Matrix multiplication is associative
Dependencies:
Let $R$ be a semiring. Let $A \in \mathbb{M}_{n_0, n_1}(R), B \in \mathbb{M}_{n_1, n_2}(R), C \in \mathbb{M}_{n_2, n_3}(R)$. Then $(AB)C = A(BC)$ and \[ (ABC)[p, q] = \sum_{i=1}^{n_1} \sum_{j=1}^{n_2} A[p, i] B[i, j] C[j, q] \]
Proof
\begin{align} & ((AB)C)[p, q] \\ &= \sum_{j=1}^{n_2} (AB)[p, j] C[j, q] \\ &= \sum_{j=1}^{n_2} \left( \sum_{i=1}^{n_1} A[p, i] B[i, j]\right) C[j, q] \\ &= \sum_{j=1}^{n_2} \sum_{i=1}^{n_1} (A[p, i] B[i, j]) C[j, q] \tag{Distributivity in $R$} \\ &= \sum_{i=1}^{n_1} \sum_{j=1}^{n_2} (A[p, i] B[i, j]) C[j, q] \tag{Additive commutativity in $R$} \\ &= \sum_{i=1}^{n_1} \sum_{j=1}^{n_2} A[p, i] (B[i, j] C[j, q]) \tag{Multiplicative associativity in $R$} \\ &= \sum_{i=1}^{n_1} A[p, i] \left( \sum_{j=1}^{n_2} B[i, j] C[j, q] \right) \tag{Distributivity in $R$} \\ &= \sum_{i=1}^{n_1} A[p, i] (BC)[i, q] \\ &= (A(BC))[p, q] \\ &\Rightarrow (AB)C = A(BC) \end{align}
Dependency for:
- Eigenpair of power of a matrix
- All eigenvalues of a hermitian matrix are real
- Square matrices form a (semi)ring
- Full-rank square matrix is invertible
- Row equivalence matrix
- Inverse of a matrix
- Bounding matrix quadratic form using eigenvalues
- Inverse of product
- AB = I implies BA = I
- Determinant of product is product of determinants
- Equations with row equivalent matrices have the same solution set
Info:
- Depth: 3
- Number of transitive dependencies: 4