Does the algebraic multiplicity of all the eigenvalues of a matrix always add up to the dimension of the matrix?

Question

If $A$ is an $n \times n$ matrix, does the algebraic multiplicities of all the eigenvalues add up to $n$? Could you give a quick proof and some intuition behind this?

Answer 1

For algebraically closed fields (Think of the complex numbers) this is the case. Say $A$ is a $n \times n$ matrix and $\chi(x)$ is the characteristic polynomial. Then it can be written as a product of linear factors. i.e. $\chi(x)=(x-\lambda_1)(x-\lambda_2)\cdots(x-\lambda_n)$ for some $\lambda_i \in \Bbb{C}$ for $i = 1,\cdots,n$. From here, you can see that the sum of the multiplicities is $n$.

If your field is not algebraically closed, this may not be the case. For example, if we are working over the real numbers and have the matrix $\begin{bmatrix} 0&1\\-1&0\end{bmatrix}$, then its characteristic polynomial is $\lambda^2+1$ which has no roots and therefore the matrix has no eigenvalues (in $\Bbb{R})$.

Answer 2

The algebraic multiplicity of an eigenvalue is its multiplicity as a root of the characteristic polynomial, so your question is equivalent to "Do the multiplicity of the roots of the characteristic polynomial sum to its degree?" or "Does the characteristic polynomial always split?".

Since moreover, every polynomial can occur as the characteristic polynomial of some matrix, the answer to your question is yes if and only if the ground field is algebraically closed.