The Transpose Map on Matrix Spaces

The Transpose as a Linear Map

The transpose changes the positions of a matrix’s entries: the entry in row \(i\), column \(j\) moves to row \(j\), column \(i\). This simple rearrangement is also a linear map between two matrix spaces. To represent it by a matrix, we must choose ordered bases for both spaces and record where each domain basis vector is sent. The order matters: the same map can have different coordinate matrices if either basis is ordered differently.

Throughout, scalars are real. Write \(M_{m\times n}(\mathbb{R})\) for the vector space of \(m\times n\) real matrices. For \(1\le i\le m\) and \(1\le j\le n\), let \(E_{ij}^{(m,n)}\) be the \(m\times n\) matrix whose entry in row \(i\), column \(j\) is \(1\), and whose other entries are \(0\). We use the row-by-row ordering \(E_{11}^{(m,n)},E_{12}^{(m,n)},\dots,E_{1n}^{(m,n)},E_{21}^{(m,n)},\dots,E_{mn}^{(m,n)}\): the row index is fixed first while the column index runs through its values.

These are bases because every matrix \(\mathbf{X}=(x_{ij})\) in \(M_{m\times n}(\mathbb{R})\) has the unique expansion \(\mathbf{X}=\sum_{i=1}^{m}\sum_{j=1}^{n}x_{ij}E_{ij}^{(m,n)}\). The coefficient of each matrix unit must be the entry in its position. The codomain has the corresponding expansion using its own matrix units. Each space has \(mn\) basis matrices, so the representation of \(T_{m,n}\) in these bases is an \(mn\times mn\) matrix.

The transpose map is linear: for \(\mathbf{X},\mathbf{Y}\in M_{m\times n}(\mathbb{R})\) and \(a,b\in\mathbb{R}\), its entry in position \((j,i)\) is \(a x_{ij}+b y_{ij}\), which is the same entry as in \(a\mathbf{X}^{\top}+b\mathbf{Y}^{\top}\). Thus \(T_{m,n}(a\mathbf{X}+b\mathbf{Y})=aT_{m,n}(\mathbf{X})+bT_{m,n}(\mathbf{Y})\).

Tracking the Matrix Units

For a linear map, the \(r\)-th column of its representation matrix is the coordinate vector of the image of the \(r\)-th domain basis vector. Here transposition sends the matrix unit in position \((i,j)\) to the matrix unit in position \((j,i)\), with the dimensions of the matrix space switched:

The domain unit \(E_{ij}^{(m,n)}\) is in position \((i-1)n+j\) in the domain list. Its image is in position \((j-1)m+i\) in the codomain list. Therefore the column indexed by \((i-1)n+j\) has a \(1\) in row \((j-1)m+i\), and zeros elsewhere.

Proof. Fix a domain basis vector \(E_{ij}^{(m,n)}\). By the definition of the representation matrix, its coordinate vector after applying the map is column \((i-1)n+j\). The image is \(E_{ji}^{(n,m)}\), whose coordinate vector in the codomain basis has a \(1\) in position \((j-1)m+i\) and zeros elsewhere. This proves the stated entry formula for that column. Every domain index occurs once because the domain basis lists every pair \((i,j)\) exactly once. Every codomain index occurs once because \((i,j)\mapsto(j,i)\) pairs the positions of the domain and codomain matrix units one-to-one. Hence each column and each row contains exactly one \(1\). \(\square\)

For a matrix \(\mathbf{X}=(x_{ij})\), its domain coordinate vector in this ordering is the list of its entries row by row. Multiplying that vector by the representation matrix rearranges the entries into the row-by-row coordinate vector of \(\mathbf{X}^{\top}\). This is a permutation of coordinates, not a change to the entries themselves.

A Rectangular Example

Square and Thin Matrix Spaces

When \(m=n\), the transpose map has the same matrix space as its domain and codomain. Its coordinate matrix need not be the identity: the off-diagonal entries exchange positions, while diagonal entries stay where they are. When one dimension is \(1\), row-by-row ordering can make the coordinate rearrangement particularly simple. These cases are useful checks on the general indexing formula.

Transposing Twice

There is also a direct way to verify that the map does not lose information. Applying transposition twice restores every entry to its original position. This fact is independent of a chosen coordinate basis and explains why transposition gives a one-to-one correspondence between the two matrix spaces.

Proof. Let \(\mathbf{X}=(x_{ij})\). The entry of \(\mathbf{X}^{\top}\) in row \(j\), column \(i\) is \(x_{ij}\), by the definition of transpose. Transposing again places that entry in row \(i\), column \(j\). Therefore every entry of \(T_{n,m}(T_{m,n}(\mathbf{X}))\) equals the corresponding entry of \(\mathbf{X}\). If \(T_{m,n}(\mathbf{X})=T_{m,n}(\mathbf{Y})\), applying \(T_{n,m}\) to both sides gives \(\mathbf{X}=\mathbf{Y}\), proving that \(T_{m,n}\) is one-to-one. For any \(\mathbf{Z}\in M_{n\times m}(\mathbb{R})\), take \(\mathbf{X}=\mathbf{Z}^{\top}\in M_{m\times n}(\mathbb{R})\). Then \(T_{m,n}(\mathbf{X})=\mathbf{Z}^{\top\top}=\mathbf{Z}\), proving that \(T_{m,n}\) is onto. \(\square\)

Common Pitfall: Mixing the Basis Orders

The formula for the representation matrix depends on the basis order in both spaces. For the \(2\times3\) example, \(E_{12}^{(2,3)}\) maps to \(E_{21}^{(3,2)}\). In the stated row-by-row codomain order, \(E_{21}^{(3,2)}\) is basis vector \(3\), not basis vector \(2\). A common wrong calculation puts a \(1\) in row \(2\) of column \(2\), treating the codomain list as though it began \(E_{11},E_{21},\ldots\). That would send the input coordinate vector \((0,1,0,0,0,0)^{\top}\) to \((0,1,0,0,0,0)^{\top}\), which represents \(E_{12}^{(3,2)}\), not \(E_{21}^{(3,2)}\).

The correction is to identify the image as a matrix first, then locate that matrix unit in the specified codomain basis. Here the correct output is \(E_{21}^{(3,2)}\), whose row-by-row coordinate vector is \((0,0,1,0,0,0)^{\top}\). If a different codomain ordering is deliberately chosen, the representation matrix changes accordingly; it is not valid to use that different ordering without changing the matrix.

The key procedure is therefore consistent: write the ordered domain basis, transpose each basis matrix, and express each image in the ordered codomain basis. The resulting coordinate vectors form the columns of the representation matrix. The method applies to rectangular matrix spaces as well as square ones, provided the dimensions and basis orders are kept explicit.