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Let us say $x$ varies over the reals and I have a matrix $T$ which is a function of $x$. (The entries of $T$ depend on $x$)

Then I can enter such a matrix to some software and ask it to calculate the rank and the size of its kernel. The software promptly gives me back some integers as answers.

It seems that for $x=0$ I get a certain value of the rank and kernel dimension and for all other non-zero values of $x$ I get a different fixed set of numbers.

  • Is the above a generic case. Won't in general the kernel size and the rank depend non-trivially on the parameter?

  • If this behaviour (i.e a certain answer for $x=0$ and another answer for all other values of $x \neq 0$) is the generic behaviour then how does one prove such a thing?

It seems that the computer software (Mathematica) that I have been using gives the answer for $x \neq 0$ when I ask it to do the calculation for general $x$ and I need to check the $x=0$ case separately.

  • Is something like this known or can be known that the rank and kernel dimension of a parametrized matrix will vary smoothly with the parameter? (..obviously since the rank and kernel dimension are integers the only way they can depend nicely on the parameter is by being constant!..)

  • I would like to know if there are other caveats in general about doing such a calculation.

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I don't understand how you entered the matrix. Are the entries polynomials? What software did you use? –  Qiaochu Yuan May 5 '11 at 18:10
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The dimension of the kernel is the multiplicity of $0$ as a root of the characteristic polynomial of the matrix (and the rank is just the dimension minus the dimension of the kernel). If your coefficients lie in some ring of functions $R$, then you can view the characteristic polynomial as a polynomial with coefficients in $R$; the multiplicity of zero cannot vary smoothly, because it can only take integer values, but rather your question would be if the set of points where zero is not a root is open (in some suitably topology), the set where multiplicity is 1, etc. –  Arturo Magidin May 5 '11 at 18:27
    
You can think of $T(x) = x$ as a $1\times 1$-matrix with kernel dimension $0$, except for $x=0$, where kernel dimension is $1$, of course. $T$ is also analytic, which should answer part two of your question. You can of course calculate a basis of the kernel of such a matrix symbolically and do some further analysis on that. –  Alexander Thumm May 5 '11 at 18:28
    
@Qiachu I just used Mathematica to enter the matrix and asked it to calculate the rank and the null space. –  Anirbit May 6 '11 at 7:15
    
@Alexander @Arturo @Qiachu I have made further edits in the question to hopefully make it clearer. –  Anirbit May 6 '11 at 9:07

1 Answer 1

Most computer algebra systems (Mathematica, Maple, etc.), when asked to solve a problem containing a parameter, will give you an answer that is valid for general values of that parameter, but not necessarily for all. For instance, if you ask the program to compute the integral of $x^n$, it will reply $\frac{1}{n+1} x^{n+1}$, which is not true for $n=-1$ but true for all other values of $n$.

The same happens with your problem. Mathematica gives you an answer that is true for most values of $x$ but not for $x=0$.

Unless you have some more information, it is impossible to know in advance that something special happens at $x=0$. Think for instance about replacing $x$ by $y-1$ everywhere in your matrix $T$. Then the special value at which the rank and kernel is different is $y=1$, and nothing happens at $y=0$.

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Thanks for the clarification. Do you have insights on this general mathematics question of how to understand this particular behaviour of the rank of the matrix with a parameter? Is there a general criteria of when would such a parametrized matrix show this behaviour? Or how generically will this behaviour be seen? Or is this a question of case by case analysis...that given a parametrized matrix one can only explicitly calculate for it and see that its rank is constant for all but one value of the parameter. –  Anirbit May 7 '11 at 6:57

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