Help with proving that the transpose of the product of any number of matrices is equal to the product of their transposes in reverse

Question

Specifically I am trying to show that (Aⁿ)^T = (A^T)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 = ((A^T)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 = (a^n-1)_1ia_j1 + (a^n-1)_2ia_j2 +...+ (a^n-1)_mia_jm

Right side of the equation:
Let a' denote the _ijth entry of A^T
((A^T)ⁿ)_ij = (a'^n-1)_i1a'_1j + (a'^n-1)_i2a'_2j +...+ (a'^n-1)_ima'_mj
= (a^n-1)_1ia_j1 + (a^n-1)_2ia_j2 +...+ (a^n-1)_mia_jm

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.

Answer 1

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.

Answer 2

Lemma 1: $(AB)^T = B^T A^T$:

\begin{array}{lll} (ab)_{ij}^T & = ab_{ji} \\ & = \sum_{k} a_{jk} b_{ki} \\ & = \sum_{k} b_{ki} a_{jk} \\ & = \sum_{k} b_{ik}^T a_{kj}^T \\ & = (b^T a^T)_{ij} \end{array}

Lemma 2: $(A^0)^T =(A^T)^0 $:

$$\forall U, \; U^0 = I \implies \begin{cases} A^0 = I \\ (A^0)^T = I \\ (A^T)^0 = I \end{cases} \implies (A^0)^T = (A^T)^0$$

Lemma 3: $(A^n)^T = (A^T)^n \implies (A^{n+1})^T = (A^T)^{n+1}$:

\begin{array}{ll} (A^n)^T = (A^T)^n & \implies A^T (A^n)^T = A^T (A^T)^n \\ & \implies (A^n A)^T = (A^T)^{n+1} \quad \text {(lemma 1)} \\ & \implies (A^{n+1})^T = (A^T)^{n+1} \end{array}

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From lemmas 2 and 3:

$$(A^n)^T = (A^T)^n \quad \forall n \in \mathbb{N}$$