More specifically, if $A$ is a matrix over some algebraically closed field and $P_A(x)$ is its characteristic polynomial, how do we know that the roots of $P_A(x)$ are the eigenvalues of $A$, with the same multiplicity as roots and as eigenvalues? (For lack of a better definition, I'm taking the multiplicity of an eigenvalue $\lambda$ of $A$ as the number of times $\lambda$ appears on the main diagonal in the Jordan normal form of $A$.)
Let $A$ be an $n\times n$ matrix. Then $\lambda$ is an eigenvalue of $A$ iff $\exists v$ such that
$$\begin{align}Av &=\lambda v\\Av-\lambda v &=0\\ (A-\lambda I_n)v&=0,\end{align}$$
and this occurs iff $\det(A-\lambda I)=0$, i.e., iff $P_A(\lambda)=0$.
This shows that the eigenvalues of the matrix are exactly the roots of the the characteristic polynomial, but it doesn't say anything about the relative multiplicities of them as eigenvalues and as roots.
Solution
Let $\lambda_1,\cdots,\lambda_n$ be the roots of $P_A(x)$. If $\lambda_1,\cdots,\lambda_m$ are the distinct roots of $P_A(x)$ (hence the distinct eigenvalues of $A$ by above), then we know for some $\alpha_1,\cdots,\alpha_m,\beta_1,\cdots,\beta_m\in\mathbb N$ with
$$\sum_{k=1}^m\alpha_k=n=\sum_{k=1}^m\beta_k,$$
we have
$$\begin{align}\prod_{k=1}^n\lambda_k&=\prod_{k=1}^m\lambda_k^{\alpha_k}\\\tag1\det(A)&=\prod_{k=1}^m\lambda_k^{\beta_k};\end{align}$$
what we want to prove is that $\alpha_k=\beta_k$ for all $k$. Since $\lambda_1,\cdots,\lambda_n$ are the roots of $P_A(x)$, by Viète's Formulas
$$P_A(0)=\prod_{k=1}^n\lambda_k.$$
On the other hand, by the definition of the characteristic polynomial, we have
$$P_A(0)=\det(A-0I_n)=\det(A),$$
so
$$\prod_{k=1}^m\lambda_k^{\alpha_k}=\prod_{k=1}^m\lambda_k^{\beta_k}.$$
Since the $\lambda_k$ are distinct and $\sum\alpha_k=\sum\beta_k$, we can conclude the desired result.
Concerns
My worry here is that I'm putting the cart before the horse, so to speak: In (1), I used the fact that the determinant of $A$ is the product of $A$'s eigenvalues, taken with appropriate multiplicity. Generally, proofs of this seem to be of the form of this answer, which makes use of the very fact I'm trying to prove! To me, this suggests that I'm fundamentally trying to do things backward, especially since the only other way I've found of showing that the determinant of $A$ is the product of $A$'s eigenvalues uses the Jordan normal form theorem:
Let $M$ be an $n\times n$ invertible matrix such that $M^{-1}AM$ is in Jordan normal form. Then $A$ and $M^{-1}AM$ have the same egenvalues, and
$$\det(M^{-1}AM)=\det(M^{-1})\det(A)\det(M)=\det(M)^{-1}\det(A)\det(M)=\det(A).$$
Thus it suffices to prove the result for matrices in Jordan normal form. But
- If a matrix is in Jordan normal form, the elements on the diagonal are the eigenvalues by the Jordan normal form theorem.
- The determinant of any upper triangular matrix (such as a matrix in Jordan normal form) is the product of the elements on the diagonal.
Thus we have the result.
This makes me even more worried, though, since I seem to recall from my linear algebra days that the Jordan normal form theorem is a very hard result, and requires us to prove pretty much everything about matrices and determinants beforehand. Therefore, I would like to avoid using it if at all possible.
TL;DR
- It seems highly improbable to me that the Jordan normal form theorem is used to prove that the determinate of a matrix is the product of its eigenvalues rather than the other way around. Which result is normally proved first, and how?
- Is there a better definition of the multiplicity of the eigenvalues of a matrix, i.e., one that doesn't refer to Jordan normal form? It seems likely that the first step to avoiding use of the Jordan normal form theorem in the proof is avoiding reference to it in the problem statement, but I'm having trouble finding an alternate definition. For example, there doesn't seem to be an easy way of defining things in terms of the dimension of various eigenspaces when the matrix is not diagonalizable.
Best Answer
The multiplicity of an eigenvalue $\lambda$ as a root of the characteristic polynomial of $A$ is called the algebraic multiplicity of $\lambda$. A more geometric characterization of the algebraic multiplicity is the number of time $\lambda$ appears on the diagonal of some upper triangular matrix $B$ which is similar to $A$. Since the characteristic polynomial of $B$ is $p_B(x) = \prod_{i=1}^n (x- b_{ii})$ (because $xI - B$ is upper triangular) and $p_A(x) = p_B(x)$ (because $A$ and $B$ are similar) we see that the multiplicity of $\lambda$ as a root of $p_A(x)$ is the same as the number of times $\lambda$ appears on the diagonal of $B$ (or any other upper triangular matrix which is similar to $A$).
One can use the Jordan form of $A$ which requires proving that any matrix is similar to a matrix in Jordan form but one can also prove directly the much easier result that any matrix with entries in a algebraically closed field is similar to some upper triangular matrix.
Finally, when we are working over an algebraically closed field, the fact that the determinant of $A$ is the product of the eigenvalues (with each eigenvalue counted according to its algebraic multiplicity) follows immediately from the definitions. Since we can factor $p_A(x)$ into linear factors, we can write $p_A(x) = \prod_{i=1}^m (x - \lambda_i)^{a_i}$ and then $$p_A(0) = \det(xI - A)|_{x = 0} = \det(-A) = (-1)^n \det(A) = (-1)^n \prod_{i=1}^m \lambda_i^{a_i}. $$