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So, help guys. How to prove that the set of all complex n×n matrices of rank not greater than $k$ is an irreducible algebraic variety of dimension $k(2n − k)$.

Some definitions here:

$A$ algebraic variety, if set of its points are common zeros of polynomials $\in K[x_1, x_2, ..., x_n]$

$A$ irreducible algebraic variety, if its cannot be represented as $A = B\cup C$. Where $B,C$ - are algebraic varieties.

duncan
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  • So, there are three things you need to show: that it's a variety, that it's irreducible, and that it has dimension $k(2n-k)$. First, can you show that it's a variety? – Qiaochu Yuan May 26 '17 at 21:08
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    yes, i can do it. we can write a system for minors of k+1. This system provide our matrixes. And by definition as zeros of polynomial system, matrices are variety – duncan May 26 '17 at 21:14
  • Each of your subquestions has been asked before on math.SE. For irreducibility, For example, see this question: https://math.stackexchange.com/questions/1488184/matrices-with-rank-exactly-r-as-variety . The dimensionality question is addressed by loup blanc (and the link in the answer) and the irreducibility question is addressed by Georges Elencwajg. – Kenny Wong May 27 '17 at 12:22
  • but actually at the end there is no proof for: if Since $GL(n)×GL(m)$ is irreducible then it's orbit irreducible too. And actually why is $GL(n)×GL(m)$ irreducible ????? – duncan May 27 '17 at 14:10
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    @duncan : product of irreducible varieties are irreducible, and $GL(n)$ is irreducible as given by the equation $\det(x)t - 1 = 0$ in $\Bbb A^{n^2 + 1}$ which is an irreducible polynomial. –  May 27 '17 at 15:36
  • send please some literature about $GL(n)$. I don't understand this object. – duncan May 27 '17 at 15:42
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    $GL(n)$ is defined by the equation $\det(x) \neq 0$ which is polynomial in the matrix coordinates $x= (x_{11}, x_{12}, \dots, x_{nn})$. But the equation $f(x) \neq 0$ in $\Bbb A^r$ define an isomorphic variety as the equation $f(x)t = 1$ in $\Bbb A^{r+1}$. This is the argument you need to understand for see why $GL(n)$ is irreducible. –  May 27 '17 at 15:49
  • ok, we have all matrixes with $n^2$ elements and their $det \neq 0$. And this is irreducible. But why orbit of $M_r$ must be irreducible ? – duncan May 27 '17 at 15:58
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    If you are talking about the proof of Georges Elencwajg, this is because if $f : X \to Y$ is surjective and $X$ is irreducible, then $Y$ is irreducible. (The proof is exactly the same as if $X$ is connected and $f : X \to Y$ continuous and surjective then $Y$ is connected). –  May 27 '17 at 16:03
  • single matrix of rank $r$ is irreducible (as single point in $C^n$) and then its orbit irreducible too (as result of irreducible action). Right? – duncan May 27 '17 at 16:08
  • Not exactly. Matrix of rank $r$ can be written as $f(X)$ where $f$ is continuous for the Zariski topology and $X = Gl(n) \times Gl(m)$ is irreducible. The definition of $f$ is : pick a matrix $A_r$ of rank $r$ and send $(g,h) \in Gl(n) \times Gl(m) \to gA_Rh^{-1}$. PS : when you are replying to someone you can write "@user" at the beginning of the comment so the user is notified you replied and can answer quickly. –  May 27 '17 at 16:54
  • @N.H. off course for single point, we can write equation f(X) = 0 over Complex numbers. – duncan May 27 '17 at 16:58
  • We don't use that a point is irreducible here. We use that $Gl(n) \times Gl(m)$ is irreducible and that $f(Gl(n) \times Gl(m))$ is exactly the set of matrices of rank $r$. Since $f$ is continuous for the Zariski topology this implies the result. –  May 27 '17 at 17:00
  • @N.H. my misunderstanding connects with orbit definition. We have group of automorphisms and have element $X$. Orbit of $X$ is all if $\phi(X)$,where $\phi$ is automorphism. right ? In the proof we consider orbit of matrix with rank r. And now you suggest to see it from other hand, don't you? – duncan May 27 '17 at 17:10
  • Ok, I suggest you to read my previous comment again and the answer of Georges Elencwajg. If something is still not clear after you can ask me again, I think you have all the detail you need now. –  May 27 '17 at 17:13
  • @N.H. Actually this is proof for fact, that matrices of rank $k$ is irreducible. But i need to show that matrices of rank $\leq k $ is irreducible. And actually these sentences in contradiction. Because you can factorise $V_k = A_0 \cup A_1 \cup ... \cup A_k$. Isn't it??? – duncan May 28 '17 at 11:27
  • You are right, but hopefully your claim is still true because the matrices of rank $\leq k$ are the closures of the set of matrix of rank exactly $k$. –  May 28 '17 at 11:38
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    $A_k$ is not closed in its closure (!) so there is no contradiction, because $A$ is not irreducible means $A = A_1 \sqcup A_2$ with $A_1, A_2$ closed. But here this is not the case in your decomposition. –  May 28 '17 at 11:40
  • @N.H. why $A_k$ is not closed ? As hence do you have any ideas for irreducible of matrices of rank ≤ k ? – duncan May 28 '17 at 11:57
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    For example, the zero matrix is in the closure of $A_k$ for all $k$. More generally, you can show that $\overline{A_k} = A_k \sqcup A_{k-1} \dots \sqcup A_0$. The closure of an irreducible set is irreducible, so if $A_k$ is irreducible then $A_{\leq k}$ is also irreducible. –  May 28 '17 at 12:00
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    @N.H. I have next definitions: $A$ - algebraic variety as common zeros of some system $F(X) = 0$. $A$ - irreductible algebraic variety, if we can't represent it as $A = A_1 \cup A_2$, where $A_i$ - algebraic variety ; So, as I understand: $A_k$ - is irreductible, but not algebraic variety. And $A_{\leq k} = \cup_{i=0}^k A_i$ - is irreducible set, as closure of irreducible (or union of disjoint irreducibles ?). – duncan May 28 '17 at 12:14
  • @N.H. I get that $\det(x)t-1$ defines an algebraic variety, but I didn't get your function $f$ for extending it to matrices $n\times n$ of rank $k$ – reuns May 28 '17 at 12:19
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    Your first comment is correct. For the second comment : fix $B \in A_k$. The claim is that $f : Gl(n) \times Gl(m) \to A_k, (g,h) \mapsto g B h^{-1}$ is surjective and continuous for Zariski topology. This show that $A_k$ is irreducible. –  May 28 '17 at 13:24
  • @N.H. Thank you so much, I understand a lot:) But my question is reduced to this: https://math.stackexchange.com/questions/2302484/closure-of-image-by-polynomial-of-irreducible-algebraic-variety-is-also-irreduci – duncan May 30 '17 at 08:23
  • @duncan : you are welcome ! it was a very good idea to open a new question. You can write an answer for this question for close it (even saying, I understood assuming that closure of irreducible subset is irreducible). –  May 30 '17 at 08:38

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