that the rank of $T:\mathbb R^2\to \mathbb R^3$ is at most $2=\operatorname{dim}\mathbb R^2$. But the dimension of the image equals the rank. This means $T$ can't be surjective. Hence neither can $T\circ S$.
$\endgroup$ $\begingroup$If the transformation $T \circ S: R^3 \longrightarrow R^3$ was the identity then $\dim(im(T \circ S)) = 3$. However since $ T : R^2 \longrightarrow R^3$ is linear transformation it cannot be surjective. The dimension of it's image must be at most $2$. So we would get $dim(im(T \circ S)) = \dim(im(T|_{im(S)})) \leq 2$ which would be contradiction with $\dim(im(T \circ S)) = 3$ .
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