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I have a question about proofs of this theorem:

Let $K$ be an algebraically closed field, $V$ be a finite-dimensional space over $K$ and $f : V → V$ be a linear operator.

Then there exists a Jordan basis for $f$, i.e. a basis of V in which the matrix of $ f$ is Jordan and the Jordan form of the matrix of f is determined uniquely up to interchanging the Jordan cells.

The proofs I've seen so far all say that:

to simplify notation we thus assume that $f$ has a unique eigenvalue $\lambda$ so that $V = V (\lambda)$. Moreover, we could also assume that $ \lambda = 0 $ since a Jordan basis for an operator $f$ is simultaneously a Jordan basis for the operator $f + c$ id where $c$ is an arbitrary constant.

Could you explain to me that is?

Thank you.

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First, I don't understand how is it possible to assume $\,f\,$ has one unique eigenvalue, when the gist of assuming an algebraically closed field is to be sure all the roots of the characteristic polynomial are there. Second, I've no idea what you mean by $\,^\perp\lambda\;,\;\lambda^\perp\,$ ... – DonAntonio May 1 '13 at 12:33
Here : page 9 – Andrew May 1 '13 at 12:43
up vote 1 down vote accepted

For any linear operator $f$ for which the characteristic (or minimal) polynomial splits into linear factors (as is always the case over an algebraically closed field), the vector space on which it acts has a (canonical) direct sum decomposition into generalised eigenspaces $V(\lambda)$ of $f$ for the different eigenvalues $\lambda$ that occur. These $V(\lambda)$ (summands of the direct sum) are invariant for $f$ (and even for any operator that commutes with $f$), which makes it possible to restrict $f$ to an operator on each of them individually, and all those restrictions together of course completely determine $f$.

The restriction of $f$ to $V(\lambda)$ by construction has $\lambda$ as unique eigenvalue; this is why one can assume that $f$ has a unique eigenvalue. Moreover, the restriction is equal to $\lambda$ times the identity on $V(\lambda)$ plus a nilpotent operator on $V(\lambda)$; a Jordan basis for the nilpotent operator will also be a Jordan basis for the restriction of $f$. Therefore one could reduce (if needed) to the nilpotent case, where $0$ is the unique eigenvalue.

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Could you explain to me why the restriction is equal to $\lambda$ times the identity plus a nilpotent operator? – Andrew May 1 '13 at 14:07
@Andrew: By definition of $V(\lambda)$, it is the kernel of some power $(f-\lambda I)^k$. This means precisely that the restriction $N$ to $V(\lambda)$ of $f-\lambda I$ satisfies $N^k=0$, and is nilpotent; the restriction of $f$ then of course is $N+\lambda I_{V(\lambda)}$. – Marc van Leeuwen May 2 '13 at 3:26

You can reduce to the case of a single eigenvalue by letting $V_\lambda :=\bigcup_{n\in\mathbb N} \ker(f-\lambda)^n$ and showing that $V=\bigoplus_\lambda V_\lambda$. The restriction $f_\lambda\colon V_\lambda\to V_\lambda$ of $f$ to $V_\lambda$ has only one eigenvalue $\lambda$. And of course the next step is to investigate the various $\ker(f_\lambda-\lambda)^n$ and note that the Jordan cell structure is determined by the sequence of the $\dim\ker(f_\lambda-\lambda)^n$.

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