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Wikipedia says that symmetric matrices are square ones, which have the property $A^T=A$. This assumes that one can have non-square $A^T=A$ and, because it does not satisfy the first property of symmetry, it is not symmetric. So, there can be non-symmetric $A^T=A$ matrices and the definition is right. Is it right? Or, definition is redundant and misleading, and it is better to define a symmetric matrix by single $A^T=A$ property and symmetric matrix squareness follows from symmetry rather than defines symmetry? The question in short: should I say that symmetric matrix is square, when define it?

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    $\begingroup$ When transposing matrix changes its dimensionality, if it's not square. You can't compare for equality matrices that have different dimensionality. $\endgroup$ – Kaster Jun 24 '13 at 18:10
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    $\begingroup$ Symmetric matrices are matrices that have the property that $A^T = A$, and all such matrices are square anyway (by counting rows and columns). The wikipedia definition is correct. Don't assume that, just because the word "square" is used, there must be non-square examples too - the word is redundant here, and is included just for clarity. $\endgroup$ – Billy Jun 24 '13 at 18:12
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    $\begingroup$ @Val: Reading mathematics is different to reading prose, and it is a separate skill that you learn, because mathematicians write in a certain way. The statement you have quoted is neither misleading nor confusing. It is precise and correct, and says nothing at all about non-square matrices with the property that $A^T = A$. In particular, it doesn't say whether they exist or not - you may try to work this out for yourself if you want to, but that's not what the definition is trying to get you to do. $\endgroup$ – Billy Jun 24 '13 at 18:33
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    $\begingroup$ @Val: Unlike in standard written English, the statement "x is true" does not hint that "not-x is false" in mathematics. Mathematics must be read literally and exactly as it was written. (There are also very good arguments for including the word "square" in the definition - for example, it's not clear what the equals sign between $A$ and $A^T$ even means unless we already know that they have the same size, so that has been hardwired into the definition.) $\endgroup$ – Billy Jun 24 '13 at 19:10
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    $\begingroup$ I do not understand the close-votes (or even the downvotes). There is a genuine question here! You are voting on a question, not on the OPs understanding of a concept! $\endgroup$ – user1729 Jun 24 '13 at 20:15
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If you consider that if $A$ is a $n$-by-$m$ matrix, then $A^T$ is a $m$-by-$n$ matrix, so it follows that if $A$ = $A^T$ then a necessary condition is that $n=m$ as the matrices need to have the same number of rows/columns if they are equal, so $A$ must be a square matrix.

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This is what Wiki says:

"In linear algebra, a symmetric matrix is a square matrix that is equal to its transpose".

Now, it certainly follows from the definition that $\,A\,$ is symmetric if $\,A^t=A\,$, and from this it follows at once that $\,A\,$ has to be square, but mentioning this in the definition above can hardly be confusing: reduntant, yes.

For example, you can check that many algebra books define a group as "a non-empty set with an binary operation...", and this "non-empty" thingy also is redudant as it follows at once from the axiom on the existence of a unity element...etc.

Things like the above ones, besides being a little boring and futile to discuss a lot about, are designed mostly to avoid misunderstandings from beginner students. That's all, imfho.

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  • $\begingroup$ I have explained you how it is confusing. In order to avoid confusion for beginners, you could simply list the squareness as the first property of symmetry. Otherwise, student is right to think that $A^T=A$ is not sufficient to establish the symmetry. Why do not you want to see the problem and, along with others, argue for confusion as clarity? $\endgroup$ – Val Jun 24 '13 at 18:43
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    $\begingroup$ @Val, what I think you've explained is what you find confusing (or misleading) in Wiki's definition, and I tried above to explain you and all why I don't think that is misleading, though I agree it is redundant. You wrote both these characteristics redundant and misleading together, as if one must appear with the other. I don't think this is right and, of course, I don't think this is so in this particular case. $\endgroup$ – DonAntonio Jun 24 '13 at 18:47
  • $\begingroup$ I have explained it objectively. If you do not know that m=n follows from $A^T=A$, you must assume that there are m≠n matrices, which satisfy $A^T=A$. In math, you must assume all the alternatives. So, all students will fall into the assumption that I did. So, it is not my, it is everybodies obligation to suppose the existence of the objects that I described. It is therefore unfair to argue that it is my personal fault and I am special and definition is ok. $\endgroup$ – Val Jun 24 '13 at 18:53
  • $\begingroup$ @Val, I've already written in my answer above that I agree the definition is redundant...no discussion about that! Now, apparently you find that "misleading", too, and I tried to explain why I believe it is not misldeading. $\endgroup$ – DonAntonio Jun 24 '13 at 18:55
  • $\begingroup$ The problem is that your "good pedagogy" is drilling of facts. It is bad pedagogy, objectively, since it raises such question as mine and does not teach the relationship between facts. Just saying that square follows from $A^T=A$ is simple and ties pieces together. By merely drilling of fragmentary, unrelated facts you neither produce masses of cattle rather than thinkers nor drilling simplifies material comprehension. I cannot stop you calling my approach "the absolutist pedagogy". $\endgroup$ – Val Jun 24 '13 at 22:34
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No, if $A$ is an $n\times m$ matrix then $A^T$ is an $m\times n$ matrix, so if $A$ is not square, then $n\neq m$ and hence $A^T$ has different dimensions to $A$ and hence can't be compared with $A$.

In that sense, $A^{T}=A$ implies $A$ square, but it also implies comparing matrices of different dimensions, which is something to avoid. (It's not technically wrong to say $A\neq B$ when $A,B$ are different dimensions, but it can cause confusion.)

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  • $\begingroup$ In a sufficiently sophisticated programming language, it is possible to arrange for matrices of different dimensions to have different "types", so that they cannot be compared. I don't know how this sort of thing is done in mathematics, but intuitively it makes sense to me to say that matrices of different dimensions are simply not equal, while a sentence like $S = M$, where $S$ is a set of integers and $M$ is a matrix of reals, just doesn't make any sense. $\endgroup$ – dfeuer Jun 24 '13 at 18:21
  • $\begingroup$ In mathematics, you define a matrix rigorously via set theory, and the sets are simply distinct in that case, so they are not equal as sets. But, as you say, there is a matter of types, which was why I suggested that it could cause confusion to write that. @dfeuer $\endgroup$ – Thomas Andrews Jun 24 '13 at 18:24

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