Help with proving that the transpose of the product of any number of matrices is equal to the product of their transposes in reverse
Specifically I am trying to show that (An)T = (AT)n where A is an mxm square matrix and n is a positive integer.
This is where I'm stuck:
To prove the theorem I would like to show that ((An)T)ij = ((AT)n)ij for all ij. All I can think of is expanding the definition of matrix multiplication.
Left side of equation:
((An)T)ij
= (An)ji = (an-1)1iaj1 + (an-1)2iaj2 +...+ (an-1)miajm
Right side of the equation:
Let a' denote the ijth entry of AT
((AT)n)ij = (a'n-1)i1a'1j + (a'n-1)i2a'2j +...+ (a'n-1)ima'mj
= (an-1)1iaj1 + (an-1)2iaj2 +...+ (an-1)miajm
Now the left and the right side of the equation are shown to be equal. In this proof, I mean for An to represent the product AAA... up to n. So if n= 3, this would represent the matrix resulting from the product of (AAA). The problem I have with this is that with my proof, determining the value in a specific position, say (AAA)ij , you must first determine the values of AA, and so on depending on the value of n. It just seems like there must be a better way of doing this proof. Can anyone help me out or show me why what I am doing is correct, or more likely, incorrect?
I am teaching myself linear algebra from Howard Anton's Elementary Linear Algebra text and unfortunately there seems to be a lot of assumed prior knowledge. I could really use detailed to the point of redundant explanations at this point in my learning! Also, since I have no one to interact with in constructing my proofs, I fear that I am missing some common practices. So feel free to be very critical of my format, so that I can get on track.
$\endgroup$1 Answer
$\begingroup$The fact that $(AB)^T=B^TA^T$ follows from the formulas $$ \begin{align} (ab)^T_{ki} &=ab_{ik}\\ &=\sum_{j=1}^na_{ij}b_{jk}\tag{1} \end{align} $$ and $$ \begin{align} (b^Ta^T)_{ki} &=\sum_{j=1}^nb^T_{kj}a^T_{ji}\\ &=\sum_{j=1}^nb_{jk}a_{ij}\\ &=\sum_{j=1}^na_{ij}b_{jk}\tag{2} \end{align} $$ To extend this to more than two matrices, use induction:
Suppose that for some $n$, we have $$ (A_1A_2\dots A_n)^T=A_n^T\dots A_2^TA_1^T\tag{3} $$ Note that we have already shown $(3)$ for $n=2$ using $(1)$ and $(2)$.
Then, using the two matrix result and $(3)$, we have $$ \begin{align} (A_1A_2\dots A_nA_{n+1})^T &=((A_1A_2\dots A_n)A_{n+1})^T\\ &=A_{n+1}^T(A_1A_2\dots A_n)^T\\ &=A_{n+1}^TA_n^T\dots A_2^TA_1^T\tag{4} \end{align} $$ Thus, the $n$ matrix result and the two matrix result imply the $n+1$ matrix result. Therefore, $(3)$ is true for two or more matrices.
$\endgroup$ 12
