Invariants of a characteristic polynomial

  • Thread starter quantum123
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  • #1
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Main Question or Discussion Point

Hi:
There are 3 invariants. The first one is a trace. The third one is a determinant. So they are invariants.
The strange thing is the 2nd one. It is a hybrid term. Why is it also an invariant?
 

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  • #2
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Hi:
There are 3 invariants. The first one is a trace. The third one is a determinant. So they are invariants.
The strange thing is the 2nd one. It is a hybrid term. Why is it also an invariant?
I guess we are talking matrices in 3-dim and you are referring to the sum of the determinants of the diagonal minors of order 2. What do you mean by why? Isn't it enough that they are coefficient of the characteristic polynomial?

You can also specifically prove to yourself that this quantity is conserved under a similarity transformation (as all other coefficients).
 
  • #3
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Its to be expected. The set of eigenvalues of a matrix is an invariant.
So, any combinations of the eigenvalues, that is invariant under permutations
will also be an invariant.

This generalises to arbitrary sized square matrices.
 

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