Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / Section B / 8 / a

False. Take the diagonal matrix $A=\mathrm{diag}(1,2)$ and its eigenvectors $\mathbf{x}=(1,0{)}^T$, $\mathbf{y}=(0,1{)}^T$. Their sum $(1,1{)}^T$ is mapped to $(1,2{)}^T$, which is not a scalar multiple of that sum. More generally, $A(\mathbf{x}+\mathbf{y})=a\mathbf{x}+c\mathbf{y}$ need not have the form $\lambda (\mathbf{x}+\mathbf{y})$ when $a\ne c$.

True. By linearity, $(A+B)\mathbf{x}=A\mathbf{x}+B\mathbf{x}=(a+b)\mathbf{x}$. Since $\mathbf{x}\ne 0$, it is an eigenvector of $A+B$ with eigenvalue

True. Applying the matrices in order gives $AB\mathbf{x}=A(b\mathbf{x})=ab\mathbf{x}$. The eigenvalue isThis conclusion only uses the shared eigenvector; it does not require $A$ and $B$ to commute on other vectors.