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

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?

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 $$\dim W_j\le \dim \ker(A-c_jI)^{r_j}=d_j.$$

• I do not know about Minimal polynomials ,can you give me an explanation that does not require any concept involving minimal polynomial. Commented Dec 24, 2019 at 12:15
• I don't have one in mind at the moment, but it's not hard to understand this notion: by Hamilton-Cayley, an endomorphism (or its matrix) is a root of its characteristic polynomial, so it is a root of a polynomial of minimum degree (just as in the case of algebraic numbers). This polynomial is a generator of the ideal $\{p\in K[X]\mid p(A)=0\}$. Is it clear? Commented Dec 24, 2019 at 12:21
• What is ideal?I am not aquainted to ring theory.......... Commented Dec 24, 2019 at 12:24
• A subgroup for addition, stable upon multiplication by elements of the fing. A principal ideal is simply the set of multiples of an element. It happens that $\mathbf Z$ and $K[X]$ (K being a field) are rings with all ideals being principal (so-called PIDs) for the simple reason that both have a Euclidean division. Commented Dec 24, 2019 at 12:28
• I have not yet studied Rings. Commented Dec 24, 2019 at 12:34