What is the relation between dimension of eigenspace corresponding to an eigenvalue and the multiplicity of the eigenvalue in characteristic equation?

Question

Suppose the Characteristic equation of a linear operator splits, i.e. $\operatorname{f}=\prod_{j=1}^{k}(x-c_j)^{d_j}$, where $c_j$ s are distinct eigenvalues of $T$. Now suppose $W_j$ is the eigenspace of the eigenvalue $c_j$. In general, is there any relation between $\dim W_j$ and $d_j$? I am having an intuition that $\dim W_j \leq d_j$. Is this correct and if it is, then why?

Answer 1

It is perfectly correct, and the equality of dimensions, $\dim W_j$ (the so-called geometric multiplicity of the eigenvalue $c_j$) and $d_j$ (its algebraic multiplicity) for all $j$ is criterion for the diagonalisability of the matrix associated to the linear operator.

One reason for this to hold is that, denoting $A$ the associated matrix in some basis , for each $j$, we have a sequence of subspaces which ultimately stabilises:

$$\{0\}\subset \underset{\textstyle=\:W_j\strut}{\ker(A-c_jI)}\subset \ker(A-c_jI)^2\subset\dots\subset\ker(A-c_jI)^{r_j}=\ker(A-c_jI)^{r_j+1}=\dotsm$$ The value $r_j$ is the multiplicity of the eigenvalue $c_j$ in the minimal polynomial of the endomorphism, which is a divisor of its characteristic polynomial.

Now one $\ker(A-c_jI)^{r_j}$ is the characteristic subspace for the eigenvalue $c_i$ and one shows its dimension is the algebraic multiplicity $d_j$ of $c_j$, so $$\dim W_j\le \dim \ker(A-c_jI)^{r_j}=d_j.$$