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I am stuck on the following homework problem: Show that if S=S(m,n,r) represents the space of m by n matrices with rank less than or equal to r (naturally isomorphic to an affine subvariety of $\mathbb{A}^{mn}$ of course) over some arbitrary field k, then if r < min(m,n), the zero matrix is a singular point of S (in the sense of tangent spaces).

Now for an affine subvariety determined by the zeros of some $f_i$, I know the definition of a tangent space in terms of the directional derivatives etc., and I guess I want to find that, but I am confused about what the $f_i$ I would be taking directional derivatives of are: our matrices have rank $\leq$ r iff every r+1 by r+1 minor of the matrix has zero determinant, so these are the "equations" for which our matrices form the vanishing set: however, there are many many such equations and clearly it is not feasible to work with all of them at once. If our r is strictly less than the minimum of the row/col size then I think this implies that there actually is such a minor for which the determinant is 0, but then to show that a point is singular I would need to show that the dimension of the tangent space does not equal the dimension of S, wouldn't I? This surely implies I know the dimension of S, which is entirely non-obvious to me. Or maybe it is just that in the case of the zero matrix, it is easy to show the dimension of the tangent space is not equal.

Otherwise, is there a better (equivalent) definition of 'singular point' I could work with here? Preferably one where the equivalence of definition is reasonably simple to establish, since it seems a bit pointless having a proof of equivalent definition longer than the actual solution. Where am I going wrong? I am only a fourth year mathematician and algebraic geometry isn't my strong point so please try not to go overboard! -Peter

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    Also apologies if anything I said/asked was unclear, please just ask for clarification if needs be.2011-06-01

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The partial derivatives of the $(r+1) \times (r+1)$ minors are, for $r > 0$, homogeneous polynomials of positive degree, so they all vanish at the zero matrix, hence the dimension of the tangent space at the zero matrix is $mn$. To show that the zero matrix is singular when $r < \text{min}(m, n)$ it therefore suffices to show that $S(m, n, r)$ has dimension less than $mn$; you don't need to compute its exact dimension.

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    Ah brilliant, that's just the sort of solution I was looking for (and evidently failing to find), thank you. Does it suffice to argue that if the set S(m,n,r) had dimension mn then t would necessarily contain a matrix of greater rank? My worry is that this doesn't work since rank obviously isn't a property preserved under addition, but since we can't just add matrices of some rank $\leq$ r and guarantee that our rank is still < r, this seems to scupper most of my arguments...2011-06-01
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    @Peter: it suffices to show that $S(m, n, r)$ is contained in a hypersurface and then to show that hypersurfaces have dimension $mn-1$.2011-06-01
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    I see, that makes sense. Many thanks again!2011-06-01