I could easily prove this using 2x2 matrices and multiplying them together, but how do you generally prove this and using letters not matrices? (this isn't homework, we haven't even taken symmetry yet I am just exploring)

EDIT: this is my attempt at proving it, I don't know whether it's correct or not.

$(AB)^{T} = B^{T}A^{T}$

And since $A$ and $B$ are symmetric, then $B^{T}A^{T} = BA = AB$ (since $AB = BA$)

doing the same for for $(BA)^{T}$ yields the same result hence showing it is in fact symmetric. It feels right but I am just hesitant about it.

EDIT 2: sorry for the bad phrasing of the question, here's the question literally: "Show that $AB$ is symmetric if and only if $AB = BA$ and both of $A$ and $B$ are symmetric."

  • 1
    $\begingroup$ Unless you're using some odd quantification, this isn't true. Let $A$ be a non-symmetric matrix (they exist) and let $B$ be the null matrix, then $AB$ is symmetric. $\endgroup$ – Git Gud Oct 13 '18 at 22:55
  • $\begingroup$ $A$ and $B$ being symmetric are both the given premises for the question, they aren't what's required to be proven. $AB$ being symmetric is what I am required to prove. $\endgroup$ – Eyad H. Oct 13 '18 at 22:57
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    $\begingroup$ If you change your "iff" to "if", then your statement is true. Saying something like "for symmetric $A,B$" instead of "and both $A$ and $B$ are symmetric" is another possible fix. As it stands we read it as "$AB$ is symmetric" being equivalent to "$AB=BA$ and $A,B$ are both symmetric", which is not true as pointed out several times. $\endgroup$ – Arthur Oct 13 '18 at 22:59
  • $\begingroup$ @Arthur fixed it $\endgroup$ – Eyad H. Oct 13 '18 at 23:05
  • $\begingroup$ I edited the titled and added an EDIT 2 that has the question literally copied and pasted here so you can read it yourself. the if and only if was emphasized. How do I approach this? $\endgroup$ – Eyad H. Oct 13 '18 at 23:08

I think the question, as expressed in the title, is quite adequately and clearly phrased.

Also, it seems clear to me that our OP Eyad H. is clearly on the right track, but didn't quite follow it far enough.

Here's my work:


$A^T = A, \; B^T = B, \tag 1$

$AB = BA \Longrightarrow (AB)^T = B^T A^T = BA = AB; \tag 2$

$(AB)^T = AB \Longrightarrow AB = (AB)^T = B^T A^T = BA. \tag 3$

Combining (2) and (3) we have

$AB = BA \Longleftrightarrow (AB)^T = AB. \tag 4$


I understand your proof that $AB=BA$ implies $AB$ is symmetric. But later you say "doing the same for $(BA)^T$". I believe that's unnecessary as you reached the desired result.

You didn't show however that $AB$ symmetric implies $AB=BA$. But yeah you established that $(AB)^T=BA$, but $AB$ is symmetric...

  • $\begingroup$ Yeah I just noticed that repeating it wasn't necessarily as it only wanted a proof for $AB$, but how do I prove it the other way around? starting with $AB$ being symmetric implying that $AB = BA$ and that $A$ and $B$ are symmetric? I don't know what to do with that one. $\endgroup$ – Eyad H. Oct 13 '18 at 23:14
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    $\begingroup$ Oh I see. Nope you don't need to prove that $A$ and $B$ are symmetric. Given that $A$ and $B$ are symmetric, all you need to show is: "$AB$ is symmetric iff $AB=BA$". So you can use the fact that $A$ and $B$ are symmetric. $\endgroup$ – Scientifica Oct 13 '18 at 23:15

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