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

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.

No. Here $A$ has more rows than columns, which does not force a nonzero kernel. For example, with $m=1$, $n=2$,the linear system has the unique solution $x=0$. With the same $A$ and $\mathbf{b}=(0,1{)}^T$ it has no solution at all.

No. Take $A=I_m$ and $\mathbf{b}=0$. The homogeneous linear system then has only the solution $\mathbf{x}=0$. Homogeneity ensures existence, but infinitely many solutions would additionally require a nonzero kernel.

No. The rank-nullity theorem givesbut this guarantees infinitely many solutions only when the linear system is consistent. For a counterexample takeThere is no solution, since the left side always vanishes.

Yes. A surjective $T_A$ has $\mathrm{im}{T}_A={\mathbb{R}}^n$, so every $\mathbf{b}$ is attained. Its rank of a linear map is $n$, and the rank-nullity theorem givesThe solution set of a linear system is therefore a translate of a nonzero vector subspace, containing infinitely many solutions as in iii.

The first two equations give $x_1=1-2x_3$ and $x_2=2-5x_3$. Substitution into the third givesThis equation is inconsistent exactly when its left side is identically zero and its right side is nonzero. Therefore

The elimination in i determines $x_3$ uniquely exactly when $\lambda \ne 5$. The other two coordinates are then also fixed. Explicitly,For $\lambda =5$, the reduced equation is either inconsistent or leaves $x_3$ free, so a unique solution is impossible.

By i, a free choice of $x_3$ is possible precisely when $\lambda =5$ and $\mu =2$. ThusThe complete solution set of a linear system isThe direction vector spans the kernel, and the displayed straight line is the entire solution set of a linear system.