Coordinates in Matrix Spaces

Matrices as Vectors

A matrix is not only an arrangement of numbers on which we perform matrix operations. Once its size is fixed, it is also a vector in a vector space: matrices of that size can be added entry by entry and multiplied by real scalars. This lets us ask familiar coordinate questions. Given an ordered basis of matrices, what numbers represent a particular matrix, and how do those numbers reconstruct it?

Throughout this tutorial, scalars are real. For positive integers \(m\) and \(n\), the set \(\mathbb{R}^{m\times n}\) of all \(m\times n\) real matrices is a vector space with entrywise addition and scalar multiplication. Its zero vector is the zero matrix \(\mathbf{O}\). A basis of this space is an ordered list of matrices that is linearly independent and spans the whole space. The order matters: it determines the order of the entries in every coordinate vector.

The definition makes coordinates depend on the chosen basis, not just on the matrix itself. A coordinate vector is therefore meaningful only together with its ordered basis. To reconstruct the matrix, multiply each basis matrix by its corresponding coordinate and add the results.

Proof. Because \(\mathcal{B}\) spans the space, \(\mathbf{A}\) is a linear combination of its members, so at least one list of coefficients \(c_1,\dots,c_p\) reconstructs \(\mathbf{A}\). Suppose another list \(d_1,\dots,d_p\) also reconstructs it. Subtracting the two expressions gives \(\sum_{k=1}^{p}(c_k-d_k)\mathbf{B}_k=\mathbf{O}\). Since the basis is linearly independent, every coefficient \(c_k-d_k\) is zero. Thus \(c_k=d_k\) for all \(k\), proving uniqueness; the reconstruction equation is the expression whose coefficients define the coordinate vector. \(\square\)

The Standard Matrix-Unit Basis

For each row index \(i\) and column index \(j\), let \(\mathbf{E}_{ij}\) be the \(m\times n\) matrix with a \(1\) in row \(i\), column \(j\), and zeros elsewhere. Fix the row-by-row order \(\mathcal{E}=(\mathbf{E}_{11},\dots,\mathbf{E}_{1n},\mathbf{E}_{21},\dots,\mathbf{E}_{mn})\): move across each row from left to right, then continue down to the next row. For example, the order for \(2\times3\) matrices is \((\mathbf{E}_{11},\mathbf{E}_{12},\mathbf{E}_{13},\mathbf{E}_{21},\mathbf{E}_{22},\mathbf{E}_{23})\).

Proof. First, any \(\mathbf{A}=[a_{ij}]\) can be written as \(\sum_{i=1}^{m}\sum_{j=1}^{n}a_{ij}\mathbf{E}_{ij}\), because the summand \(a_{ij}\mathbf{E}_{ij}\) places \(a_{ij}\) in position \((i,j)\) and zero in every other position. Adding all these summands gives exactly \(\mathbf{A}\), so the matrix units span the space. Next, suppose \(\sum_{i=1}^{m}\sum_{j=1}^{n}c_{ij}\mathbf{E}_{ij}=\mathbf{O}\). The entry in position \((i,j)\) on the left is \(c_{ij}\), while that entry on the right is \(0\). Thus \(c_{ij}=0\) for every pair \((i,j)\), proving linear independence. Hence the list is a basis. In this basis, the coefficient of \(\mathbf{E}_{ij}\) is precisely \(a_{ij}\), so following the stated order gives the entries row by row. \(\square\)

Coordinates in a Different Ordered Basis

For a basis other than the matrix-unit basis, the coordinates are still the coefficients in a linear combination, but they no longer need to equal the entries of the matrix. A direct method is to write an unknown linear combination of the basis matrices and match its entries with the target matrix. The matching equations must be read using the basis matrices in their given order.

The same entry-matching equations show why the displayed list is indeed a basis, rather than merely a list that happens to represent this one matrix. For any \(\begin{bmatrix}a&b\\c&d\end{bmatrix}\), the equations are \(x+y=a\), \(z=b\), \(w=c\), \(x-y=d\), with the unique solution \(x=(a+d)/2\), \(y=(a-d)/2\), \(z=b\), \(w=c\). Thus the list spans every \(2\times2\) matrix. The unique solution for each target also shows that a linear combination equalling the zero matrix has only zero coefficients, so the list is linearly independent.

Entry Matching with a Reordered, Combined Basis

A basis matrix can have several nonzero entries, so one coordinate can affect several positions of the reconstructed matrix. This is not a complication in the definition; it is a reason to calculate the entire linear combination before reading off coefficients. The next example has basis matrices that overlap in the first row and in the second row.

Reconstruction and a Common Ordering Pitfall

Coordinates are useful only if they preserve the chosen order. In the standard matrix-unit basis, the convention here is row by row. In another basis, the first coordinate always multiplies the first listed matrix, the second coordinate multiplies the second listed matrix, and so on. If a basis is reordered, the coordinate vector must be reordered with it; the matrix represented by a list of numbers can change when the basis order changes.

Common mistake: silently switching to column-by-column order. For \(\mathbf{A}=\begin{bmatrix}1&2\\3&4\end{bmatrix}\), the row-by-row coordinate vector is \(\begin{bmatrix}1&2&3&4\end{bmatrix}^{\top}\), corresponding to \((\mathbf{E}_{11},\mathbf{E}_{12},\mathbf{E}_{21},\mathbf{E}_{22})\). A wrong calculation lists the entries column by column and reports \(\begin{bmatrix}1&3&2&4\end{bmatrix}^{\top}\) while keeping that same basis order. Reconstructing with those coefficients gives \(\begin{bmatrix}1&3\\2&4\end{bmatrix}\), not the original matrix. The correction is to match each coefficient to the stated basis order: use \(\begin{bmatrix}1&2&3&4\end{bmatrix}^{\top}\). Column-by-column listing can also be used, but only if the basis is explicitly ordered column by column as well.

For an unfamiliar ordered basis, a reliable check is to take the proposed coordinate vector, multiply its entries by the basis matrices in order, and add entry by entry. If the result is not the target matrix, either a coefficient or the basis order has been mishandled. For the standard basis, this check reduces to placing each coordinate back in its designated matrix position.