Identity matrix is identity of matrix product

Dependencies:

1. Semiring
2. Matrix
3. Identity matrix
4. Comparing coefficients of a polynomial with disjoint variables

Let $R$ be a semiring with unity. \[ (\forall A \in \mathbb{M}_{m, n}(R), AB = A) \iff B = I_n. \] \[ (\forall A \in \mathbb{M}_{m, n}(R), BA = A) \iff B = I_m. \]

Proof

Part 1

\begin{align} & (AI_n)[i, j] \\ &= \sum_{k=1}^n A[i, k] I_n[k, j] \\ &= \left(\sum_{k=1}^{j-1} A[i, k] I_n[k, j]\right) + (A[i, j] I_n[j, j]) + \left(\sum_{k=j+1}^n A[i, k] I_n[k, j] \right) \\ &= \left(\sum_{k=1}^{j-1} A[i, k] 0\right) + (A[i, j] 1) + \left(\sum_{k=j+1}^n A[i, k] 0 \right) \\ &= A[i, j] \end{align}

Therefore, $AI_n = A$.

Similarly, \[ (I_mA)[i, j] = \sum_{k=1}^m I_m[i, k] A[k, j] = A[i, j] \implies I_mA = A \]

Part 2

$AB = A \Rightarrow B \in \mathbb{M}_{n, n}(R)$.

\begin{align} & (AB)[i, j] = A[i, j] \\ &\Rightarrow \sum_{k=1}^n A[i, k]B[k, j] = A[i, j] \\ &\Rightarrow B[k, j] = \begin{cases} 1 & k = j \\ 0 & k \neq j \end{cases} \tag{comparing coefficients of $A[i, *]$} \end{align}

Therefore, $B = I_n$.

Similarly, when $BA = A$, $B = I_m$.

Dependency for: None

Info:

• Depth: 4
• Number of transitive dependencies: 8

Transitive dependencies:

1. Group
2. Ring
3. Polynomial
4. Integral Domain
5. Comparing coefficients of a polynomial with disjoint variables
6. Semiring
7. Matrix
8. Identity matrix