Algebraic Laws Apply Entry by Entry
When two matrices have the same size, they can be added by adding corresponding entries. A real scalar multiplies a matrix by multiplying each entry by that scalar. These definitions make matrix calculations manageable, but they also raise a useful question: which familiar rules of arithmetic remain valid for matrices? For addition and scalar multiplication, the answer is that the expected laws hold, provided the matrix dimensions and scalar field are respected.
Throughout, \(m,n\in\mathbb{N}\), matrices belong to \(\mathbb{R}^{m\times n}\), and scalars belong to \(\mathbb{R}\). If \(\mathbf{A}=[a_{ij}]\) and \(\mathbf{B}=[b_{ij}]\) have the same size, their sum is the matrix with entry \(a_{ij}+b_{ij}\) in position \((i,j)\). For \(a\in\mathbb{R}\), the scalar multiple \(a\mathbf{A}\) has entry \(aa_{ij}\) in that position. The zero matrix \(\mathbf{O}_{m\times n}\) has every entry equal to \(0\).
The zero matrix is an additive identity: adding \(0\) to each entry leaves that entry unchanged. For any \(\mathbf{A}=[a_{ij}]\), define \(-\mathbf{A}\) to be the matrix with entries \(-a_{ij}\). Then \(\mathbf{A}+(-\mathbf{A})\) has entries \(a_{ij}+(-a_{ij})=0\), so \(-\mathbf{A}\) is an additive inverse. These objects are not merely convenient notation: the laws below show that the identity and inverse are unique.
The Addition Laws
Proof. Write \(\mathbf{A}=[a_{ij}]\), \(\mathbf{B}=[b_{ij}]\), and \(\mathbf{C}=[c_{ij}]\). At each position \((i,j)\), the entry of \(\mathbf{A}+\mathbf{B}\) is \(a_{ij}+b_{ij}\), which equals \(b_{ij}+a_{ij}\) by commutativity of real addition. Thus \(\mathbf{A}+\mathbf{B}=\mathbf{B}+\mathbf{A}\). The \((i,j)\) entries of the two associative expressions are \((a_{ij}+b_{ij})+c_{ij}\) and \(a_{ij}+(b_{ij}+c_{ij})\); these are equal by associativity of real addition. Therefore the matrices are equal.
The \((i,j)\) entry of \(\mathbf{A}+\mathbf{O}_{m\times n}\) is \(a_{ij}+0=a_{ij}\), so \(\mathbf{O}_{m\times n}\) is an additive identity. If \(\mathbf{Z}\) is any additive identity, then, because \(\mathbf{Z}\) is an identity, \(\mathbf{Z}+\mathbf{O}_{m\times n}=\mathbf{O}_{m\times n}\). But because \(\mathbf{O}_{m\times n}\) is also an identity, \(\mathbf{Z}+\mathbf{O}_{m\times n}=\mathbf{Z}\). Hence \(\mathbf{Z}=\mathbf{O}_{m\times n}\), proving uniqueness.
For each \(\mathbf{A}=[a_{ij}]\), the entries of \(\mathbf{A}+(-\mathbf{A})\) are \(a_{ij}+(-a_{ij})=0\), so \(-\mathbf{A}\) is an inverse. If \(\mathbf{B}\) and \(\mathbf{C}\) are both inverses of \(\mathbf{A}\), then, using the identity, associativity, and the inverse equations, we obtain \(\mathbf{B}=\mathbf{B}+\mathbf{O}_{m\times n}=\mathbf{B}+(\mathbf{A}+\mathbf{C})=(\mathbf{B}+\mathbf{A})+\mathbf{C}=\mathbf{O}_{m\times n}+\mathbf{C}=\mathbf{C}\). Thus the inverse is unique. \(\square\)
Equality of matrices means equality at every corresponding entry. That is why these proofs reduce to familiar laws of real numbers. The matrix dimensions matter: addition is defined only when the two matrices have the same number of rows and the same number of columns. Associativity concerns three matrices of one common size, so both groupings are defined.
