Show that $M^m + A = M^{m+1}$ with $M,A$ matrices?

Question

Let $M,A$ denote matrices with $$ M = \left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right), ~~~~ A = \left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right). $$

Show that $$M^m + A = M^{m+1}.$$

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Context: I encountered this problem when trying to prove that $M^m = E_2 + mA$ (1) by induction (where $E_2$ is the $2\times 2$ identity matrix. Adding $A$ on both sides yields $M^m + A = E_2 + mA +A =E_2 + (m+1)A$. Equation (1) would follow if $M^m + A = M^{m+1}$.

Answer 1

Since $MA=A$, we prove by induction.

When $m=0$, check that:

$$M^1=M^0+A$$

where $M^0=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I.$

Now, if $M^{m+1}=M^m+A$ then multiply both sides by $M$, and you get $$M^{m+2}=M^{m+1}+MA=M^{m+1}+A$$

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More generally, if $A^2=0$ and $M=I+A$ then $MA=A+A^2=A$ and you have again that $M^{m+1}=M^{m}+A$.

Answer 2

By the Cayley-Hamilton Theorem, $M$ satisfies its characteristic polynomial, so $(M-I)^2 = 0$ where $I$ is the identity matrix. Since $A = M - I$, then \begin{align*} M^{m+1} - M^m - A &= M^m(M-I) - (M-I) = (M^m - I)(M-I)\\ &= (M^{m-1} + M^{m-2} + \cdots + M + I)(M-I)^2 = 0 \end{align*} so $M^{m+1} = M^m + A$.