The transpose of a product of matrices is one of the most elegant and frequently used properties in linear algebra. On the flip side, it states that the transpose of a product equals the product of the transposes taken in reverse order. Also, mathematically, for matrices $A$ and $B$ where the product $AB$ is defined, $(AB)^T = B^T A^T$. This reversal rule extends to any finite number of matrices and serves as a cornerstone for proofs, computational algorithms, and theoretical derivations across mathematics, physics, and computer science Surprisingly effective..
Understanding the Core Concept
Before diving into the mechanics of the product rule, You really need to recall what a transpose operation actually does. Because of that, given a matrix $A$ with dimensions $m \times n$, its transpose, denoted $A^T$, is an $n \times m$ matrix where the rows of $A$ become the columns of $A^T$, and vice versa. Formally, the element at position $(i, j)$ in $A$ moves to position $(j, i)$ in $A^T$.
When we multiply two matrices, say $A$ (size $m \times p$) and $B$ (size $p \times n$), the resulting matrix $C = AB$ has dimensions $m \times n$. In real terms, this new entry is the dot product of the $j$-th row of $B^T$ (which was the $j$-th column of $B$) and the $i$-th column of $A^T$ (which was the $i$-th row of $A$). Which means if we were to transpose $C$, the entry $c_{ij}$ moves to position $(j, i)$. Even so, the entry $c_{ij}$ is the dot product of the $i$-th row of $A$ and the $j$-th column of $B$. This structural symmetry is precisely why the order must reverse.
Not obvious, but once you see it — you'll see it everywhere.
The General Rule: Reversal Law
The property $(AB)^T = B^T A^T$ is often called the reversal law for transposes. It is critical to note that the dimensions must align for the multiplication to be valid on both sides Practical, not theoretical..
- Left Side $(AB)^T$: $A$ is $m \times p$, $B$ is $p \times n$. Product $AB$ is $m \times n$. Transpose is $n \times m$.
- Right Side $B^T A^T$: $B^T$ is $n \times p$, $A^T$ is $p \times m$. Product $B^T A^T$ is $n \times m$.
The dimensions match perfectly, confirming the algebraic validity of the rule. If one mistakenly writes $(AB)^T = A^T B^T$, the dimensions would generally be incompatible (yielding $m \times n$ vs $n \times m$), serving as an immediate dimensional check for errors.
Extending to Multiple Matrices
This rule generalizes naturally to a product of three or more matrices. For matrices $A_1, A_2, \dots, A_k$ where the product is defined:
$(A_1 A_2 \dots A_k)^T = A_k^T \dots A_2^T A_1^T$
The order is completely reversed. This is exceptionally useful when dealing with long chains of transformations, such as in computer graphics pipelines or Kalman filtering equations, where transposing a composite transformation requires applying the inverse transpose logic to each component in reverse sequence.
Step-by-Step Verification with a Concrete Example
Let us verify this property with a numerical example to build intuition.
Let matrix $A$ be $2 \times 3$ and matrix $B$ be $3 \times 2$:
$A = \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \end{bmatrix}, \quad B = \begin{bmatrix} 7 & 8 \ 9 & 1 \ 2 & 3 \end{bmatrix}$
Step 1: Compute the product $AB$. $AB$ will be a $2 \times 2$ matrix. $AB = \begin{bmatrix} (1\cdot7 + 2\cdot9 + 3\cdot2) & (1\cdot8 + 2\cdot1 + 3\cdot3) \ (4\cdot7 + 5\cdot9 + 6\cdot2) & (4\cdot8 + 5\cdot1 + 6\cdot3) \end{bmatrix}$ $AB = \begin{bmatrix} 7+18+6 & 8+2+9 \ 28+45+12 & 32+5+18 \end{bmatrix} = \begin{bmatrix} 31 & 19 \ 85 & 55 \end{bmatrix}$
Step 2: Transpose the result $(AB)^T$. $(AB)^T = \begin{bmatrix} 31 & 85 \ 19 & 55 \end{bmatrix}$
Step 3: Compute individual transposes $A^T$ and $B^T$. $A^T = \begin{bmatrix} 1 & 4 \ 2 & 5 \ 3 & 6 \end{bmatrix} \quad (3 \times 2), \quad B^T = \begin{bmatrix} 7 & 9 & 2 \ 8 & 1 & 3 \end{bmatrix} \quad (2 \times 3)$
Step 4: Compute the reverse product $B^T A^T$. Note the order: $B^T$ ($2 \times 3$) times $A^T$ ($3 \times 2$) yields a $2 \times 2$ matrix. $B^T A^T = \begin{bmatrix} (7\cdot1 + 9\cdot2 + 2\cdot3) & (7\cdot4 + 9\cdot5 + 2\cdot6) \ (8\cdot1 + 1\cdot2 + 3\cdot3) & (8\cdot4 + 1\cdot5 + 3\cdot6) \end{bmatrix}$ $B^T A^T = \begin{bmatrix} 7+18+6 & 28+45+12 \ 8+2+9 & 32+5+18 \end{bmatrix} = \begin{bmatrix} 31 & 85 \ 19 & 55 \end{bmatrix}$
Conclusion: $(AB)^T = B^T A^T$. The entries match perfectly, confirming the reversal law.
Algebraic Proof Using Summation Notation
For a rigorous understanding, we can prove this using index notation (Einstein summation convention or explicit sums). Let $A = [a_{ik}]$ be $m \times p$ and $B = [b_{kj}]$ be $p \times n$. The product $C = AB$ has entries:
$c_{ij} = \sum_{k=1}^{p} a_{ik} b_{kj}$
The transpose $(AB)^T$ has entries at position $(j, i)$: $[(AB)^T]{ji} = c{ij} = \sum_{k=1}^{p} a_{ik} b_{kj}$
Now consider the right-hand side $B^T A^T$. On the flip side, the entries of $B^T$ are $(B^T){jk} = b{kj}$. The entries of $A^T$ are $(A^T){ki} = a{ik}$ Most people skip this — try not to. That alone is useful..
Since scalar multiplication is commutative ($a_{ik}b_{kj} = b_{kj}a_{ik}$), the summation terms are identical. Because of this, $d_{ji} = [(AB)^T]_{ji}$ for all $i, j$, proving the equality Easy to understand, harder to ignore..
