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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.

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    $\begingroup$ 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 $A$ $\endgroup$ – Henry Sep 18 '15 at 0:15
  • $\begingroup$ Check a very simple case, say 1*2 & 2*3, you will find the problem. $\endgroup$ – Stan Sep 18 '15 at 0:20
  • $\begingroup$ @Henry Can you promote your comment to an answer? $\endgroup$ – Jeff Jan 4 '16 at 0:49
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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 $\;{}^{\mathrm t}\mkern-1mu C={}^{\mathrm t}\!B\:{}^{\mathrm t}\!A$.

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  • $\begingroup$ where has the notation of the $t$ in the top left come from? It's not in the original post. $\endgroup$ – baxx Dec 30 '19 at 23:45
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    $\begingroup$ 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. $\endgroup$ – Bernard Dec 30 '19 at 23:48
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$$(AB)_{ij} = \sum_{k=1}^n a_{ik} b_{kj}$$

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


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

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

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    $\begingroup$ -1: Lack of words and spurious equalities between matrices and entries do not make a good answer. $\endgroup$ – darij grinberg Feb 6 at 12:52
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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 $$(AB)^T = B^TA^T.$$

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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:$$(A \cdot B)^T=B^T \cdot A^T.$$

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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 $<x,y>_r=\sum_{i\leq r}x_iy_i$ -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 $x\in K^q,y\in K^n$, $<Ax,y>_n=<x,A^Ty>_q$. 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.

$\textbf{Proof}$. $<ABx,y>=<Bx,A^Ty>=<x,B^TA^Ty>=<x,(AB)^Ty>$.

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