Transpose and Matrix Arithmetic
Transposition changes the arrangement of a matrix without changing its entries: rows become columns, and columns become rows. When a matrix expression also involves addition or scalar multiplication, it is important to know how transposition interacts with those operations. The rules are simple, but keeping track of dimensions and entry positions makes them easier to use correctly.
Throughout this tutorial, matrices have real entries. If \(\mathbf{A}\) has \(m\) rows and \(n\) columns, its transpose \(\mathbf{A}^{\top}\) has \(n\) rows and \(m\) columns. The entry in row \(i\), column \(j\) of \(\mathbf{A}\) moves to row \(j\), column \(i\) of \(\mathbf{A}^{\top}\). In symbols, \((\mathbf{A}^{\top})_{ji}=a_{ij}\). Transposition does not change the value of an entry; it changes its position.
Recall that two matrices are equal when they have the same dimensions and the same entry in every corresponding position. This gives a direct way to prove identities involving transposes: determine the dimensions on both sides, then compare the entries.
The Transpose of a Sum
Two matrices can be added only when they have the same dimensions. If \(\mathbf{A}\) and \(\mathbf{B}\) are both \(m\times n\), then their sum is also \(m\times n\). Its entry in row \(i\), column \(j\) is \(a_{ij}+b_{ij}\). Transposing that sum gives an \(n\times m\) matrix. The individual transposes \(\mathbf{A}^{\top}\) and \(\mathbf{B}^{\top}\) are also both \(n\times m\), so they can be added, and their sum has the same dimensions.
Proof. Both sides are \(n\times m\) matrices. Fix any row index \(j\) with \(1\le j\le n\) and column index \(i\) with \(1\le i\le m\). By the definition of transpose and matrix addition, the entry in row \(j\), column \(i\) on the left is \((\mathbf{A}+\mathbf{B})_{ij}=a_{ij}+b_{ij}\). The entry in row \(j\), column \(i\) on the right is \((\mathbf{A}^{\top})_{ji}+(\mathbf{B}^{\top})_{ji}=a_{ij}+b_{ij}\). The entries agree for every allowed \(i,j\), and the dimensions agree, so the matrices are equal. \(\square\)
Worked Example: Transposing a Sum of Two Rectangular Matrices
Let \(\mathbf{A}=\begin{bmatrix}1&-2&0\\3&4&5\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}-1&6&2\\0&-3&1\end{bmatrix}\). Both matrices are \(2\times3\), so they can be added. Compute the transpose of their sum and compare it with the sum of their transposes.
Add corresponding entries, showing each row of the result:
The sum is \(2\times3\), so its transpose is \(3\times2\). Swapping rows and columns gives \((\mathbf{A}+\mathbf{B})^{\top}=\begin{bmatrix}0&3\\4&1\\2&6\end{bmatrix}\). Now transpose each matrix separately:
Adding corresponding entries gives \(\mathbf{A}^{\top}+\mathbf{B}^{\top} =\begin{bmatrix}1+(-1)&3+0\\-2+6&4+(-3)\\0+2&5+1\end{bmatrix} =\begin{bmatrix}0&3\\4&1\\2&6\end{bmatrix}\). Thus both calculations give the same \(3\times2\) matrix, as the theorem predicts.
The Transpose of a Scalar Multiple
For a real scalar \(c\) and an \(m\times n\) matrix \(\mathbf{A}\), the scalar multiple \(c\mathbf{A}\) is formed by multiplying each entry of \(\mathbf{A}\) by \(c\). The result is still \(m\times n\), and its transpose is \(n\times m\). Multiplying \(\mathbf{A}^{\top}\) by \(c\) also gives an \(n\times m\) matrix.
Proof. Both sides have dimensions \(n\times m\). For any \(1\le i\le m\) and \(1\le j\le n\), the entry in row \(j\), column \(i\) of the left side is \((c\mathbf{A})_{ij}=ca_{ij}\). The entry in the same position on the right side is \(c(\mathbf{A}^{\top})_{ji}=ca_{ij}\). Since the entries match in every position and the dimensions match, the two matrices are equal. \(\square\)
Worked Example: A Negative Scalar and a Rectangular Matrix
Take \(\mathbf{C}=\begin{bmatrix}2&-1\\0&3\\-4&5\end{bmatrix}\) and \(c=-2\). Compute both \((c\mathbf{C})^{\top}\) and \(c\mathbf{C}^{\top}\).
First multiply every entry by \(-2\):
Transpose this \(3\times2\) matrix to obtain a \(2\times3\) matrix: \((-2\mathbf{C})^{\top}=\begin{bmatrix}-4&0&8\\2&-6&-10\end{bmatrix}\). For the other order, first transpose \(\mathbf{C}\): \(\mathbf{C}^{\top}=\begin{bmatrix}2&0&-4\\-1&3&5\end{bmatrix}\). Then multiply every entry by \(-2\):
The two results agree. In particular, the negative scalar changes each entry's sign as well as its magnitude; the transpose then places each resulting entry in the corresponding swapped position.
Combining the Two Laws
The sum rule and scalar rule can be used together, including when the matrices are rectangular. Each matrix in a sum must have the same dimensions, and each side of the resulting transpose identity then has the dimensions with rows and columns exchanged. For instance, if \(\mathbf{A}\) and \(\mathbf{B}\) are both \(m\times n\) and \(c,d\) are real, then transposing \(c\mathbf{A}+d\mathbf{B}\) gives \(c\mathbf{A}^{\top}+d\mathbf{B}^{\top}\). The next example checks this calculation entry by entry.
