Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics I / 2 / a

By definition, the image of a linear map $T_A$ isConsequently $\mathbf{b}\in \mathrm{im}{T}_A$ means exactly that some $\mathbf{x}\in {\mathbb{R}}^m$ satisfies $A\mathbf{x}=\mathbf{b}$. This is precisely the existence of a solution of the linear system.

Suppose ${\mathbf{x}}_0$ is one solution. By linearity,Thus the solution set of a linear system is exactly ${\mathbf{x}}_0+\ker T_A$. It has one member if the kernel is $\{0\}$. Conversely, if there is a unique solution and $\mathbf{z}\in \ker T_A$, then ${\mathbf{x}}_0+\mathbf{z}$ is also a solution, forcing $\mathbf{z}=0$. Combining this with i gives

If $\mathbf{b}$ is outside the image of a linear map $T_A$, there is no solution. Otherwise choose ${\mathbf{x}}_0$ with $A{\mathbf{x}}_0=\mathbf{b}$. By ii, a zero kernel gives a unique solution. If the kernel contains a nonzero vector $\mathbf{z}$, allare solutions. They are distinct, since $(t-s)\mathbf{z}=0$ forces $t=s$. Hence there are infinitely many. These alternatives exhaust all possibilities.