# How to prove $(AB)^T=B^T A^T$

Given an $$m\times n$$-matrix $$A$$ and an $$n\times p$$-matrix $$B$$. Prove that $$(AB)^T = B^TA^T$$.

Here is my attempt:

Write the matrices $$A$$ and $$B$$ as $$A = [a_{ij}]$$ and $$B = [b_{ij}]$$, meaning that their $$\left(i,j\right)$$-th entries are $$a_{ij}$$ and $$b_{ij}$$, respectively.

Let $$C=AB=[c_{ij}]$$, where $$c_{ij} = \sum_{k=1}^n a_{ik}b_{kj}$$, the standard multiplication definition.

We want $$(AB)^T = C^T = [c_{ji}]$$. That is the element in position $$j,i$$ is $$\sum_{k=1}^n a_{ik}b_{kj}$$. For instance, if $$i=2, j=3$$, then the element in $$2,3$$ of $$C$$ is that sum, but the element in position $$3,2$$ of the transpose is that sum.

I need to get the same value for the element in position $$3,2$$ of the right side.

The transpose matrices are $$B^T=[b_{ji}], A^T=[a_{ji}]$$. They are size $$p \times n$$ and $$n \times m$$. That is, they switch rows and columns.

Let $$D = B^T A^T = [d_{ji}]$$. I write the indices backwards because if I want the element in position $$3,2$$, that is, $$i=2, j=3$$ just like on the other side.

So I need the summation for $$d_{ji}$$. But I get as $$d_{ji} = \sum_{k=1}^n b_{jk}a_{ki}$$, which does not match.

• Why do you not have $d_{ji} = \sum_{k=1}^n b_{kj}a_{ik}$? Remember this is the multiplying the transposes of $B$ and $A$ – Henry Sep 18 '15 at 0:15
• Check a very simple case, say 1*2 & 2*3, you will find the problem. – Stan Sep 18 '15 at 0:20
• @Henry Can you promote your comment to an answer? – Jeff Jan 4 '16 at 0:49

I would write it this way: denoting $a'$ and $b'$ the coefficients of $\;{}^{\mathrm t}\!A$ and $\;{}^{\mathrm t}\!B$, we have:

$$d_{ij}=c_{ji}=\sum_{1\le k\le n}a_{jk}b_{ki}=\sum_{1\le k\le n}b'_{ik}a'_{kj},$$ hence $\;{}^{\mathrm t}\mkern-1mu C={}^{\mathrm t}\!B\:{}^{\mathrm t}\!A$.

• where has the notation of the $t$ in the top left come from? It's not in the original post. – baxx Dec 30 '19 at 23:45
• This was the standard notation for the transpose when I was a student. I think it comes from Bourbaki, and it is available in LaTeX. – Bernard Dec 30 '19 at 23:48

$$(AB)_{ij} = \sum_{k=1}^n a_{ik} b_{kj}$$

$$[(AB)_{ij}]^T = \sum_{k=1}^n a_{jk}b_{ki}$$

$$(BA)_{ij} = \sum_{k=1}^n b_{ik}a_{ki}$$

$$B^{T}A^{T} = (B_{ik})^{T}(A_{kj})^{T} = \sum_{k=1}^n b_{ki}a_{jk} = \sum_{k=1}^n a_{jk}b_{ki}$$

• -1: Lack of words and spurious equalities between matrices and entries do not make a good answer. – darij grinberg Feb 6 at 12:52

We know on the one hand that $$(AB)_{ji} = (AB)^T_{ij}$$, hence $$(AB)^T_{ij} = (AB)_{ji} = \sum\limits_{k=1}^nA_{jk}B_{ki},$$ on the other hand $$(B^TA^T)_{ij}= \sum\limits_{k=1}^nB^T_{ik}A^T_{kj}= \sum\limits_{k=1}^nB_{ki}A_{jk}= \sum\limits_{k=1}^nA_{jk}B_{ki},$$ so, since $$(AB)^T_{ij} = (B^TA^T)_{ij}$$ for all $$i=1,...,p$$ and $$j=1,...,m$$ we have $$(AB)^T = B^TA^T.$$

The product of $$A$$ and $$B$$ matrix is $$(A\cdot B)_{ij}=\text{row}(A)_i \cdot \text{column}(B)_j$$ $$(A\cdot B_{ij})^T=C_{ji} = \text{row}(A)_j \cdot \text{column}(B)_i$$ $$\text{row}(A)_i=\text{column}(A^T)_i$$ $$\text{column}(B)_j=\text{row}(B^T)_j$$ $$(A\cdot B)_{ij}=\text{row}(A)_i \cdot \text{column}(B)_j=\text{column}(A^T)_i \cdot \text{row}(B^T)_j=\text{row}(B^T)_j \cdot \text{column}(A^T)_i=(B^T \cdot A^T)_{ji}.$$ So, we have: $$(A \cdot B)_{ij}=(B^T \cdot A^T)_{ji}.$$ We can rewrite last statement in following form: $$(A \cdot B)=(B^T \cdot A^T)^T.$$ Now, lets transpose both sides and we will get what we want:$$(A \cdot B)^T=B^T \cdot A^T.$$

To show, with hand, that the entries of $$(AB)^T$$ and $$B^TA^T$$ are the same is easy but has no theoretical interest. Moreover, a young student, to whom the transposed matrix is ​​defined as $$A^T_{i,j}=A_{j,i}$$, may think that the above considered formula is a pure miracle while it's only a consequence of the natural definition that follows -cf. also the Ted Shifrin's answer in

Transpose of product of matrices

Let $$K$$ be a field and $$_r=\sum_{i\leq r}x_iy_i$$ -a non-degenerate bilinear symmetric form over $$K^r$$-. Then we define the transpose by duality (exercise)

$$\textbf{Proposition 1}$$. Let $$A\in M_{n,q}$$; then $$A^T\in M_{q,n}$$ is uniquely defined by, for every $$x\in K^q,y\in K^n$$, $$_n=_q$$. Moreover, for every $$i,j$$, $$(A^T)_{i,j}=A_{j,i}$$.

$$\textbf{Proof}$$. Choose $$x=e_i,y=e_j$$.

$$\textbf{Proposition 2}$$. $$(AB)^T=B^TA^T$$ when the product $$AB$$ is defined.

$$\textbf{Proof}$$. $$===$$.