MHB Proving Symmetry of Matrix Multiplication with Transpose | Step-by-Step Guide

  • Thread starter Thread starter Yankel
  • Start date Start date
  • Tags Tags
    Matrix Transpose
Click For Summary
SUMMARY

The discussion centers on proving that the product of a matrix 'A' and its transpose, denoted as \(AA^T\), results in a symmetric matrix. The proof is established through element-wise definitions and properties of matrix multiplication, specifically showing that \((AA^T)^T = AA^T\). Participants emphasize the importance of understanding both direct proofs and the underlying concepts of symmetry in matrices, particularly in the context of commutative rings.

PREREQUISITES
  • Understanding of matrix multiplication
  • Familiarity with matrix transpose properties
  • Knowledge of symmetric matrices
  • Basic concepts of commutative rings in linear algebra
NEXT STEPS
  • Study the properties of symmetric matrices in linear algebra
  • Learn about matrix operations in commutative rings
  • Explore proofs of matrix transpose properties, particularly \((AB)^T = B^T A^T\)
  • Investigate the implications of matrix symmetry in various applications
USEFUL FOR

Mathematicians, students of linear algebra, and anyone interested in understanding matrix properties and proofs related to symmetry.

Yankel
Messages
390
Reaction score
0
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 !
 
Physics news on Phys.org
my favorite proof:

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

any questions?
 
Deveno said:
my favorite proof:

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

any questions?
The proof suits me to a "T."

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

-Dan
 
Yankel said:
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 !

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

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

CB

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).
 
Deveno said:
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).

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.

Before lecturing people, who if you thought about it for 30 seconds you would realize 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:
heh. i KNOW why you did it your way. methinks "lecturing" is a strong word.

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.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
Replies
7
Views
7K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
13
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
20K