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Let $X$ be an $m\times n$ matrix.For a given positive integer $t ≤ \min(m,n)$, we denote the determinantal ideal $I_t = I_t(X)$ generated by the t-minors.What is the relationship between $I_1,I_2,I_3,\cdots$?

Can I get $I_{3}$ through some operations on $I_2$?

I know that $V(I_2)$ is Segre variety and it can be seen as the product of two projective spaces.I hope that I can find "$?$" such that $I_2\ ?\ I_2 =I_3$.

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Set $M_k:=V(I_k) \subset \operatorname{Mat}(m \times n) \simeq \mathbb{A}^{mn}$.

(1) If $k < \min\{m, \, n\}$ then $M_{k-1}$ is precisely the singular locus of $M_k$. See

E. Arbarello, M. Cornalba, P. A. Griffiths, J. Harris: Geometry of Algebraic Curves, p. 69.

(2) The determinantal variety $M_k$ is the $k$th secant variety to $M_1$, see MSE question 4384405.

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  • $\begingroup$ Only for $k < \operatorname{min}(m,n)$. $\endgroup$
    – Sasha
    Commented May 9 at 10:06
  • $\begingroup$ Thanks for your answer. I know the result, but I don't understand how it can help me. I hope that smaller k can lead to larger k, but this seems to be the opposite. $\endgroup$
    – zhjzwlys
    Commented May 9 at 10:11
  • $\begingroup$ Then you must consider secant varieties. I will add it in the answer. $\endgroup$
    – Francesco Polizzi
    Commented May 9 at 11:13
  • $\begingroup$ @Sasha: fixed, thanks (it seems that [ACGH] forget to specify this, too...) $\endgroup$
    – Francesco Polizzi
    Commented May 9 at 11:18
  • $\begingroup$ For the second statement you should put $M_2$ instead of $M_1$. In this notation $M_k$ are the matrices of rank $k-1$. $M_{k+1}$ is the $k$ secant variety of $M_2$. Other than that, aren't these statements true when $k=\min(m,n)$? $\endgroup$
    – Zach Teitler
    Commented May 10 at 12:06

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