0
$\begingroup$

I'm learning about elementary linear algebra and I am confused on a specific point related to minimal polynomial.

When we have non degenerate eigenvalues it is just equal to the characteristic polynomial ($\mathrm{Det}(xI-A)$), when the eigenvalues span "eigenspaces" with dimension equal to the degeneracy of the eigenvalue we can just lift the multiplicity from the characteristic polynomial (which I understand by going to the diagonal basis and checking that it is enough to "substract once" the eigenvalue to get zeros on the diagonal and a null matrix in the end).

But I don't understand why, when the geometric degeneracy is strictly lower than than the algebraic degeneracy I cannot do the same thing and I do have to keep the multiplicity. Is there an intuitive argument for that?

$\endgroup$
  • 2
    $\begingroup$ It isn't clear what you mean by "keep the multiplicity". An eigenvalue has both an algebraic and a geometric multiplicity, and the geometric multiplicity is less than or equal to an eigenvalue's algebraic multiplicity. Probably the best way to understand this is by looking at some small examples. Would that help? $\endgroup$ – hardmath Nov 19 '14 at 11:57
  • 1
    $\begingroup$ I made a minor edit to highlight the run-on sentence. It seems to have a few misunderstandings. E.g. the "eigenspaces" are not spanned by eigenvalues but rather by eigenvectors. The minimal polynomial divides the characteristic polynomial, so we don't "lift the multiplicity from the characteristic polynomial" of an eigenvalue to get the minimal polynomial, but rather the opposite: the multiplicity of an eigenvalue in the minimal polynomial is less than or equal to its multiplicity in the characteristic polynomial. $\endgroup$ – hardmath Nov 19 '14 at 12:16
  • 1
    $\begingroup$ The last sentence of the Question refers to "geometric degeneracy" versus "algebraic degeneracy". I would rather use the terms "geometric multiplicity" and "algebraic multiplicity". The multiplicity of an eigenvalue in the characteristic polynomial is its algebraic multiplicity. The dimension of the eigenspace (number of linearly independent eigenvectors for an eigenvalue) is its geometric multiplicity. This avoids confusion with the deficiency of an eigenvalue, meaning the difference between the algebraic and geometric multiplicities. $\endgroup$ – hardmath Nov 19 '14 at 12:30
0
$\begingroup$

Supposing the characteristic polynomial$~\chi$ is split with distinct roots $\lambda_1,\ldots,\lambda_k$, the minimal polynomial$~\mu$ is so as well, with only the multiplicities of the roots possibly differing. If one writes $$ \mu=(X-\lambda_1)^{m_1}\ldots(X-\lambda_k)^{m_k} $$ one always has a direct sum decomposition of the whole space $$ V=\ker((A-\lambda_1I)^{m_1})\oplus\cdots\oplus \ker((A-\lambda_kI)^{m_k}) $$ and the summands are the generalised eigenspaces of $A$. Replacing an exponent by$~1$, one gets the definition of an (ordinary) eigenspace, which is smaller in general than the generalised eigenspace. In fact, by the definition of the minimal polynomial, one cannot lower the exponents without changing the kernels (making them smaller), since then the corresponding product in the place of $\mu$ would already be an annihilating polynomial of $A$. So the case where $\mu=(X-\lambda_1)\ldots(X-\lambda_k)$ (i.e., all exponents $m_i$ are already $1$) is exactly the diagonalisable case, where all generalised eigenspaces are the same as the eigenspaces. This means each geometric multiplicity (dimension of the eigenspace) equals the corresponding algebraic multiplicity (dimension of the generalised eigenspace).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.