What is a generic matrix

Question

What is a generic matrix? I try to google this but even can not find a definition. And what is a generic nilpotent matrix? Does generic has some canonical meaning?

Answer 1

That follows is the standard definition of a couple (for example) of "generic matrices" $A=[a_{ij}],B=[b_{ij}]\in M_n$. We assume that $A,B$ are fixed known matrices in $M_n$.

Def: Let $(a_{ij})_{ij},(b_{ij})_{ij}$ be independent commuting indeterminates over $\mathbb{Q}$; in other words, the $(a_{ij})_{ij},(b_{ij})_{ij}$ are elements of transcendental extension of $\mathbb{Q}$ and they are mutually transcendental over $\mathbb{Q}$. Then the matrices $(A,B)$ are called "generic matrices" over $\mathbb{Q}$.

In particular, $A,B$ have distinct non-zero eigenvalues in the algebraic closure $K$ of $\mathbb{Q}((a_{ij})_{ij},(b_{ij})_{ij})$, $AB\not= BA$, $A,B$ have no common eigenvectors,... More generally, if $P((x_{i,j})_{ij},(y_{ij})_{ij})$ is a non-identically zero polynomial, then, when $(A,B)$ is generic, $P((a_{ij})_{ij},(b_{ij})_{ij})\not=0$; the last inequality implies that $(A,B)$ is in an open dense Zariski set.

However we may want to stay in (for example) in $M_n(\mathbb{C})$. We can define a generic property of a system of equations. Let $S$ be a polynomial system of $n$ equations in $n$ unknowns, over a field $K$ st its coefficients are polynomial functions of $\tau$ parameters $(u_i)_i\in K^{\tau}$. Let $P$ be a property that may possibly have the solutions of $S$. $S$ is said to have "generically the property $P$" if there is a Zariski open dense $U\subset K^{\tau}$ st if $(u_i)_i$ is chosen in $U$, then the property is fulfilled by the associated solutions.

For example, consider the equation in $X$: $X^2=A$. The set of matrices $A$ that has (P) non-zero distinct eigenvalues is Zariski open dense; then this equation has generically exactly $2^n$ solutions. In particular, to choose $A$ as a nilpotent Jordan block is not generic. (if you randomly choose $A$, the $a_{ij}$ following a normal law, then the probability that $A$ has property (P) is ONE).

Now what is a nilpotent generic matrix $A$ ? It is a matrix that satisfies $A^n=0$; moreover the non-identically zero polynomials killed by $A$ are ALL in the ideal generated by the equations $A^n=0$. It is not difficult to see that such a matrix satisfies $A^{n-1}\not= 0$. Then, such a $A$ is always similar to the nilpotent Jordan block of dimension $n$.

Answer 2

Based on your choice of words I presume that you are working on the problem of extracting information about the Jordan normal form of matrix, possibly using orthogonal transformations to convert the matrix to staircase form in an effort to determine the size of the Jordan blocks in a numerically reliable manner.

In this context a generic matrix is one for which all eigenvalues are distinct. If you were to choose the entries of a matrix at random, then this is what you are likely to get. A matrix is less generic if two eigenvalues identical and so on. There is an entire hierachy, with the least generic matrices at top and the most generic matrices at the other bottom. Small perturbations of the matrix will move you downwards, whereas it is not necessarily true that a small perturbation will allow you to move up. These structural hiearchies can be very complex, when you are dealing with matrix pencils.

As for the generic nilpotent matrix I suspect that this is simply a Jordan block corresponding to the eigenvalue $0$, but that is entirely speculation on my part.