Eigenvalues of matrix and its transpose.
A matrix and its transpose have the same set of eigenvalues/other version: $A$ and $A^T$ have the same spectrum
If $A$ commutes with $B$ and $B$ commutes with $C$, then $A$ commutes with $C$ $f$ is irreducible $\iff$ $G$ act transitively on the roots Let $F$ be a field and $f\in F[X]$ a separable polynomial. Let $K_f$ be the splitting field of $f$ and $G_f=\text{Gal}(K_f/F)$ its Galois group. Show that $f$ is irreducible $\iff$ $G_f$ acts transitively on the roots on $f$. My work $\implies :$ Let $f\in F[x]$ irreducible, $\alpha \in K_f$ a root and $\alpha _1,...,\alpha _n$ the other roots of $f$ (so $K_f=F(\alpha ,\alpha _1,...,\alpha _n)$). What I have to show (I think) is that for all $i$ there is a $\sigma \in G_f$ s.t. $\sigma \alpha =\alpha _i.$ Q1) Is it right ? Q2) Is $f$ the minimal polynomial of $\alpha $ ? (To me it is, but my teacher said that it can be not the case, but I don't understand why). Since $\alpha $ and $\alpha _1$ have the same minimal polynomial, there is an isomorphism $\sigma :\alpha \longmapsto \alpha _1$. We can observe that $\sigma : F(\alpha )\longrightarrow K_f$. Now $F(\alpha ,\alpha _1)/F(\alpha )$ is algebraic, then we can extend $\sigma $ to a morphism $F(\alpha ,\alpha _1)\longrightarrow K_f$. Now, we can continue like that and extend $\sigma $ as $\psi: K_f\longrightarrow K_f$ s.t. $\psi(\alpha )=\alpha _1$. Q3) The fact that $\psi(\alpha )=\alpha _1$ comes from the fact that $\psi|_{F(\alpha )}=\sigma $, right ? The bijectivity is obvious since it's a field homomorphism and that $\sigma (K_f)=K_f$. Q4) How to conclude that $G_f$ act transitively on the roots ? I don't have particular problem for the converse. I had created the xml document with xml version="1.0".  In that document I need to use the greater than symbol &gt; and less than symbol &lt;.  How should I include those symbols? It's not working.  &amp;gt; and &amp;lt; are not working for me.  Is there any special encoder for this? how to prove an element is non-zero in a tensor-product I was studying the following example from Atiyah &amp; MacDonald's Introduction to Commutative Algebra:   Let $x$ be the non-zero element in $N := \mathbf{Z}/ 2\mathbf{Z}$, $M := \mathbf{Z}$, and $M' := 2 \mathbf{Z}$. The element $2 \otimes x$ is zero in $M \otimes N$, but non-zero in $M' \otimes N$. I suppose this can be seen by the fact that $2 \otimes x$ generates $M' \otimes N$, and the tensor-product $M' \otimes N$ is non-zero.   However, I was wondering, what one would do if that element weren't a generator of $M' \otimes N$. So my question is:   What are the methods to prove that an element is non-zero in a tensor-product of modules?   Thanks a lot! What does the phrase "Underfull \hbox (badness 10000) in paragraph" actually mean? What does the actual phrase above mean? Prove that the additive group $ℚ$ is not isomorphic with the multiplicative group $ℚ^*$. Prove that $ℚ^*_{&gt;0}$ is not isomorphic with $ℚ$. Suppose that $X_1,X_2,\ldots,X_n$ is a random sample from a distribution with probability function:  $$p_X(x)=\begin{cases} 1/2 &amp;x=-1,1 \\ 0 &amp; \text{otherwise} \end{cases}$$  Now if we define the average   $\bar{X}= \frac{1}{n} \sum X_i$, I need to show that that for an odd $n$ the probability function for $\bar{X}$ is:  $$p_{\bar{X}} (x)= \frac{\binom{n}{\frac{n}{2} (x+1)}}{2^n}$$ for $x=\pm 1/n,\pm 3/n,\ldots,\pm 1$, $0$ otherwise.    Since $P[\sum X_i=k] =P [\bar{X}=k/n ]$, I thought it would be easier to first derive the probability function for the sum. Since they are still $2^n$ n-tuples and each one is equilikely, I need to count the ones whose sum is $k$ for:  $k=\pm 1,\pm 3,\ldots \pm n-2,\pm n$ (since even numbers are not a feasible combination)  The problem here though is that I do not immediately recognize how I could count all these tuples. Could you please help me with this? If it is easier to proceed another way, I am of course all ears. Moment of inertia: why $\mathrm{dI}=r^2\mathrm{dm}$ instead of $\mathrm{dI}=m\mathrm{dr^2}$? What is serial voting and how does it affect me? I just noticed that I lost a bunch of points from my reputation score, and I used the "reputation" tab on my user profile page to try and track down the cause.  During my investigation, I noticed there was an unusual event of type "reversal". In the normal place of a question title, it says "voting corrected".     What does this mean, and what caused it?  Did I do something wrong? Why did I lose all of that reputation? Is the system punishing me for leaving too many good answers?  Is there any way to earn it back? New messages flagged blue bubble does not go away When someone flags a message in chat all 10k+ users in chat get a little blue bubble to vote on. It shows up in the upper left hand corner of your gravatar (which is in the bottom left of the page).  When someone flags a message and it shows up, if I wait for a second to click on it, and then click on it, and it has already been handled, it shows the message  "There are no flags to display."  In the past, after showing this message, the blue (1) would disappear. However, lately it has not been going away and requires a page refresh in order to remove. Can this be fixed?  