Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 1 / 7B (Vectors and Matrices)

Expanding the characteristic polynomial givesThus the eigenvalues are $1,2,3$. Solving the corresponding homogeneous linear systems givesEvery nonzero vector in the displayed eigenspace is an eigenvector. Direct multiplication verifies the three eigenvalue equations. The distinct eigenvalues imply these eigenvectors are linearly independent and form a basis of ${\mathbb{R}}^3$.

The equation is solvable exactly when $b\in \mathrm{im}A$, the image of a linear map, equivalently when the rank of a matrix satisfies $\mathrm{rank}A=\mathrm{rank}[A\mid b]$. If it is solvable and $x_0$ is one solution, every solution is preciselyIndeed differences of solutions lie in the kernel of a linear map, and adding a kernel vector preserves the equation. Hence there are no solutions when $b\notin \mathrm{im}A$; one solution when $A$ is invertible; and infinitely many solutions when $b\in \mathrm{im}A$ and $A$ is singular. In the last case the rank-nullity theorem gives a nonzero kernel vector $v$, and $x_0+tv$ gives distinct solutions for every real $t$. This proves that a finite number greater than one is impossible.

Use the eigenvector basis $v_1=(-1,0,1{)}^T$, $v_2=(1,1,0{)}^T$, $v_3=(1,1,1{)}^T$. The forcing has decomposition $b=-v_1+3v_3$. If $x=c_1{v}_1+c_2{v}_2+c_3{v}_3$, the linear system reduces toFor $\lambda =0$, all coefficients are determined, givingFor $\lambda =1$, the first equation would say $0=-1$, so there are no solutions. For $\lambda =2$, $c_1=1$, $c_3=3$, and $c_2$ is arbitrary. Thus all solutions areThese outcomes illustrate respectively an invertible matrix, incompatible forcing outside its image of a linear map, and a compatible forcing with a nontrivial kernel of a linear map.