definition of generalized eigenspace If $G_\lambda$ is the generalized eigenspace corresponding to the eigenvalue $\lambda$, and $E_\lambda$, the eigenspace, then why is both of the following true?

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*$G_\lambda = \ker(T-\lambda I)^{\dim V}$

*$G_\lambda = \ker(T-\lambda I)^m$
Where $m$ is the algebraic multiplicity? $\dim V$ is not necessarily equal to the algebraic multiplicity an eigenvalue?
This is really confusing since Axler's Linear Algebra Done Right book on pg. 255 defines the algebraic multiplicity $m = \dim(G_\lambda) = \dim(\ker(T-\lambda I)^{\dim V})$. On the other hand, Friedberg's Linear Algebra on pg. 486 theorem 7.2 defines $G_\lambda$ as in $2.$
Am I missing something? I know that dimension of kernel of any power of $T$ is at most $\dim V$. I may have gone as far as thinking it has something to do with the restriction of $T$ to $G_\lambda$.

Both 1 and 2 in the original question are true, and the two definitions of the generalized eigenspaces given by 1 and 2 lead to the same subspace. The two definitions arise from the two different approaches taken by the two books mentioned in the question.
Suppose $T$ is a linear operator on a finite-dimensional vector space $V$. In Linear Algebra Done Right, the generalized eigenspace $G_\lambda$ corresponding to an eigenvalue $\lambda$ is defined to be the set of vectors $v \in V$ such that $(T - \lambda I)^j v= 0$ for some positive integer $j$. The (algebraic) multiplicity of $\lambda$ as an eigenvalue of $T$ is then defined to be $\dim G_\lambda$. This definition is equivalent to the usual definition but it is much more geometric, defining the (algebraic) multiplicity as the dimension of a natural subspace instead of defining the (algebraic) multiplicity to be the multiplicity of $\lambda$ as the root of a mysterious polynomial.
