# can the dimension of the eigenspace corresponding to c of a Linear operator be less than d where d is the exponent of (x-c) in the minimal polynomial?

I know that if we are in an algebraic closed field then the characteristic polynomial of a linear operator $T=(x-c_1)^{d_1}...(x-c_n)^{d_n}$ and for each $c_i$ the corresponding eigenspace has dimension less than or equal to $d_i$. Now I wondering if the minimal polynomial associated to $T=(x-c_1)^{a_1}...(x-c_n)^{a_n}$ does it follow that the dimension of the eigenspace associated with the characteristic value $c_i$ is greater than or equal to $a_i$. I have tried a few examples and that seem to be true but I can't know for sure since I can't prove it. So is it true?

Consider $A=\left(\array{0 & 0\\1 & 0}\right)$. Its minimal polynomial $p(A)=A^2=(A-0)^2$ has degree $2$, and the eigenspace asociated to the eigenvalue $0$ has dimension $1$.