0.1 Fundamental Properties

First, when are two matrices “equal” to each other? Consider the following definition for a formal explanation.

Theorem 0.1.1

Two matrices \(A\) and \(B\) with orders \(m \times n\) and \(p \times q\) respectively are said to be equal to each other if they retain the same order and corresponding elements, i.e. they satisfy the following properties.

1.
\(m = p\)
2.
\(n = q\)
3.
For elements \(a_{ij}\) in \(A\) and \(b_{ij}\) in \(B\), the following equation satisfies. \[ a_{ij} = b_{ij} \, \forall \, i \in [1, m] \text { and } j \in [1, n] \]

Although this is very trivial and intuitive, it is always nice to formally define concepts!

Theorem 0.1.2

For matrices \(A\) and \(B\) with same order, the element in the \((i, j)\)-entry of their sum \(A + B\) is equivalent to the sum of elements in the \((i, j)\)-entry of \(A\) and \(B\).

Note that the orders of matrices being added must be identical. We do not consider additions with matrices of different orders. Below is an example of addition. \[ \begin {bmatrix} 1 & 3 & 5 \\ 2 & \pi & 4 \end {bmatrix} + \begin {bmatrix} -2 & 1 & 3 \\ 3 & \pi & -4 \end {bmatrix} = \begin {bmatrix} -1 & 4 & 8 \\ 5 & 2\pi & 0 \end {bmatrix} \] Similarly, we could see that subtractions are defined in similar way. Subtracting the matrices above, we get the following results. \[ \begin {bmatrix} 1 & 3 & 5 \\ 2 & \pi & 4 \end {bmatrix} - \begin {bmatrix} -2 & 1 & 3 \\ 3 & \pi & -4 \end {bmatrix} = \begin {bmatrix} 3 & 2 & 2 \\ -1 & 0 & 8 \end {bmatrix} \] Now, as we look at the additions and subtractions of matrices, you probably wondered about the application of additive properties. Intuitively, the fundamental properties of addition can be applied to matrices addition.

Theorem 0.1.3

For matrices \(A\) and \(B\) with order \(m \times n\), \(A + B = B + A\).

Proof.

Let \(A = [a_{ij}]_{m \times n}\) and \(B = [b_{ij}]_{m \times n}\). By definition, the sum \(A + B = [a_{ij} + b_{ij}]\). Because addition is commutative, the following equation holds. \[ A + B = [a_{ij} + b_{ij}] = [b_{ij} + a_{ij}] = B + A \] Thus, the commutative property of matrix addition holds.

In addition to the commutative property, the associative property is also applicable.

Theorem 0.1.4

For matrices \(A\), \(B\), and \(C\) with order \(m \times n\), \(A + (B + C) = (A + B) + C\).

Proof.

Let \(A = [a_{ij}]_{m \times n}\), \(B = [b_{ij}]_{m \times n}\), and \(C = [c_{ij}]_{m \times n}\). Consider the following equation. \begin{align*} A + (B + C) &= [a_{ij}] + [b_{ij} + c_{ij}] = [a_{ij} + (b_{ij} + c_{ij})] = [a_{ij} + b_{ij} + c_{ij}] \\ &= [(a_{ij} + b_{ij}) + c_{ij}] = [a_{ij} + b_{ij}] + [c_{ij}] \\ &= (A + B) + C \end{align*}

Thus, the associative property of addition applies to the matrix addition.

One obvious property in addition is that when you add \(0\) to a number, you get the number itself. The same applies in matrix addition.

Theorem 0.1.5

The sum of matrices \(A_{m \times n}\) and \(0_{m \times n}\) yields the matrix \(A\), i.e. the zero matrix is the identity elements of matrix addition.

Proof.

Let \(A = [a_{ij}]_{m \times n}\). By definition, the sum \(A + 0 = [a_{ij} + 0] = [a_{ij}] = A\). Therefore, the zero matrix is the identity element of matrix addition.

Now what if we want to add the same matrix multiple times? Fortunately, we already resolved the issue. For instance, we write \(x + x + x\) as \(3x\). We can multiply matrices with scalars.

Theorem 0.1.6

For a complex number \(\lambda \) and \(A = [a_{ij}]\), \(\lambda A = [\lambda a_{ij}]\).

