# Jordan blocks and the characteristic polynomial.

For $A \in \mathbb{C}^{n,n}$ and $\{ \lambda_1, \dots , \lambda_r\}$ are the eigenvalues of $A$.

Then my lecture notes say that the characteristic polynomial of $A$ is $$(-1)^n\prod_{i=1}^r(x-\lambda_i)^{a_i}$$ where $a_i$ is the sum of the degrees of the Jordan blocks of $A$ of eigenvalue $\lambda_i$.

But I thought that there was only one Jordan block for a particular eigenvalue? I could understand it if it said that $a_i$ was just the degree of each block for each eigenvalue, because then for the JCF of an $n\times n$ matrix surely the characteristic equation would be raised to power $n$? If anyone could explain this it would be great!

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It is because you could have different eigenvalue blocks for the same eigenvalue. Just think of this. A diagonal case with the same $\lambda$ appearing on the diagonal, the characteristic polynomio is $(-1)^n(x-\lambda)^n$, but this $n$ does not represent the size of one jordan block, it represents the sum of all $1\times 1$ blocks, which in this case is $n$. – Daniel Montealegre Mar 21 '12 at 16:27
Ok, I feel like I'm asking stupid questions here but why does the characteristic poly have $(-1)^n$ at the start? I could have sworn for a diagonal matrix it was just the product of the diagonal when you've added in the '$- \lambda$' bit.. – user26069 Mar 21 '12 at 16:40
Because say $A$ is upper triangular, and in this case determinant is still the product of the diagonal, then the char polynomial is just $\mbox{det}(A-xI)$, here the $x$ is the one with the negative sign, and in a diagonal case you would have $(a_1-x)....(a_n-x)$ , but by convention, we want our characteristic polynomial to have a $1$ in the front, so we have to take care of this. – Daniel Montealegre Mar 21 '12 at 16:44
Or, you could calculate $\det(xI-A)$ instead and avoid the $(-1)^n$ completely. – Martin Argerami Mar 21 '12 at 17:24
Why is this question being proposed for closure? – user1729 Jun 17 '13 at 12:26

Consider $$\begin{bmatrix} 2&1 &0&0&0\\ 0&2&\mathbf{0}&0&0\\ 0&0&2&1&0\\ 0&0&0&2&1\\ 0&0&0&0&2 \end{bmatrix}$$
That matrix has only $2$ as an eigenvalue, but two Jordan blocks, one of size $2$ and the other of size $3$.