Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 4 / 40B (Numerical Analysis)

Jordan form shows each block of $A^k$ is a polynomial in $k$ times ${\lambda }^k$. These tend to zero exactly when every $|\lambda |<1$, i.e. $\rho (A)<1$.

The same Jordan calculation gives convergence of $\sum A^k$ exactly for $\rho (A)<1$. Finite sums satisfy $(I-A){\sum }_{k=0}^m{A}^k=I-A^{m+1}$; taking limits gives $(I-A{)}^{-1}$. Conversely convergence forces $A^k\to 0$, so part (a) applies.

If $B\ge 0$ and $\rho (B)<1$, then $(I-B{)}^{-1}={\sum }_{k\ge 0}{B}^k$, an entrywise nonnegative limit.

Put $G=M^{-1}N\ge 0$ and note $A=M(I-G)$. If $\rho (G)<1$, then $A^{-1}=(I-G{)}^{-1}{M}^{-1}\ge 0$. Conversely, if $A^{-1}\ge 0$, take $x\ge 0$, $x\ne 0$, with $Gx=\rho x$. Since $M^{-1}=A^{-1}(I-G)$, $M^{-1}x=(1-\rho )A^{-1}x$. The left side is nonnegative and nonzero, forcing $1-\rho >0$. Finally the iteration error is $e_{k+1}=G{e}_k$, so part (a) gives convergence for every start exactly under this condition.