Let’s take a look at an example.

Exercise 0.1.7

Compute \(2A - 3B\) for \(A\) and \(B\) defined as the following. \[ A = \begin {bmatrix} 2 & 3 & -1 \\ 1 & 7 & 3 \end {bmatrix}, \quad B = \begin {bmatrix} -2 & 1 & 0 \\ 5 & 3 & 8 \end {bmatrix} \]

Solution.

Before subtracting, we could perform scalar multiplication first. \begin{align*} 2A - 3B &= 2\begin {bmatrix} 2 & 3 & -1 \\ 1 & 7 & 3 \end {bmatrix} - 3\begin {bmatrix} -2 & 1 & 0 \\ 5 & 3 & 8 \end {bmatrix} \\ &= \begin {bmatrix} 4 & 6 & -2 \\ 2 & 14 & 6 \end {bmatrix} - \begin {bmatrix} -6 & 3 & 0 \\ 15 & 9 & 24 \end {bmatrix} \end{align*}

From here we could subtract two matrices. Therefore, \(2A - 3B\) can be represented as following. \[ 2A - 3B = \begin {bmatrix} 10 & 3 & -2 \\ -13 & 5 & -18 \end {bmatrix} \]

With the scalar multiplication in mind, we can expand the application of foundational concepts here with matrices.

Theorem 0.1.8

For \(A = [a_{ij}]_{m \times n}\), \(B = [b_{ij}]_{m \times n}\), and \(\lambda _1, \lambda _2 \in \mathbb {C}\), the following properties hold.

1.
\(\lambda _1 (A \pm B) = \lambda _1 A \pm \lambda _1 B\)
2.
\((\lambda _1 \pm \lambda _2) A = \lambda _1 A \pm \lambda _2 A\)
3.
\(\lambda _1 (\lambda _2 A) = (\lambda _1 \lambda _2) A\)

Let’s prove them! Below is the proof for the first property.

Proof.

Consider the following equation. \[ \lambda _1 (A \pm B) = \lambda _1 ([a_{ij}] \pm [b_{ij}]) = \lambda _1 ([a_{ij} \pm b_{ij}]) \] Note that individual elements are multiplied in scalar multiplication. Moreover, because \(\lambda _1 (a_{ij} \pm b_{ij}) = \lambda _1 a_{ij} \pm \lambda _1 b_{ij}\), the following equation holds. \[ \lambda _1 ([a_{ij} \pm b_{ij}]) = [\lambda _1 a_{ij} \pm \lambda _1 b_{ij}] = [\lambda _1 a_{ij}] \pm [\lambda _1 b_{ij}] = \lambda _1 A \pm \lambda _1 B \] Therefore, \(\lambda _1 (A \pm B) = \lambda _1 A \pm \lambda _1 B\).

Now let’s prove the second property.

Proof.

By definition of scalar multiplication, \((\lambda _1 \pm \lambda _2) A\) can be represented as the following. \[ (\lambda _1 \pm \lambda _2) A = (\lambda _1 \pm \lambda _2) [a_{ij}] = [(\lambda _1 \pm \lambda _2) a_{ij}] \] Using the distributive property, the following equation holds. \[ [(\lambda _1 \pm \lambda _2) a_{ij}] = [\lambda _1 a_{ij} \pm \lambda _2 a_{ij}] = [\lambda _1 a_{ij}] \pm [\lambda _2 a_{ij}] = \lambda _1 A \pm \lambda _2 A \] Thus, the equation \((\lambda _1 \pm \lambda _2) A = \lambda _1 A \pm \lambda _2 A\) holds.

Finally, here is the proof to the third property.

Proof.

Consider the following equation. \begin{align*} \lambda _1 (\lambda _2 [a_{ij}]) &= \lambda _1 [\lambda _2 a_{ij}] = [\lambda _1 \lambda _2 a_{ij}] = [(\lambda _1 \lambda _2) a_{ij}] \\ &= (\lambda _1 \lambda _2) [a_{ij}] = (\lambda _1 \lambda _2) A \end{align*}

Therefore, the property holds.

We have discussed fundamental properties so far. In the next section of the note, let’s discuss what matrix multiplication with intuitions and computations.