Zero Matrix is Identity for Matrix Entrywise Addition

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Theorem

Let $\map \MM {m, n}$ be a $m \times n$ matrix space over one of the standard number systems.

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


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


Proof 1

From:

Integers form Ring
Rational Numbers form Ring
Real Numbers form Ring
Complex Numbers form Ring

the standard number systems $\Z$, $\Q$, $\R$ and $\C$ are rings whose zero is the number $0$ (zero).

Hence we can apply Zero Matrix is Identity for Matrix Entrywise Addition over Ring.

$\Box$


The above cannot be applied to the natural numbers $\N$, as they do not form a ring.

However, from Natural Numbers under Addition form Commutative Monoid, the algebraic structure $\struct {\N, +}$ is a commutative monoid whose identity is $0$ (zero).

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

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

$\blacksquare$


Proof 2

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

Then:

\(\ds \mathbf A + \mathbf 0\) \(=\) \(\ds \sqbrk a_{m n} + \sqbrk 0_{m n}\) Definition of $\mathbf A$ and $\mathbf 0_R$
\(\ds \) \(=\) \(\ds \sqbrk {a + 0}_{m n}\) Definition of Matrix Entrywise Addition
\(\ds \) \(=\) \(\ds \sqbrk a_{m n}\) Identity Element of Addition on Numbers
\(\ds \leadsto \ \ \) \(\ds \mathbf A + \mathbf 0\) \(=\) \(\ds \mathbf A\) Definition of Zero Matrix


Similarly:

\(\ds \mathbf 0 + \mathbf A\) \(=\) \(\ds \sqbrk 0_{m n} + \sqbrk a_{m n}\) Definition of $\mathbf A$ and $\mathbf 0_R$
\(\ds \) \(=\) \(\ds \sqbrk {0 + a}_{m n}\) Definition of Matrix Entrywise Addition
\(\ds \) \(=\) \(\ds \sqbrk a_{m n}\) Identity Element of Addition on Numbers
\(\ds \leadsto \ \ \) \(\ds \mathbf 0 + \mathbf A\) \(=\) \(\ds \mathbf A\) Definition of Zero Matrix

$\blacksquare$


Also see


Sources