I have a question regarding the proof (seen below) of Putzers algorithm for matrix exponentiation. It's written by our danish lecturer at the university, so I translated the important parts into somewhat readable english.
What I don't understand is the line:
$$\sum^{n-1}_{k = 0} (\lambda_{k+1} r_{k+1}(t) + r_r(t))P_k - A \sum^{n-1}_{k = 0} r_{k+1}(t)P_k = \sum^{n-1}_{k = 0} (-r_{k+1}(t)(A - \lambda_{k+1}I)P_k + r_k(t)P_k)$$
Can anyone tell me how to go from the left hand side to the right hand side?
Definition: Let $A \in Mat_{n, n}(\mathbb{C})$. Let $\lambda_1, \dotsc, \lambda_n$ be the eigenvalues of $A$ (counting algebraic multiplicity). Let $P_0 = I$ and for $k = 1, \dotsc, n$, $P_k = \Pi^{k}_{j = 1}(A - \lambda_j I)$. Define \begin{equation*} Q(t) = \sum^{n-1}_{k = 0} r_{k+1}(t)P_k \end{equation*} where $r_1(t) = e^{\lambda_1 t}$ og for $k = k, \dotsc, n$ \begin{equation} r_k(t) = e^{\lambda_k t} \int^{t}_{0} e^{-\lambda_k s} r_{k-1}(s) \, d s \end{equation} It is now true that $Q(0) = I$ and $Q'(t) = A Q(t)$
Proof: We observe that $r_1(0) = 1$ and $r_k(0) = 0$ for $k = 2, \dotsc, n$, so $Q(0) = P_0 = I$.
For $k > 1$ we now have that \begin{align*} r'_k(t) &= \lambda_k e^{\lambda_k t} \int^{t}_{0} e^{-\lambda_k s} r_{k-1}(s) \, ds + e^{\lambda_k t} e^{-\lambda_k t} r_{k-1}(t)\\ &= \lambda_k r_k(t) + r_{k-1}(t) \end{align*} Defining $r_0(t) = 0$ we also see that this is true for $k = 1$.
We now have \begin{equation*} Q'(t) = \sum^{n-1}_{k = 0}r'_{k+1}(t) P_k = \sum^{n-1}_{k = 0} (\lambda_k r_k(t) + r_{k-1}(t))P_k \end{equation*} And therefore \begin{align*} Q'(t) - A Q(t) &= \sum^{n-1}_{k = 0} (\lambda_{k+1} r_{k+1}(t) + r_r(t))P_k - A \sum^{n-1}_{k = 0} r_{k+1}(t)P_k\\ &= \sum^{n-1}_{k = 0} (-r_{k+1}(t)(A - \lambda_{k+1}I)P_k + r_k(t)P_k)\\ &= \sum^{n-1}_{k = 0} (-r_{k+1}(t)P_{k+1} + r_k(t)P_k)\\ &= -r_n(t)P_n\\ &= 0 \end{align*} Since $(-1)^n P_n = p_A(A) = 0$ from the Cayley-Hamilton theorem.
Put everything into one sum, rearrange the summands (we omit $\sum$s and $P_k$s below):
$$ (\color{Blue}{\lambda_{k+1}} \color{Teal}{r_{k+1}} + \color{Purple}{r_k}) - \color{Blue}{A}\color{Teal}{r_{k+1}} = -\color{Teal}{r_{k+1}}\color{Blue}{(A-\lambda_{k+1}I)} + \color{Purple}{r_k}.$$