Zero Matrix is Identity for Matrix Entrywise Addition over Ring

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {R, +, \circ}$ be a ring.

Let $\map {\MM_R} {m, n}$ be a $m \times n$ matrix space over $R$.

Let $\mathbf 0_R = \sqbrk {0_R}_{m n}$ be the zero matrix of $\map {\MM_R} {m, n}$.


Then $\mathbf 0_R$ is the identity element for matrix entrywise addition.


Proof 1

Let $\mathbf A = \sqbrk a_{m n} \in \map {\MM_R} {m, n}$.

Then:

\(\ds \mathbf A + \mathbf 0_R\) \(=\) \(\ds \sqbrk a_{m n} + \sqbrk {0_R}_{m n}\) Definition of $\mathbf A$ and $\mathbf 0_R$
\(\ds \) \(=\) \(\ds \sqbrk {a + 0_R}_{m n}\) Definition of Matrix Entrywise Addition
\(\ds \) \(=\) \(\ds \sqbrk a_{m n}\) Ring Axiom $\text A3$: Identity for Addition is $0_R$
\(\ds \leadsto \ \ \) \(\ds \mathbf A + \mathbf 0_R\) \(=\) \(\ds \mathbf A\) Definition of Zero Matrix over Ring


Similarly:

\(\ds \mathbf 0_R + \mathbf A\) \(=\) \(\ds \sqbrk {0_R}_{m n} + \sqbrk a_{m n}\) Definition of $\mathbf A$ and $\mathbf 0_R$
\(\ds \) \(=\) \(\ds \sqbrk {0_R + a}_{m n}\) Definition of Matrix Entrywise Addition
\(\ds \) \(=\) \(\ds \sqbrk a_{m n}\) Ring Axiom $\text A3$: Identity for Addition is $0_R$
\(\ds \leadsto \ \ \) \(\ds \mathbf 0_R + \mathbf A\) \(=\) \(\ds \mathbf A\) Definition of Zero Matrix over Ring

$\blacksquare$


Proof 2

By definition, matrix entrywise addition is the Hadamard product with respect to ring addition.

We have from Ring Axiom $\text A3$: Identity for Addition that the identity element of ring addition is the ring zero $0_R$.

The result then follows directly from Zero Matrix is Identity for Hadamard Product.

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Zero_Matrix_is_Identity_for_Matrix_Entrywise_Addition_over_Ring&oldid=540077"