If $S\colon \mathbb{R}^3\to\mathbb{R}^4$ and $T\colon \mathbb{R}^4\to \mathbb{R}^3$ satisfy $TS=\operatorname{id}$, then is $ST=\operatorname{id}$? Let $S\colon \mathbb{R}^3\to \mathbb{R}^4$ and $T\colon \mathbb{R}^4\to \mathbb{R}^3$ be two linear transformations. If $TS$ is an identity map then can we say that $ST$ is also an identity map? I think it should  be, because $ST\colon \mathbb{R}^3\to \mathbb{R}^3$. Please guide me.
 A: There are two conventions about the order of multiplication. The more common one is that $AB$ means do $B$ first and then $A$. 
With that convention, let $S$ be the map from $\mathbb{R^3}$ to $\mathbb{R}^4$ that takes $(a,b,c)$ to $(a,b,c,0)$. Let $T$ be the map from $\mathbb{R}^4$ to $\mathbb{R}^3$ that takes $(a,b,c,d)$ to $(a,b,c)$. Then $TS$ is the identity map from $\mathbb{R}^3$ to $\mathbb{R}^3$. But $ST$ is certainly not the identity map from $\mathbb{R}^4$ to $\mathbb{R}^4$.   
If we use the opposite convention, the result is vacuously true. Any linear map from $\mathbb{R}^4$ to $\mathbb{R}^3$ has image of dimension $\le 3$. If this is followed by a linear map from $\mathbb{R}^3$ to $\mathbb{R}^4$, the image remains of dimension $\le 3$. Thus the composition cannot be the identity mapping. 
A: Not necessarily. For example, consider 
$S\colon \mathbb{R}^3\to \mathbb{R}^4:(x,y,z)\mapsto (x,y,z,0)$ & $T\colon \mathbb{R}^4\to \mathbb{R}^3:(x,y,z,w)\mapsto (x,y,z)$. Then $TS:\mathbb R^3\to\mathbb R^3$ is the identity map. But $ST:\mathbb{R}^4\to \mathbb{R}^4:(x,y,z,w)\mapsto(x,y,z,0)$ is not the identity map.
Note: 


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*$S(\alpha(x_1,y_1,z_1)+(x_2,y_2,z_2))=(\alpha x_1+x_2,\alpha y_1+y_2,\alpha z_1+z_2,0)=(\alpha x_1,\alpha y_1,\alpha z_1,0)+(x_2,y_2,z_2,0)=\alpha S(x_1,y_1,z_1)+S(x_2,y_2,z_2).$

*$T(\alpha(x_1,y_1,z_1,w_1)+(x_2,y_2,z_2,w_2))=(\alpha x_1+x_2,\alpha y_1+y_2,\alpha z_1+z_2)=\alpha(x_1,y_1,z_1)+(x_2,y_2,z_2)=\alpha T(x_1,y_1,z_1,w_1)+T(x_2,y_2,z_2,w_2).$