This issue is addressed here:  from September 12th 2012. But since it is still occurring 10 months later, I believe the cause is different. How to add a gif file to my LaTeX file? Related:          I have an image file -- a gif file, which is the graph of a function. How do I add it to a LaTeX file? Will it come out OK in the subsequent pdf file? Why are effect size estimates (such as $R^2$) difficult in mixed-effects models? How to show that $\mathfrak{sl}_n(\mathbb{R})$ and $\mathfrak{sl}_n(\mathbb{C})$ are simple? Articles: When do I use "a", "the", or "__"? How do I know whether to use the definite (the) or indefinite (a, an) article, and when to omit it altogether? divides, not divides and cardinalities In group theory it happens very often that you want to say: "The order of $H$ divides / does not divide the order of $G$." But the length of | is then always the same:  $ |H| | |G| $ produces |H|||G|  which is not ideal at all. (The negation \nmid has just this length too.) The length of "divides" should be longer than the "cardinality length". I could use   $ |H| \big| |G|$  but then there is no good way to say "does not divide" with the same length. Also, \big| is no solution when the expression has to be used in subscripts.   I thought it was just me, but I asked other group theorists, and nobody had a solution. I hope there is one.  I am seeking a division sign which is a little longer than the usual | , allows negation (with the same length) and proportionally correct use in subscripts (also under \sum etc.). It does not look professional if the "order sign" has the same length as the "divides" sign. (And what about combinations with absolute values, norms - although personally I don't use them as often as group orders.) Are token creatures considerd as a creature spell? Are token creatures considerd creature spells? For instince when I use  tap effect combined with  would it allow me to draw a card? How do ideal quotients behave with respect to localization? Suppose $R$ is commutative ring with unity. For ideals $I$, $J \subseteq R$, the ideal quotient $(J:I)$ is $$(J:I) := \{x\in R \, : \, xI \subseteq J\}$$ Let $S\subset R$ be a multiplicative set. When does localization at $S$ commute with taking quotients, i.e. when does the equality $$(S^{-1} J : S^{-1} I) = S^{-1} (J:I)$$ hold? More generally, when is it true that $$S^{-1} \text{Ann}_{R} (M) = \text{Ann}_{S^{-1} R} (S^{-1} M)$$ for an $R$-module $M$? After startup for 5 minutes my system is very slow. The Task Manager shows that the disk usage is at 100%. (Later this number is around 2-4 %, but sometimes for 1-2 minutes go up again at 100%.)  I searched a lot on Google. For example I read , and also other pages. I opened two topic on Microsoft answer.  and one in Hungarian.  But I didn't found the solution yet. So I would like to ask Community's help. :)  Here the steps I already done:   I disabled everything at startups (including services) and the issue is the same.  I disabled Prefetch and Superfetch, the Baground Intellingent Transmission, Windows Search and Windows Defender but the issue is still persists. I made a health check on HDD, I made defrag and file system error check, but there are no errors. Also I tried to reinstall the HDD's driver. I made a full hardware check up, with no errors. The Windows and all the drivers are up-to-date. I tried a clean start. The issue is the same. I deleted Dropbox and all antivirus. I turned of the MSI mode.      I using also Linux, where this error is not occurs. Good to know, that Windows had 4 partitions, but I deleted 2 of them for the Linux. I don't know what partitions was these but both was only 10 - 20 MB.  I using a Dell Inspiron 15R SE 7520 with Windows 10 Home.  Personally I think it can be the Intel Rapid Store Technology "driver", because the problem started around when I installed it. (However I'm not sure.) And my hardware not supported yet by Intel, so I have no the newest driver. However I deleted Intel's software and driver and the issue is the same.    Thank you in advance for your assistance! Kinetic energy conservation in collision.  In any type of collision type of collision if we take the system to comprised of both bodies then the net external force is zero. So the work done by external force is also zero. By the relation $W=K_f-K_i$ as $W$ is zero hence kinetic energy is conserved. But this not the case.   Can you tell me what is the flaw in my argument? Some people say that the argument has neglected the fact that some kinetic energy is transformed into heat. But heat is created due to the interactions between bodies during collision which is thus an internal force and thus shouldn't affect kinetic energy. I know how to use the subfig package to fit many subfigures (each with its own caption) into a larger figure (with a general caption).  I know how to use the sidecap package, the put a figure's caption on the side of the figure.  What I do not know is how to display a caption inside the figure, as if it were a subfigure.  More precisely, I have three subfigures that I would like to fit on a 2x2 layout, with the remaining quadrant used for the caption.    Is that easy to achieve?