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

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.

If $A\mathbf{v}=\lambda \mathbf{v}$ for a nonzero eigenvector, then $A^2\mathbf{v}={\lambda }^2\mathbf{v}=A\mathbf{v}=\lambda \mathbf{v}$. Thus $\lambda (\lambda -1)=0$, soBoth values need not occur: the zero matrix has only $0$, and the identity matrix only $1$. More precisely, an idempotent linear map whose rank of a linear map is $r$ has $r$ copies of $1$ and $n-r$ copies of $0$. Indeed, its image of a linear map and kernel of a linear map form a direct sum, giving an eigenbasis without assuming an orthogonal projection.

On an eigenvector of $A$ with eigenvalue $\lambda$, the linear operator $B=I-A$ has eigenvalue $1-\lambda$. ThereforeThe multiplicities are exchanged: if the matrix rank of $A$ is $r$, then $B$ has $n-r$ copies of $1$ and $r$ copies of $0$. Again, a value with zero multiplicity is absent.

Computing the characteristic polynomial with the convention $\det (\lambda I-H)$ givesThus the three eigenvalues areThey can also be checked directly on the respective eigenvectors $(1,0,-1{)}^T$, $(1,-1,1{)}^T$, and $(1,2,1{)}^T$.

For the lowest eigenvalue, $-1$, solving $(H+I)\mathbf{v}=0$ gives $v_2=0$ and $v_3=-v_1$. A unit eigenvector is thereforeAny nonzero scalar multiple is an equally valid eigenvector.