Matrix Spaces Under Entrywise Operations

Matrices as Vectors

A matrix is often introduced as a rectangular array used to store numbers or represent a linear system. It can also be treated as a single vector-like object. To do that, we fix its number of rows and columns, then define addition and scalar multiplication by applying the corresponding real-number operation to each entry. With these operations, matrices of that fixed size satisfy the vector space axioms.

Throughout, let \(m,n\in\mathbb{N}\). The set of all \(m\times n\) real matrices is written \(\mathbb{R}^{m\times n}\). If \(\mathbf{A}\in\mathbb{R}^{m\times n}\), write its entry in row \(i\), column \(j\) as \(a_{ij}\), where \(1\le i\le m\) and \(1\le j\le n\). The dimensions are fixed: matrices of different sizes are not elements of the same \(\mathbb{R}^{m\times n}\).

These operations preserve the fixed size: each position \((i,j)\) in the output comes from that same position in the input matrices. Addition therefore requires matching dimensions, while multiplication by a real scalar does not change the dimensions. The zero matrix also has a specified size; for instance, \(\mathbf{O}_{2\times3}\) and \(\mathbf{O}_{3\times2}\) are different matrices.

Why the Vector Space Axioms Hold

A real vector space is a set with an addition and a real scalar multiplication satisfying the usual rules: addition is closed, associative, and commutative; there is an additive identity and every element has an additive inverse; scalar multiplication distributes over vector addition and scalar addition, is associative with real multiplication, and has identity scalar \(1\). For matrices, each rule can be checked at one entry at a time. The reason this works is that the real numbers already satisfy the corresponding arithmetic rules.

Proof. Let \(\mathbf{A},\mathbf{B},\mathbf{C}\in\mathbb{R}^{m\times n}\), with entries \(a_{ij},b_{ij},c_{ij}\), and let \(r,s\in\mathbb{R}\). The entries of \(\mathbf{A}+\mathbf{B}\) are real numbers, so \(\mathbf{A}+\mathbf{B}\in\mathbb{R}^{m\times n}\); likewise, every entry \(ra_{ij}\) is real, so \(r\mathbf{A}\in\mathbb{R}^{m\times n}\). At each position \((i,j)\), the real-number laws give \((\mathbf{A}+\mathbf{B})_{ij}=a_{ij}+b_{ij}=b_{ij}+a_{ij}=(\mathbf{B}+\mathbf{A})_{ij}\), and \(((\mathbf{A}+\mathbf{B})+\mathbf{C})_{ij}=(a_{ij}+b_{ij})+c_{ij}=a_{ij}+(b_{ij}+c_{ij})=(\mathbf{A}+(\mathbf{B}+\mathbf{C}))_{ij}\). The matrix \(\mathbf{O}_{m\times n}\) is an additive identity because \((\mathbf{A}+\mathbf{O}_{m\times n})_{ij}=a_{ij}+0=a_{ij}\), and the matrix with entries \(-a_{ij}\) is an additive inverse because \(a_{ij}+(-a_{ij})=0\). For scalar rules, entrywise calculation gives \((r(\mathbf{A}+\mathbf{B}))_{ij}=r(a_{ij}+b_{ij})=ra_{ij}+rb_{ij}=(r\mathbf{A}+r\mathbf{B})_{ij}\), \(((r+s)\mathbf{A})_{ij}=(r+s)a_{ij}=ra_{ij}+sa_{ij}=(r\mathbf{A}+s\mathbf{A})_{ij}\), \((r(s\mathbf{A}))_{ij}=r(sa_{ij})=(rs)a_{ij}=((rs)\mathbf{A})_{ij}\), and \((1\mathbf{A})_{ij}=1a_{ij}=a_{ij}\). Since each required equality holds at every entry, the matrices on its two sides are equal. Thus all the vector space axioms hold. \(\square\)

The proof uses no special property of a particular number of rows or columns. Fixing \(m\) and \(n\) makes every operation well-defined on one set; the entrywise real-number laws then establish the axioms for that set.

Calculating Entrywise Operations

A linear combination of matrices is formed in the same way as a linear combination of vectors: multiply each matrix by its scalar, then add the results. All matrices in the expression must have the same size so that each addition is defined.

Proof. Scalar multiplication and addition each preserve the number of rows and columns, so \(r\mathbf{A}+s\mathbf{B}\) belongs to \(\mathbb{R}^{m\times n}\). By the definitions of the operations, its entry at \((i,j)\) is \((r\mathbf{A}+s\mathbf{B})_{ij}=(r\mathbf{A})_{ij}+(s\mathbf{B})_{ij}=ra_{ij}+sb_{ij}\). This holds for every allowed \(i,j\), proving the claim. \(\square\)

The Zero Matrix and Additive Inverses

For a matrix space, the additive identity is not a single number but a matrix of the same size as its members. Its entries must all be zero. The additive inverse of a matrix is obtained by negating each entry; adding the original and the inverse then gives zero in every position.

Proof. At each position, \((\mathbf{A}+\mathbf{O}_{m\times n})_{ij}=a_{ij}+0=a_{ij}\), so \(\mathbf{O}_{m\times n}\) is an additive identity. Also, \((\mathbf{A}+(-\mathbf{A}))_{ij}=a_{ij}+(-a_{ij})=0\), so \(-\mathbf{A}\) is an additive inverse. These are the identity and inverse for the entrywise addition because each entry must satisfy the corresponding real-number equation: if \(z_{ij}\) satisfies \(a_{ij}+z_{ij}=a_{ij}\), then \(z_{ij}=0\); if \(a_{ij}+y_{ij}=0\), then \(y_{ij}=-a_{ij}\). Finally, \((0\mathbf{A})_{ij}=0a_{ij}=0\) and \((r\mathbf{O}_{m\times n})_{ij}=r\cdot0=0\). Each equality holds at every position, proving all the stated identities. \(\square\)

Common Pitfalls

Do not add a scalar to a matrix as though it were a matrix. For \(\mathbf{C}=\begin{bmatrix}1&0\\-2&3\end{bmatrix}\), the calculation \(\mathbf{C}+2=\begin{bmatrix}3&2\\0&5\end{bmatrix}\) is not an instance of the defined matrix addition: \(2\) is a real number, not a \(2\times2\) matrix. If the intention is to add \(2\) to every entry, that is a different operation and has not been defined here. Scalar multiplication is defined, but it gives \(2\mathbf{C}=\begin{bmatrix}2&0\\-4&6\end{bmatrix}\), not the incorrect array above. If the intention is to add the zero matrix, then \(\mathbf{C}+\mathbf{O}_{2\times2}=\mathbf{C}\).

Check dimensions before adding. The expression \(\begin{bmatrix}1&2\end{bmatrix}+\begin{bmatrix}3\\4\end{bmatrix}\) is undefined under entrywise addition: the first matrix has size \(1\times2\), while the second has size \(2\times1\). Having the same number of entries is not enough; corresponding entries must occupy the same row and column positions in matrices of matching dimensions.

Finally, entrywise addition and scalar multiplication are not matrix multiplication. For example, \(2\mathbf{A}\) means multiply every entry of \(\mathbf{A}\) by \(2\), while a product of two matrices is a separate operation with different dimension conditions and entry formulas. The vector space structure here uses only the two operations defined at the start.