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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.

Martin Sleziak
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1west
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2 Answers2

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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.

robjohn
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  • Thanks,that's right. I am trying to figure out how to apply the exact formula you provided to the product of more than 2 matrices. Any ideas? – 1west May 14 '14 at 23:12
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    @1west Try using induction. We have $A^TB^TC^T=(BA)^TC^T=(C(BA))^T=(CBA)^T$. Proceed in the same manner. – Alex Becker May 14 '14 at 23:18
  • @1west: as Alex suggests, use induction. I have added this to my answer. – robjohn May 14 '14 at 23:23
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    An important point is that the matrix multiplication is associative! – Berci May 15 '14 at 00:11
  • @Berci: indeed! – robjohn May 15 '14 at 00:37
  • Proof by induction is a new concept to me and I haven't learned about it until now. But having read up on it, I don't understand how showing that the original equality also holds true for n+1 is a proof of my original equality. @robjohn: how would I extend your original sigma notation to include A^n? – 1west May 15 '14 at 00:53
  • If you are willing, please explain in detail to the point where you feel you are being redundant. Trust me, I likely won't think it is redundant! – 1west May 15 '14 at 01:00
  • @1west: the sums at the beginning only show the two matrix result. Trying to do more than that leads to unnecessary complication. We use the two matrix result and induction to show the case with an arbitrary number of matrices. – robjohn May 15 '14 at 01:02
  • @1west: I have added a bit of explanation to my answer. I have to go out for a bit, but read it again and see if the extra explanation helps. – robjohn May 15 '14 at 01:04
  • @1west: Mathematical induction is an extremely important tool, worth taking a detour to understand correctly. There are two essential parts of an inductive proof: prove the "base" case (usually $n=1$), and prove the case $n+1$ on the assumption it's true for $n$. In fact, the second part by itself is not a proof! But if you have both parts, you have a theorem that's true for $n=1$, therefore also true for $n=2$, therefore also for $n=3$, and so forth. – David K May 15 '14 at 03:25
  • @1west: For more guidance on mathematical induction, try http://math.stackexchange.com/questions/106404/mathematical-induction-question-why-can-we-assume-pk-holds and/or http://math.stackexchange.com/questions/505297/mathematical-induction-what-makes-it-true or ask your own follow-up question about what specifically troubles you about it. – David K May 15 '14 at 03:26
  • @robjohn Thank you for your help. After reading the logic behind proof by induction, I understand the reasoning. I think I have learned about this sometime in the past. David K, thank you for the links, I will take the detour you recommend to ensure I have proper mastery of the concept – 1west May 15 '14 at 20:07
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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}

From lemmas 2 and 3:

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

mins
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