I am trying to learn the mindset behind constructing proofs, and may have been given some bad advice along the way, but I'm really not sure - hence, I would like to check.

I did Calculus I-III last year, and we did very basic proofs. I was told by the instructor that we shouldn't solve one side, then the other - that we need to transform one side into the other. I'm taking Linear Algebra now and got stuck proving that $(AB)^{T} = B^{T}A^{T} $ because I couldn't think of any way beyond showing what $(AB)^{T}$ is, and then showing what $B^{T}A^{T} $ is, and then showing that they're the same. Which incidentally, was exactly what Proof Wiki did when I checked.

Was the advice inaccurate? Or is the random proof I found invalid or poor form?

Proof copied from Proof Wiki:

Let $A = [a]_{mn}, B = [b]_{np}$

Let $AB = [c]_{mp}$

Then from the definition of matrix product:

$∀i∈[1..m],j∈[1..p]:c_{ij}=\sum_{k=1}^{n} a_{ik}∘b_{kj}$

So, let $(AB)^{T}=[r]_{pm}$

The dimensions are correct, because AB is an $m×p$ matrix, thus making $(AB)^{T}$ a $p×m$ matrix. Thus:


Now, let $B^{T}A^{T}=[s]_{pm}$

Again, the dimensions are correct because $B^{T}$ is a $p×n$ matrix and $A^{T}$ is an $n×m$ matrix. Thus:


As the underlying structure of A and B is a commutative ring, then $a_{ik}∘b_{kj}=b_{kj}∘a_{ik}$.

Note the order of the indices in the term in the summation sign on the right hand side of the above. They are reverse what they would normally be because we are multiplying the transposes together.

Thus it can be seen that $r_{ji}=s_{ji}$ and the result follows.

  • $\begingroup$ The advice you were given ("need to transform one side into the other") is UTTER RUBBISH. $\endgroup$
    – user21820
    Commented Jan 29, 2020 at 3:10

1 Answer 1


The proof is valid.

To show that the two matrices are equal, what we need is just to show that the entries are equal to each other, which is what the proof has illustrated.

In general, if you want to prove that $A=B$, we can illustrate that $A=C$ and $B=C$ and conclude that $A=C$ since equality is an equivalence relation.

Perhaps your teacher was talking about proving an "If $A$ then $B$" question, then we can't assume that we have $B$ since that is what we are trying to prove.

  • $\begingroup$ Thank you - I really appreciate you taking the time to answer a basic question that nonetheless was really muddying my understanding. It's hard to gain traction at the beginning because little misunderstandings can throw you off. $\endgroup$
    – HFBrowning
    Commented Jan 28, 2020 at 16:34

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