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

In summary, the proof suits Dan to a "T." He provides a summary of the content and ends with a humorous remark.
  • #1
Yankel
395
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
  • #2
my favorite proof:

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

any questions?
 
  • #3
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
 
  • #4
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
 
  • #5
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).
 
  • #6
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:
  • #7
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.
 

1. What is the definition of the transpose of a matrix?

The transpose of a matrix is a new matrix that is obtained by interchanging the rows and columns of the original matrix. This means that the first row of the original matrix becomes the first column of the transpose matrix, the second row becomes the second column, and so on.

2. Why do we need to find the transpose of a matrix?

The transpose of a matrix is useful in various mathematical operations, such as matrix multiplication and solving systems of linear equations. It also helps in simplifying calculations and finding patterns in data.

3. How do you find the transpose of a matrix?

To find the transpose of a matrix, simply interchange the rows and columns of the original matrix. This can be done by writing the first row of the original matrix as the first column of the transpose matrix, the second row as the second column, and so on.

4. What is the difference between a matrix and its transpose?

The main difference between a matrix and its transpose is the arrangement of elements. In a matrix, the rows and columns represent different entities or variables, while in its transpose, the rows and columns are switched, and the variables are now represented in a different order.

5. Is the transpose of a matrix always possible?

Yes, the transpose of a matrix is always possible as long as the matrix is a square matrix, meaning it has the same number of rows and columns. For non-square matrices, the transpose is not defined.

Similar threads

Replies
3
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
11
Views
2K
  • Linear and Abstract Algebra
Replies
2
Views
1K
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
9
Views
1K
  • Linear and Abstract Algebra
Replies
14
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
614
  • Linear and Abstract Algebra
Replies
3
Views
2K
  • Linear and Abstract Algebra
Replies
5
Views
2K
Back
Top