3

The problem here is straight forward. Let $a_{\lambda}$ be the algebraic multiplicity corresponding to $\lambda$. Prove $null (A - \lambda I)^{a_{\lambda}} = a_{\lambda}$

I know the following bits: $a_\lambda$ is the highest power of $(x-\lambda)$ that divides evenly, the characteristic polynomial of $A$. Is this any use at all, or am I thinking about this the wrong way?

Calculus08
  • 783
  • The easiest proof I know uses the Jordan normal form of the matrix. I don't know if something useful comes from the properties of characteristic polynomials, though (I think it does). – TZakrevskiy Feb 11 '14 at 14:22
  • I didn't think it would necessarily be easy, and it only makes sense that it arises from the Jordan normal form. Could you direct me to the proof, please? – Calculus08 Feb 11 '14 at 14:24
  • 1
    Related, very interesting: Down with determinants (http://www.axler.net/DwD.html). – Martín-Blas Pérez Pinilla Feb 11 '14 at 16:01

1 Answers1

2

First, each matrix has a unique structure of Jordan normal form. It consists of Jordan cells of the form $$J_\lambda=\begin{pmatrix}\lambda&1&0&\dots&\dots\\0&\lambda&1&0&\dots\\ \vdots&0&\ddots&\ddots&\ddots\\0&\dots&\dots&\lambda&1\\0&\dots&\dots&0&\lambda\end{pmatrix}$$ The size of this cell is called its order - denote it $ord(J_\lambda)$. Easy to see that $(J_\lambda-\lambda I)^{ord(J_\lambda)}=0$.

If $\mu\ne \lambda$, then $\det(J_\lambda-\mu I)^l\ne0$ for all $l$.

Finally, we take the whole JNF and study $(J-\lambda)^{a_\lambda}$. Clearly, $$a_\lambda = \text {sum of orders of all cells corresponding to $\lambda$}$$ by definition of algebraic multiplicity. Therefore, each such cell to the power $a_\lambda$ is zero and all other cells are non-singular. Thus, $$\dim\ker (J-\lambda)^{a_\lambda} =\dim\ker (A-\lambda)^{a_\lambda} = \text {sum of orders of all cells corresponding to $\lambda$}=a_\lambda.$$

TZakrevskiy
  • 22,980
  • Very nice, thank you so much. It makes a lot of sense. Our professor had only briefly discussed Jordan forms. – Calculus08 Feb 11 '14 at 17:38