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In linear algebra I've seen that there are two major different ways to write the characteristic polynomial of a mapping $f$: As $$ p_{f}=\pm\left(T-\lambda_{1}\right)^{m_{1}}\cdot\ldots\cdot\left(T-\lambda_{k}\right)^{m_{k}} $$ with distinct $\lambda_{i}$ or as $$ p_{f}=\pm\left(T-\lambda_{1}\right)\cdot\ldots\cdot\left(T-\lambda_{n}\right) $$ where the $\lambda_{i}$ are not necessarily distinct.

The only advantage I can see the latter expression has over the former is that the eigenvalues are already given in an order from $1$ to $n$, so it's easier to user the latter expression when doing something like induction over the eigenvalues.

But this argument is not very convincing to me...what kind of situations (theorems) do you know, where it is definitely better to use one expression over the other ? What are the disadvantages of each of the expressions ?

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I usually prefer the first representation, because that way I can know exactly how many eigenvectors I have to find for every eigenvalue, in order for the map to be diagonalizable. –  Noy Soffer May 22 '13 at 11:43
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I agree with @NoySoffer. I also wouldn't use \cdot\ldots\cdot. Instead I'd use only \cdots. –  Sigur May 22 '13 at 11:46
    
The characteristic polynomial is usually taken to be monic, so there is no need to use $\pm$. –  lhf May 22 '13 at 12:16
    
@lhf I've seen in different texts both version appear. –  temo May 22 '13 at 15:30
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1 Answer 1

In bringing the matrix to Jordan canonical form (see http://en.wikipedia.org/wiki/Jordan_canonical_form) it is crucial to have the multiplicities you denoted $m_i$ so as to describe the blocks into which the matrix decomposes (up to conjugation).

On the other hand, computationally speaking the coincidence of eigenvalues is not stable, and from this point of view the significance of the Jordan canonical form is purely theoretical. From the viewpoint of applications the second form may be more relevant.

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If the coincidence of eigenvalues is not stable, then I'd guess any method of ordering the eigenvalues runs into the same problem (the ordering is not stable). So the second method has the same difficulty, since giving it explicitly requires ordering the eigenvalues. –  Marc van Leeuwen May 22 '13 at 12:02
    
In computational approaches they don't order the eigenvalues. The "standard form" they get is more complicated than Jordan canonical form in that it has nonzero entries in the line both below and above the main diagonal. –  user72694 May 22 '13 at 13:14
    
Could you be more specific concerning the viewpoint of application ? (For example, for the theorem that tells you that any matrix, whose characteristic polynomial splits, is triangularisable, it is better to use the second form I think...) –  temo May 22 '13 at 13:54
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