19

How do you prove the following fact about the transpose of a product of matrices? Also can you give some intuition as to why it is so.

$(AB)^T = B^TA^T$

4 Answers4

28

Here's an alternative argument. The main importance of the transpose (and this in fact defines it) is the formula $$Ax\cdot y = x\cdot A^\top y.$$ (If $A$ is $m\times n$, then $x\in \Bbb R^n$, $y\in\Bbb R^m$, the left dot product is in $\Bbb R^m$ and the right dot product is in $\Bbb R^n$.)

Now note that $$(AB)x\cdot y = A(Bx)\cdot y = Bx\cdot A^\top y = x\cdot B^\top(A^\top y) = x\cdot (B^\top A^\top)y.$$ Thus, $(AB)^\top = B^\top A^\top$.

Ted Shifrin
  • 125,228
  • How can we talk of a dot product as lying in a vector space? Isn't it just a number? And are there any pages where this definition is expanded upon? I haven't seen this way of defining the transpose anywhere. – harry May 15 '21 at 01:07
  • 1
    @HarryHolmes The dot product is an operation on two vectors living in a vector space, with output (here) real numbers. My “in” referred to the space the vectors live in, not to the output. This is a totally standard result, to be found in every lineR algebra book. – Ted Shifrin May 15 '21 at 01:17
6

When you multiply $A$ and $B$, you are taking the dot product of each ROW of $A$ and each COLUMN of $B$.

The resulting dimension is $A_{\#col}\times B_{\#row}$, and after transposing, you have $B_{\#row}\times A_{\#col}$.

When you multiply $B^T$ and $A^T$, you take the dot product of each row of $B^T$ (column of B) and column of $A^T$, or row of $A$.

Your resulting dimension is $B^T_{\#col}\times A^T_{\#row}$ which is just $B_{\#row}\times A_{\#col}$

This formula ensures that each entry is correct, and that the dimensions are identical.

Saketh Malyala
  • 13,897
  • 6
    Just to make up some notation to express your first + third sentence: let $\operatorname{row}i(M)$ and $\operatorname{col}_j(M)$ denote the $i^{\text{th}}$ row and $j^{\text{th}}$ column of $M$, respectively. Then $(AB){ij} = \operatorname{row}i(A) \cdot \operatorname{col}_j(B)$, and $(B^T A^T){ji} = \operatorname{row}j(B^T) \cdot \operatorname{col}_i(A^T) = \operatorname{col}_j(B) \cdot \operatorname{row}_i(A)$, so $(AB){ij} = (B^T A^T)_{ji}$. – Misha Lavrov May 01 '19 at 01:15
5

I marked this as community wiki since it so close to Saketh Malyala's answer.

For the intuition/background, please read this site answer.

We will now prove the assertion.

For any matrix $C$ let $\text{Row}(C,i)$ denote the $i^\text{th}$ row of $C$ represented in a natural way as vector.

For any matrix $C$ let $\text{Col}(C,j)$ denote the $j^\text{th}$ column of $C$ represented in a natural way as vector.

The $(i,j)^\text{th}$ entry of $AB$ is equal to $\langle \text{Row}(A,i), \text{Col}(B,j)\rangle$

The $(j,i)^\text{th}$ entry of $B^tA^t$ is equal to $\langle \text{Row}(B^t,j), \text{Col}(A^t,i)\rangle$

But $\text{Row}(B^t,j) = \text{Col}(B,j)$ and $\text{Col}(A^t,i) = \text{Row}(A,i)$, so indeed,

$$ {(AB)}^t = B^t A^t$$

CopyPasteIt
  • 11,870
2

If you know about dual spaces and maps, a conceptual proof can be obtained by observing that $A^T$ corresponds to the dual map of $A$ and that taking the dual is contravariant with respect to composition. That is, $(T \circ S)^* = S^* \circ T^*$.

Aniruddh Agarwal
  • 344
  • But then you're just delaying the actual argument until you prove that taking duals is a contravariant functor – Josselin Poiret May 31 '19 at 23:26
  • Actually, my bad, the fact that $ (-)^* = \mathrm{Hom}(-, k) $ is enough. But it still is a lot of work (the term "corresponds" actually hiding equivalences of categories). – Josselin Poiret May 31 '19 at 23:29
  • Well, proving that taking the dual corresponds to transposing a matrix only takes 3--4 lines.

    Let $T : V \rightarrow W$ be a linear map and $(v_i)$ and $(w_i)$ be basis for $V$ and $W$ respectively. Let $A$ be the matrix for $T$ and $A'$ be the matrix for $T^*$. It is enough to show that $A_{ij} = A'_{ji}$.

    Well $A_{ij} = w_i^(T(v_j))$ and similarly $A'_{ji} = v_j^{}(w_j^ \circ T)$ so it is enough to show that $v_j^{*}(w_j^ \circ T) =w_i^*(T(v_j))$. But this calculation is very simple.

     – Aniruddh Agarwal Jun 02 '19 at 20:32