Is the sum of the algebraic multiplicities of every eigenvalue of a matrix equal to the dimension of the matrix?

Question

If this is true, that means if the sum of the geometric multiplicities of all the eigenvalues are strictly less than the sum of the algebraic multiplicities, then the matrix is not diagonalizable, right?

Answer 1

The algebraic multiplicity of an eigenvalue is by definition its multiplicity as root of the characteristic polynomial$~\chi$, and of course all roots of$~\chi$ are eigenvalues. So your question amounts to whether the sum of the multiplicities of all roots of$~\chi$ equals$~\deg(\chi)$ (which is the size of the matrix). This is true if and only if $\chi$ (which is always monic) is the product over its roots $r_i$ (repeated in case of multiplicity) of factors $X-r_i$, in other words if $\chi$ is split. This depends on which field you allow the roots (eigenvalues) to be in; every monic polynomial splits over a sufficiently large field (for instance over an algebraically closed field like$~\Bbb C$). So if (like is most common) you allow for complex eigenvalues, the answer is yes.