Matrix multiplication subtraction

Published:

Theorem

Let Mi,Ni∈Rn×nM_i, N_i \in \mathbb{R}^{n \times n} for i=1,2,…,ki = 1, 2, \dots, k be some matrices. Then, for all k∈Nk \in \mathbb{N}, the following holds:

M1M2…Mk−N1N2…Nk==(M1−N1)N2…Nk+M1(M2−N2)N3…Nk+⋯+M1M2…Mk−1(Mk−Nk)\begin{align*} &M_1 M_2 \dots M_k - N_1 N_2 \dots N_k = \\ &= (M_1 - N_1) N_2\dots N_k + M_1(M_2 - N_2)N_3\dots N_k + \dots + M_1 M_2 \dots M_{k-1}(M_k - N_k) \end{align*}

Proof

The proof is by induction on kk.

Base case

For k=1k = 1, the statement is trivially true:

M1−N1=M1−N1M_1 - N_1 = M_1 - N_1

Inductive step

Assume the statement holds for k=nk = n:

M1M2…Mn−N1N2…Nn==(M1−N1)N2…Nn+M1(M2−N2)N3…Nn+⋯+M1M2…Mn−1(Mn−Nn)\begin{align*} &M_1 M_2 \dots M_n - N_1 N_2 \dots N_n = \\ &= (M_1 - N_1) N_2 \dots N_n + M_1 (M_2 - N_2) N_3 \dots N_n + \dots + M_1 M_2 \dots M_{n-1} (M_n - N_n) \end{align*}

Then, for k=n+1k = n + 1, we have:

M1M2…MnMn+1−N1N2…NnNn+1==(M1M2…Mn−N1N2…Nn)⏟Inductive hypothesisNn+1+M1M2…Mn(Mn+1−Nn+1)==(M1−N1)N2…NnNn+1+M1(M2−N2)N3…NnNn+1++⋯+M1M2…Mn−1(Mn−Nn)Nn+1+M1M2…Mn(Mn+1−Nn+1)□\begin{align*} &M_1 M_2 \dots M_{n}M_{n+1} - N_1 N_2 \dots N_{n}N_{n+1} = \\ &= \underbrace{(M_1 M_2 \dots M_n - N_1 N_2 \dots N_n)}_{\text{Inductive hypothesis}} N_{n+1} + M_1 M_2 \dots M_n(M_{n+1} - N_{n+1}) = \\ &= (M_1 - N_1) N_2 \dots N_n N_{n+1} + M_1 (M_2 - N_2) N_3 \dots N_n N_{n+1} + \\ &+\dots + M_1 M_2 \dots M_{n-1} (M_n - N_n)N_{n+1} + M_1 M_2 \dots M_n(M_{n+1} - N_{n+1}) \quad \square \end{align*}