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Let $M$ be a $n\times n$ matrix whose coefficients depend on some parameters (the number of parameters does not really matter). For some values of the parameters, $M$ is non-singular. Enforcing $\mathrm{rank}(M)\leq n-1$ requires solving a single equation: $\det M =0$.

My question is, how many equations are needed to enforce $\mathrm{rank}(M)\leq n-2$ (and which ones)? I would say that necessarily, the four $(n-1)\times (n-1)$ submatrices should be singular, which gives four equations. But it does not seem to be a sufficient condition.

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For real matrices, the answer is one, because any set of $k$ polynomial equations $p_1=\cdots=p_k=0$ is equivalent to the single equation $p_1^2+\cdots+p_k^2=0$. Yet, we can explicitly construct the polynomial equation we need here:

When $n\ge2$, a real $n\times n$ matrix $A$ has rank $\le n-2$ if and only if its entries satisfy the polynomial equation $$ \operatorname{tr}\left(\operatorname{adj}(A^TA)\right) = 0. $$

This follows easily from the following facts:

  • $A$ and $A^TA$ have identical ranks.
  • A matrix has rank $\le n-2$ if and only if its adjugate is zero.
  • The adjugate matrix of a positive semidefinite matrix is positive semidefinite.
  • A real positive semidefinite matrix is zero if and only if it has zero trace.
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    But then you loose the "information" of how many parameters are needed for the condition to be satisfied, right? For example, if there's only one free parameter, it's not possible to satisfy the rank $\geq n-2$ condition, even though there's a single equation (there's no bijection around the roots, because of the squares). – 2017-02-18
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    @anderstood Yes, but if you wish, you can always rewrite the equation in my answer as $n$ polynomial equations, so that each of the principal $(n-1)\times(n-1)$ minor of $A^TA$ is zero. – 2017-02-18
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    And there are $2n$ $(n-1)\times (n-1)$ minors of $A^\top A$, so that gives $2n$ equations. As in the other answer! **Edit**: there are $2n$ first minors, but you mention _principal_ minors... there are $n$. – 2017-02-18
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    Right, I just noticed it (see edit in comment). It's not very clear to me why it differs from the $\mathbb{C}$ case. I'll think about it. It "might" be because $A^\top A$ squares some terms, which would come back to my first comment (non bijectivity around roots). – 2017-02-18
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    @anderstood In the complex case, elements of $A^\ast A$ are not polynomials of entries of $A$, so the real case doesn't apply. – 2017-02-18
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In the paper

W. Bruns and R. Schwänzl. The number of equations defining a determinantal variety. Bull. London Math. Soc. 22 (1990), no. 5, 439-445.

you will find that over an algebraically closed field, the set of $n \times n$ matrices of rank $< t$, for $1 \le t \le n$, can be defined by $n^{2} - t^{2} + 1$ equations, and no less.

So in your case, with $t = n-1$, you need $2 n$ equations.

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    What if $M$ is a real matrix ($\mathbb R$ is not an algebraically closed field)? – 2017-02-17