Worked Example: Transposing a Linear Combination
Let \(\mathbf{A}=\begin{bmatrix}1&2&-1\\0&3&4\end{bmatrix}\), \(\mathbf{B}=\begin{bmatrix}2&-1&5\\1&0&-2\end{bmatrix}\), \(c=2\), and \(d=-1\). Compare \((c\mathbf{A}+d\mathbf{B})^{\top}\) with \(c\mathbf{A}^{\top}+d\mathbf{B}^{\top}\).
First calculate \(2\mathbf{A}\) and \(-\mathbf{B}\), then add their corresponding entries:
Transposing yields \((2\mathbf{A}-\mathbf{B})^{\top} =\begin{bmatrix}0&-1\\5&6\\-7&10\end{bmatrix}\). For the other order, \(\mathbf{A}^{\top}=\begin{bmatrix}1&0\\2&3\\-1&4\end{bmatrix}\) and \(\mathbf{B}^{\top}=\begin{bmatrix}2&1\\-1&0\\5&-2\end{bmatrix}\). Therefore
Both calculations give the same \(3\times2\) matrix. The check also confirms the signs in the subtraction: for example, the entry in row 3, column 2 is \(8-(-2)=10\), not \(8-2\).
Proof. For \(p=1\), the statement is exactly the scalar multiplication law. Suppose it holds for a sum of \(p\) matrices. For a sum of \(p+1\) matrices, group the first \(p\) terms as \(\mathbf{S}=c_1\mathbf{A}_1+\cdots+c_p\mathbf{A}_p\). The sum rule and scalar rule give \((\mathbf{S}+c_{p+1}\mathbf{A}_{p+1})^{\top} =\mathbf{S}^{\top}+c_{p+1}\mathbf{A}_{p+1}^{\top}\). By the assumption for \(p\) terms, \(\mathbf{S}^{\top}=c_1\mathbf{A}_1^{\top}+\cdots+c_p\mathbf{A}_p^{\top}\). Substitution gives the claimed formula for \(p+1\) terms. By induction, it holds for every integer \(p\ge1\). \(\square\)
Common Pitfall: Transposing Changes the Shape
A frequent error is to multiply a matrix by a scalar correctly and then forget to swap its rows and columns. Consider \(\mathbf{A}=\begin{bmatrix}1&0&-2\\3&4&1\end{bmatrix}\) and the scalar \(-2\). Multiplying first gives \(-2\mathbf{A}=\begin{bmatrix}-2&0&4\\-6&-8&-2\end{bmatrix}\). A wrong calculation might report this same \(2\times3\) matrix as \((-2\mathbf{A})^{\top}\). That cannot be right: the transpose of a \(2\times3\) matrix must be \(3\times2\).
The correct result is \((-2\mathbf{A})^{\top}=\begin{bmatrix}-2&-6\\0&-8\\4&-2\end{bmatrix}\). For a position check, the entry \(-2\) in row 1, column 3 of \(\mathbf{A}\) is multiplied by \(-2\), giving \(4\), and then moves to row 3, column 1. The entry \(4\) in row 2, column 2 is multiplied by \(-2\), giving \(-8\); because it is on the diagonal, it remains in row 2, column 2. On the other side, \(\mathbf{A}^{\top}=\begin{bmatrix}1&3\\0&4\\-2&1\end{bmatrix}\), and multiplying it by \(-2\) gives \(\begin{bmatrix}-2&-6\\0&-8\\4&-2\end{bmatrix}\), as required.
Two checks are useful in routine calculations: first, confirm the dimensions on both sides; second, follow at least a representative entry from its original position through the scalar operation and the transposition. These checks catch different kinds of errors. Matching dimensions alone does not guarantee that entries are correct, while a correctly computed entry in the wrong-sized matrix cannot establish the identity.
Why These Rules Are Useful
The transpose laws let us reorganize matrix expressions without recalculating every entry from the beginning. The sum rule applies to matrices of the same size, and the scalar rule applies to any real scalar and matrix. Together, they show that a finite linear combination can be transposed either before or after its terms are combined. The order of operations matters for reliable bookkeeping, but the final entries agree because each is formed from the same original entries with the same coefficients.
These identities concern addition and scalar multiplication only. They do not provide a rule for transposing a matrix product; that operation requires separate attention to dimensions and the order of the factors. For the laws established here, remember the basic procedure: verify that the sum is defined, record the dimensions after transposing, and compare corresponding entries.
Check Your Understanding
For each question, keep track of both the entries and the dimensions.
- If \(\mathbf{A},\mathbf{B}\in\mathbb{R}^{3\times2}\), what are the dimensions of \((\mathbf{A}+\mathbf{B})^{\top}\) and \(\mathbf{A}^{\top}+\mathbf{B}^{\top}\)?
- For \(\mathbf{C}=\begin{bmatrix}1&-3\\2&0\\-1&4\end{bmatrix}\), compute \((3\mathbf{C})^{\top}\) and \(3\mathbf{C}^{\top}\).
- State the entry in row 2, column 1 of \((\mathbf{A}+\mathbf{B})^{\top}\) in terms of the entries of \(\mathbf{A}\) and \(\mathbf{B}\), assuming both are \(2\times3\).
- Explain why the scalar multiplication law remains valid when the scalar is negative.
- Use the sum and scalar laws to rewrite \((-2\mathbf{A}+5\mathbf{B})^{\top}\) when \(\mathbf{A}\) and \(\mathbf{B}\) have the same dimensions.
