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Hi all please help me with the following questions.

Let $A$ be a $n \times n$ matrix with complex entries and let $f(t)=\det(A-tI)$ be its characteristic polynomial. Prove that $f(A)=0$. Show the proof using Jordan form.

user1551
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Chloe
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    What have you tried so far? If you know how to show Cayley-Hamilton for diagonalizable matrices, then it's virtually the same proof when generalized to Jordan form. – EuYu Apr 15 '13 at 08:08
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    Well, when you define your $f:t\mapsto det(A-tI)$, what is your $t$ ? I believe $t\in\Bbb{C}$ but then you write $f(A)$ where $A\in\Bbb{C}^{n\times n}$ ... – Dolma Apr 15 '13 at 08:14
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    @Dolma That's not the issue here. It's two different maps with the same underlying formal polynomial. – EuYu Apr 15 '13 at 08:18
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    @EuYu Yes I know that, but it's not exactly the same mapping. Let me explain what I mean: In the first case, $t$ is actually mapped within the matrix (before the determinant is evaluated). However in the second case (where you map a matrix), $A$ is mapped after you evaluate the determinant (i.e. in the resulting polynomial). You're right saying it's not the issue here but it's important to note (imo ;)) – Dolma Apr 15 '13 at 08:25
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    The issue being slightly debated in these comments is, imo, an important conceptual one, and it boils down to ask: why cannot we ""simply"" substitute $,f(A)=\det(A-AI)=\det (0)=0\ldots,$ and we're done?! Well, this is, perhaps, what @Dolma was trying to convey: one can not do that. – DonAntonio Apr 15 '13 at 11:58
  • @DonAntonio Thank you, yes that's exactly what I was trying to say. The "matrix form" of the function has to be defined in order to be formal. If it's not, then one might think that you can "simply" substitute $t$ by $A$ as you said, and that would be a mistake. – Dolma Apr 15 '13 at 15:39

2 Answers2

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Throwing the Jordan normal form at the Cayley-Hamilton theorem seems overkill to me. In my opinion proving the Cayley-Hamilton theorem (even just over the complex numbers) requires a bit of work, but not nearly as much as proving the existence of a Jordan normal form. If you want to do it using an upper triangular form (which is just one of the many, many, approaches to proving the theorem), it suffices that each complex matrix is similar to a block diagonal matrix with each block (and therefore the whole matrix) upper triangular, and with all diagonal entries within any one block equal. (To obtain this form you still need to first find the generalised eigenspaces, and then find a basis of triangularisation in each such subspace; this already seems more work to me than needed for some other proofs of the C-H theorem.)

Given such a block decomposition, you need to show that (1) if $A$ is of this form, and has an upper triangular block of size $d$ and diagonal entries $\lambda$ as one of those blocks, then $(A-\lambda I)^d$ has a zero block at the corresponding (diagonal) position, and (2) a product of block diagonal matrices (with matching block sizes) such that for each diagonal block position at least one of them has a null block is entirely zero (this is easy).

Marc van Leeuwen
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Since $f$ is a complex polynomial of degree $n$, by the fundamental theorem of algebra, it has $n$ complex roots, probably with multiplicity, say $\lambda_1,...,\lambda_n$, anyway these are all the eigenvalues of $A$.

A Jordan block corresponds to an $A$-invariant subspace $U$, such that $A|_U=\lambda\cdot I+N$ where $\lambda$ is among the roots of $f$, $\ I$ is the identity (on $U$), and $N$ is nilpotent; and $A$ is identified with the linear map $x\mapsto A\cdot x$.

Say, the dimension of $U$ is $k$ (this also means that there are at least $k$ pieces of $\lambda$ among $\lambda_1,..,\lambda_n$). Then, as $N$ is nilpotent on $U$, we have $N^k=0$ (since by nilpotency $N$ must decrease the dimension in each step), that is, $(A-\lambda\cdot I)^k\,|_U=0$. Now, as $f(t)=\prod_i(\lambda_i-t)$, is a multiple of $(t-\lambda)^k$, and each $\lambda_i I-A$ is interchangable with each $\lambda_j I-A$, so we also have $$f(A)u=(\lambda_1 I-A)\dots(\lambda_n I-A)u =(\lambda_{s_1} I-A)(\lambda_{s_2} I-A)\dots(\lambda I-A)^ku =0\,,$$ if $u\in U$.

Finally, the whole space can be split as the direct sum of $A$-invariant subspaces.

Berci
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