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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?

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

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  • $\begingroup$ I do not know about Minimal polynomials ,can you give me an explanation that does not require any concept involving minimal polynomial. $\endgroup$ Commented Dec 24, 2019 at 12:15
  • $\begingroup$ 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? $\endgroup$
    – Bernard
    Commented Dec 24, 2019 at 12:21
  • $\begingroup$ What is ideal?I am not aquainted to ring theory.......... $\endgroup$ Commented Dec 24, 2019 at 12:24
  • $\begingroup$ 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. $\endgroup$
    – Bernard
    Commented Dec 24, 2019 at 12:28
  • $\begingroup$ I have not yet studied Rings. $\endgroup$ Commented Dec 24, 2019 at 12:34

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