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I've been stuck on this problem for a while. I found a way to do but it seems inappropriate. I'm very grateful if anyone can help me solve this:

Let $A$ be an $n \times n$ matrix with entries in the field $F$ and let $f_{1}, f_{2}, ..., f_{n}$ be the diagonal entries of the normal form of $xI - A$. For which matrices $A$ is $f_{1} \neq 1?$

The way I did it is: The normal form of $A$ has diagonal entries including $1,1,..., 1, p_{r}, p_{r-1}, ..., p_{1}$. Here $r$ is number of factors in cyclic decomposition of $A$. So $f_{1} \neq 1 $ where the number of factors in the cyclic decomposition of $A$ is $n$. In this case, $A$ has $n$ characteristic vectors.

Thanks so much for your help.

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Even though your way to do it might seem inappropriate, it is still appropriate to put your approach here, so that we understand what you have tried. Thanks. –  awllower Feb 25 '13 at 14:27
    
Ok. I've already added my solution to this problem. I think it's wrong but I can't find why? Can you post your solution, please –  le duc quang Feb 25 '13 at 14:33
    
I take it we are thinking of the Smith normal form of $xI-A$, and the $f_i$ are elementary divisors (whose product will be the characteristic polynomial of $A$)? –  hardmath Feb 25 '13 at 14:33
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Sorry, but how can a function of the form $x-a$ be equal to $1$ for all $x$? –  Sh3ljohn Feb 25 '13 at 14:38
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What if $A$ is the zero matrix? $A$ does not have distinct eigenvalues (or characteric values, as you put it). But the diagonal entries of $xI-A$ are all $x \neq 1$, and I take it this would be the "normal form". –  hardmath Feb 25 '13 at 14:38

1 Answer 1

up vote 2 down vote accepted

Here's a link discussing invariant factors of a matrix, which here are (apparently) the $f_i$, polynomials whose product is the characteristic polynomial of $A$ (necessarily of degree $n$) and which sequentially divide one another:

$$ f_i | f_{i+1} $$

Now by a degree argument, the only way to avoid $f_1 = 1$ is for all the invariant factors to be equal. That is, if $f_1 \neq 1$, then $\deg(f_1) > 0$ and by the divisibility conditions, all $deg(f_i) > 0$. But $\sum_i \deg(f_i) = n$ because their product is the characteristic polynomial of $A$ (determinant of $xI-A$), and this means all $deg(f_i) = 1$. But monic first-degree polynomials $f_i|f_{i+1}$ only if they are equal. Hence all the invariant factors are equal.

Now the minimal polynomial of $A$ is the least common multiple of its invariant factors, which by the divisibilty condition means $f_n$. Since $deg(f_n) = 1$, $A$ must have a single "characteristic value" $r$ whose geometric multiplicity is $n$. Equivalently $A = rI$ for some scalar $r \in F$.

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The $f_i$'s are called invariant factors. –  user26857 Feb 25 '13 at 15:34
    
@YACP: Yes, a synonym I noted in a comment on the Question. –  hardmath Feb 25 '13 at 15:36
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Wrong: the elementary divisors are the irreducible factors of invariant factors! –  user26857 Feb 25 '13 at 15:38
    
@YACP: Okay, I'm convinced! I changed that... –  hardmath Feb 25 '13 at 15:50
    
@hardmath: I don't think if $A$ has single "character value" means $A = rI$. It's not clearly convincible. Can you prove that??? Maybe $A$ just needs to be similar to $rI$ –  le duc quang Feb 25 '13 at 16:01

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