Worked Example: Check Commutativity and Associativity
Let \(\mathbf{A}=\begin{bmatrix}1&-2\\0&3\end{bmatrix}\), \(\mathbf{B}=\begin{bmatrix}4&1\\-1&2\end{bmatrix}\), and \(\mathbf{C}=\begin{bmatrix}-2&0\\5&-3\end{bmatrix}\). Compute both sides of the addition laws.
First, add \(\mathbf{A}\) and \(\mathbf{B}\) entry by entry, and then add \(\mathbf{C}\):
For the other grouping, first compute \(\mathbf{B}+\mathbf{C}\):
The two groupings agree. For commutativity, the two orders give \(\mathbf{A}+\mathbf{B}=\begin{bmatrix}5&-1\\-1&5\end{bmatrix}\) and \(\mathbf{B}+\mathbf{A}=\begin{bmatrix}4+1&1+(-2)\\-1+0&2+3\end{bmatrix}=\begin{bmatrix}5&-1\\-1&5\end{bmatrix}\). Both computations check the stated laws.
Scalar Multiplication Laws
Scalar multiplication has two distributive laws: one distributes a scalar across a matrix sum, and the other distributes a sum of scalars across a matrix. It also has a compatibility law describing successive scalar multiplications. In each law, the real-number operations on individual entries supply the justification.
Proof. Let \(\mathbf{A}=[a_{ij}]\) and \(\mathbf{B}=[b_{ij}]\). At position \((i,j)\), the entries of \(a(\mathbf{A}+\mathbf{B})\) and \(a\mathbf{A}+a\mathbf{B}\) are respectively \(a(a_{ij}+b_{ij})\) and \(aa_{ij}+ab_{ij}\). They are equal by distributivity in \(\mathbb{R}\). The entries of \((a+b)\mathbf{A}\) and \(a\mathbf{A}+b\mathbf{A}\) are \((a+b)a_{ij}\) and \(aa_{ij}+ba_{ij}\), equal by the same real-number law.
The entries of \(a(b\mathbf{A})\) and \((ab)\mathbf{A}\) are \(a(ba_{ij})\) and \((ab)a_{ij}\), equal by associativity and commutativity of real multiplication. The entry of \(1\mathbf{A}\) at \((i,j)\) is \(1a_{ij}=a_{ij}\). The entry of \(0\mathbf{A}\) is \(0a_{ij}=0\), and that of \(a\mathbf{O}_{m\times n}\) is \(a\cdot0=0\); both matrices therefore equal \(\mathbf{O}_{m\times n}\). Finally, the entries of \((-1)\mathbf{A}\) are \((-1)a_{ij}=-a_{ij}\), which are the entries of \(-\mathbf{A}\). Each stated matrix equality follows because its entries agree at every position. \(\square\)
Worked Example: Verify Both Distributive Laws
Take \(\mathbf{A}=\begin{bmatrix}1&-2\\0&3\end{bmatrix}\), \(\mathbf{B}=\begin{bmatrix}2&1\\-1&4\end{bmatrix}\), \(a=2\), and \(b=-3\). We will check both distributive laws and compatibility.
First, \(\mathbf{A}+\mathbf{B}=\begin{bmatrix}1+2&-2+1\\0+(-1)&3+4\end{bmatrix}=\begin{bmatrix}3&-1\\-1&7\end{bmatrix}\). Therefore
On the other side, \(2\mathbf{A}=\begin{bmatrix}2&-4\\0&6\end{bmatrix}\) and \(2\mathbf{B}=\begin{bmatrix}4&2\\-2&8\end{bmatrix}\), so
Thus \(2(\mathbf{A}+\mathbf{B})=2\mathbf{A}+2\mathbf{B}\). Next, \(a+b=2+(-3)=-1\), and \((-1)\mathbf{A}=\begin{bmatrix}-1&2\\0&-3\end{bmatrix}\). Meanwhile, \((-3)\mathbf{A}=\begin{bmatrix}-3&6\\0&-9\end{bmatrix}\), so
For compatibility, multiplying \(\mathbf{A}\) first by \(-3\) and then by \(2\) gives \(2((-3)\mathbf{A})=\begin{bmatrix}-6&12\\0&-18\end{bmatrix}\). Since \(2(-3)=-6\), multiplying \(\mathbf{A}\) directly by the product gives \((-6)\mathbf{A}=\begin{bmatrix}-6&12\\0&-18\end{bmatrix}\). The two calculations agree, as required.
