Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 1 / 6

Direct matrix multiplication givesConsequently $\boxed{A^2-6A+5I=0}$, the Cayley-Hamilton theorem for its characteristic polynomial $t^2-6t+5$. Rearranging gives $A(6I-A)=5I$, so

For $\mathbf{u}=(1,1{)}^T$ and $\mathbf{v}=(1,-1{)}^T$, the outer products areTheir inner product is zero and both squared norms are two. Thus $P_u=\mathbf{u}{\mathbf{u}}^T/2$ and $P_v=\mathbf{v}{\mathbf{v}}^T/2$ are orthogonal spectral projectors, with $P_u+P_v=I$. Since $A\mathbf{u}=5\mathbf{u}$ and $A\mathbf{v}=\mathbf{v}$, the eigenvalues are $\lambda =5$, $\mu =1$, andMultiplying these decompositions uses $P_u{P}_v=0$ and $P_u^2=P_u$, $P_v^2=P_v$, giving the identity matrix.