Why does the transpose of a matrix product equal B^T A^T rather than A^T B^T?

## What transpose does The transpose `A^T` of a matrix `A` reflects it across the main diagonal: row `i` of `A` becomes column `i` of `A^T`, so `(A^T)[i,j] = A[j,i]`. An `m x n` matrix transposes to an `n x m` matrix. Applying transpose twice returns the original, `(A^T)^T = A`, and transpose distributes over addition without any surprise: `(A + B)^T = A^T + B^T`. Multiplication is where the twist appears. ## The shape argument Start with `A` of shape `m x n` and `B` of shape `n x p`, so `AB` is `m x p` and therefore `(AB)^T` is `p x m`. Now look at what the transposed factors offer: ``` A^T is n x m B^T is p x n ``` Try the naive order first: `A^T B^T` is `(n x m)(p x n)`. The inner numbers are `m` and `p`, which have no reason to agree, so in general this expression does not even exist. Try the reversed order: `B^T A^T` is `(p x n)(n x m)`. The inner numbers are `n` and `n`; they match, and the result is `p x m` — precisely the shape `(AB)^T` must have. The shapes alone tell you that if any product of the transposed factors can equal `(AB)^T`, it has to be the reversed one. This is a good habit in general: when you are unsure of an identity, check whether the two sides can even have the same shape. It rules out the wrong version immediately, without arithmetic. ## The entrywise proof Shapes show which candidate is possible; the entry computation shows the identity is true. Take any valid indices `i` (from 1 to `p`) and `j` (from 1 to `m`): ``` [(AB)^T][i,j] = (AB)[j,i] by the definition of transpose = sum over k of A[j,k]*B[k,i] by the definition of the product = sum over k of B[k,i]*A[j,k] numbers commute = sum over k of (B^T)[i,k]*(A^T)[k,j] = (B^T A^T)[i,j] ``` The only step doing real work is the third: the individual entries are ordinary numbers, so they may be swapped inside the sum. The matrices themselves are not swapped — their indices are — and the effect of rewriting both index pairs in transposed form is that `B` moves to the front. ## A tiny worked check Let `A = [[1, 2]]`, a `1 x 2` row, and `B = [[3],[4]]`, a `2 x 1` column. Then `AB = [[11]]`, a `1 x 1` matrix, and `(AB)^T = [[11]]` as well. Now `A^T` is `2 x 1` and `B^T` is `1 x 2`, so `B^T A^T = [[3, 4]] [[1],[2]] = [[11]]`. Correct. Meanwhile `A^T B^T` is `(2 x 1)(1 x 2) = [[3, 4],[6, 8]]`, a `2 x 2` matrix — a perfectly legal product here, since `m = p = 1` happens to hold, but a completely different object from `(AB)^T`. It is a useful reminder that the wrong version is not always undefined; sometimes it exists and is simply wrong. ## Chaining and relatives The rule extends to longer products by applying it repeatedly: ``` (ABC)^T = C^T B^T A^T ``` The whole chain reverses. The same reversal governs inverses of products, `(AB)^-1 = B^-1 A^-1` whenever both inverses exist, and for the same underlying reason: both transpose and inversion are order-reversing operations on products. A memorable way to hold it is the socks-and-shoes picture — to undo "socks then shoes" you take off shoes first, then socks. Other facts worth having ready alongside the identity: a matrix is called **symmetric** when `A^T = A`, which forces it to be square; `A^T A` and `A A^T` are always defined for any `A` and are always symmetric, since `(A^T A)^T = A^T (A^T)^T = A^T A`. That last derivation is itself a one-line application of the reversal rule, which is why interviewers like the question — it is a small identity that shows up constantly the moment you start manipulating expressions. ## Common mistakes The dominant error is writing `(AB)^T = A^T B^T` by analogy with `(A + B)^T = A^T + B^T`. Addition is entrywise and order-blind, so it transposes term by term; multiplication mixes rows with columns, so it does not. The second common error is applying the rule but forgetting it chains — reversing only the outer two factors of a triple product. Check with shapes and both errors surface immediately.