Inverses, Subtraction, and a Common Sign Error
Because every matrix has a unique additive inverse, subtraction can be defined without introducing a new entry operation: \(\mathbf{A}-\mathbf{B}\) means \(\mathbf{A}+(-\mathbf{B})\). The law \((-1)\mathbf{B}=-\mathbf{B}\) explains why multiplying by \(-1\) changes every entry’s sign. In particular, parentheses matter when negating a sum: the negative applies to every entry of the sum.
Worked Example: Find and Check a Matrix Inverse
Let \(\mathbf{A}=\begin{bmatrix}2&-1&0\\3&4&-2\end{bmatrix}\). Its inverse under addition is \(-\mathbf{A}\):
Adding corresponding entries verifies the claim:
Reversing the order gives the same result: \((-\mathbf{A})+\mathbf{A}=\begin{bmatrix}-2+2&1+(-1)&0+0\\-3+3&-4+4&2+(-2)\end{bmatrix}=\mathbf{O}_{2\times3}\). This checks the inverse in both orders and illustrates that its size must match the original matrix.
Common mistake: negating only part of a matrix sum. A tempting but incorrect calculation is \(-(\mathbf{A}+\mathbf{B})=-\mathbf{A}+\mathbf{B}\). For example, take \(\mathbf{A}=\begin{bmatrix}1&0\\0&0\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}0&2\\0&0\end{bmatrix}\). Then \(\mathbf{A}+\mathbf{B}=\begin{bmatrix}1&2\\0&0\end{bmatrix}\), so the left side is \(\begin{bmatrix}-1&-2\\0&0\end{bmatrix}\). The proposed right side is \(\begin{bmatrix}-1&0\\0&0\end{bmatrix}+\begin{bmatrix}0&2\\0&0\end{bmatrix}=\begin{bmatrix}-1&2\\0&0\end{bmatrix}\), which is different. The correction follows from scalar distributivity: \((-1)(\mathbf{A}+\mathbf{B})=(-1)\mathbf{A}+(-1)\mathbf{B}=-\mathbf{A}-\mathbf{B}\).
What the Laws Do—and Do Not—Permit
These laws allow regrouping and reordering sums of same-sized matrices, inserting or removing the zero matrix, and distributing real scalars across sums. They do not make matrices interchangeable with numbers in every operation. For example, matrix addition is defined only for equal sizes, and a scalar must be real in the setting used here. Multiplication of matrices is a different operation with its own dimension conditions and laws; the results in this tutorial concern addition and scalar multiplication only.
When checking a claimed identity, a reliable method is to compare the \((i,j)\) entries on its two sides. If the sizes match, write down those entries and reduce the claim to a real-number identity. This method both explains why the laws hold generally and helps locate errors in a numerical calculation. A single entry where the two sides disagree is enough to disprove an incorrectly stated law.
Check Your Understanding
Use entrywise reasoning for the proof questions, and exact calculations for the checks.
- For \(\mathbf{A},\mathbf{B}\in\mathbb{R}^{m\times n}\), explain why \(\mathbf{A}+\mathbf{B}\) is defined, and why the same argument does not establish that \(\mathbf{A}+\mathbf{C}\) is defined for a matrix \(\mathbf{C}\) of a different size.
- Prove that \(a(\mathbf{A}+\mathbf{B})=a\mathbf{A}+a\mathbf{B}\) by comparing the \((i,j)\) entries.
- Find the additive inverse of \(\begin{bmatrix}-2&3\\1&0\end{bmatrix}\), then add the two matrices to verify your answer.
- For \(\mathbf{A}=\begin{bmatrix}1&-1\end{bmatrix}\), check whether \((-2)(3\mathbf{A})\) equals \((-6)\mathbf{A}\).
- Identify and correct the error in the claim \((a+b)\mathbf{A}=a\mathbf{A}+b\), where \(a,b\in\mathbb{R}\) and \(\mathbf{A}\) is a matrix.
