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- Thread starter Yankel
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- Feb 15, 2012

- 1,967

my favorite proof:

$(AA^T)^T = ((A^T)^T)(A^T) = AA^T$

any questions?

$(AA^T)^T = ((A^T)^T)(A^T) = AA^T$

any questions?

- Aug 30, 2012

- 1,123

The proof suits me to a "T."my favorite proof:

$(AA^T)^T = ((A^T)^T)(A^T) = AA^T$

any questions?

Hahahahahahahaha...(runs out the door laughing maniacally)

-Dan

- Jan 26, 2012

- 890

\[\begin{aligned}\left( AA^T \right)_{i,j}&=\sum_k A_{i,k} \left(A^T\right)_{kj}\\Hello

I need to prove that for all matrices 'A', the multiplication of A with it's transpose, is a symmetric matrix.

How should I do it ?

Thanks !

&=\sum_k A_{i,k}A_{j,k}\\

&=\sum_k A_{j,k}A_{i,k}\\

&=\sum_k A_{j,k}\left(A^T\right)_{k,i}\\

&=\left(AA^T\right)_{j,i}

\end{aligned}\]

CB

- Feb 15, 2012

- 1,967

but:\[\begin{aligned}\left( AA^T \right)_{i,j}&=\sum_k A_{i,k} \left(A^T\right)_{kj}\\

&=\sum_k A_{i,k}A_{j,k}\\

&=\sum_k A_{j,k}A_{i,k}\\

&=\sum_k A_{j,k}\left(A^T\right)_{k,i}\\

&=\left(AA^T\right)_{j,i}

\end{aligned}\]

CB

\[\begin{aligned}(AB)^T_{i,j}&=(AB)_{j,i}\\

&=\sum_k A_{j,k} B_{k,i}\\

&=\sum_k B_{k,i}A_{j,k}\\

&=\sum_k (B^T)_{i,k}(A^T)_{k,j}\\

&=((B^T)(A^T))_{i,j}\\

\end{aligned}\]

that is: $(AB)^T = (B^T)(A^T)$ alllowing for my "economical" proof. (it should be obvious that $(A^T)^T = A$).

(the third equality shows why one should only study matrices over commutative rings...over non-commutative rings things get...ugly).

- Jan 26, 2012

- 890

The purpose of my post was to demonstrate the result using only the element-wise definition of transpose and of matrix products, and nothing else derived from them. Under this restriction you will have to prove the transpose of the product result and you have something more fiddly as it is indirect, as in having to prove a subsidiary result to get there.but:

\[\begin{aligned}(AB)^T_{i,j}&=(AB)_{j,i}\\

&=\sum_k A_{j,k} B_{k,i}\\

&=\sum_k B_{k,i}A_{j,k}\\

&=\sum_k (B^T)_{i,k}(A^T)_{k,j}\\

&=((B^T)(A^T))_{i,j}\\

\end{aligned}\]

that is: $(AB)^T = (B^T)(A^T)$ alllowing for my "economical" proof. (it should be obvious that $(A^T)^T = A$).

(the third equality shows why one should only study matrices over commutative rings...over non-commutative rings things get...ugly).

Before lecturing people, who if you thought about it for 30 seconds you would realise could have proven it your way if they had wanted to, on the relative merits of your proof over theirs you should consider why they have done it their way rather than yours.

CB

Last edited:

- Feb 15, 2012

- 1,967

one can consider the analogy with programming languages: it is possible to reduce a computer program to pure binary, but such a form is often unintelligible to most. on the other hand, an algorithm chart doesn't even say how the computer program actually does what it does, everything is "buried under the hood".

to reduce tedious computation to a minimum is part of why mathematicians use theorems, instead of reducing everything to a logical derivation from the basic tenets of an axiomatic system. detours can be longer, but nevertheless easier travelling.

of course, neither demonstration by either of us illustrates the "why" of this:

transposing means exchanging rows for columns (or the reverse). there is a certain "left-ness" to rows, and a certain "right-ness" to columns (at least in the conventional way we define matrix multiplication). so when you transpose a product, you get the "backwards" product of the transposes (what was on the left, is now on the right).

if the two matrices are mirror-images (in the row versus column way) of each other, this swapping doesn't do anything (and its this kind of symmetry, symmetry of reflection about the diagonal that we mean by "symmetric matrix").

personally i strive for clarity of exposing the underlying ideas, rather than the "undeniable truthfulness" of what i post. there are both advantages and drawbacks to this (such as i may be asked to "provide the details" if my exposition is too "packaged"). often, yes, this is indirect, but more economical. all other things being equal, i'd much rather remember things that give me "the most bang for the buck" then the nitty-gritty details. my memory only holds so much.

now *this* qualifies as a lecture.