To prove that the generalized eigenspace is indeed a linear subspace of $V$, we first notice that $\vec{0}$ belongs to $E_\lambda^*$. Next, if $\vec{v}$ lies in $E_\lambda^*$ then according to the definition there is an integer $k$ that satisfies $(L-\lambda\, I_V)^k(\vec{v}) = \vec{0}$. If $\alpha$ is a scalar, then we have $$(L-\lambda\, I_V)^k(\alpha \vec{v}) = \alpha (L-\lambda\, I_V)^k(\vec{v}) =\alpha\,\vec{0} = \vec{0}$$ showing that $\alpha\, \vec{v}$ belongs to $E_\lambda^*$. Finally, let $\vec{w}$ also be a vector in $E_\lambda^*$. Then, there must be a natural number $\ell$ satisfying $(L-\lambda\, I_V)^\ell(\vec{w}) = \vec{0}$. We now have for $m=\max(k,\ell)$ $$\begin{array}{rcl}(L-\lambda\, I_V)^m(\vec{v}+\vec{w})& =& (L-\lambda\, I_V)^m(\vec{v})+(L-\lambda\, I_V)^m(\vec{w})\\&&\phantom{xxx}\color{blue}{(L-\lambda\, I_V)^m\text{ is a linear map}}\\& =& (L-\lambda\, I_V)^{m-k}(L-\lambda\, I_V)^{k}(\vec{v})+(L-\lambda\, I_V)^{m-\ell}(L-\lambda\, I_V)^{\ell}(\vec{w})\\&&\phantom{xxx}\color{blue}{\text{composition of linear maps rewritten}}\\ & =& (L-\lambda\, I_V)^{m-k}(\vec{0})+(L-\lambda\, I_V)^{m-\ell}(\vec{0})\\&&\phantom{xxx}\color{blue}{\text{ choice of }k,\, \ell }\\ &=&\vec{0}+\vec{0} \\&&\phantom{xxx}\color{blue}{(L-\lambda\, I_V)^n\text{ is a linear map for all }n\in\mathbb{N}\text{ including }n=0}\\ &=& \vec{0}\end{array}$$ from which it follows that $\vec{v}+\vec{w}$ belongs to $E_\lambda^*$.

With this we have proven that $E_\lambda^*$ is a linear subspace of $V$.

To prove the invariance of $E_\lambda^*$ under $L$, we let $\vec{v}$ be a random vector in $E_\lambda^*$, and prove that $L(\vec{v})$ also belongs to $E_\lambda^*$. From the definition of $E_\lambda^*$ it follows that there exists a natural number $k$ such that $(L - \lambda\, I_V)^k(\vec{v}) = \vec{0}$. Hence, $\vec{v}$ belongs to $\ker{(L - \lambda\, I_V)^k}$. Because $M=(L - \lambda\, I_V)^k$ commutes with $L$, we can apply theorem Invariance of kernel and image under commuting linear maps to see that $\ker{M}$ is invariant under $L$, showing that $L(\vec{v})$ belongs to $E_\lambda^*$, which is what we needed to prove.

Let $\ell$ be the multiplicity of $\lambda$ in the minimal polynomial $m_L(x)$. From the definition of generalized eigenspace it is clear that $\ker{(L-\lambda\, I_V)^{\ell}}$ is enclosed in $E_\lambda^*$. Suppose that $\vec{w}$ is a vector in ${E_\lambda^*}$. Then there exists a natural number $m$, such that $\vec{w}$ belongs to $\ker{(L-\lambda\, I_V)^{m}}$. We will show that we can choose $m\le\ell$. Assume $m\gt\ell$. Because $\ell$ is the multiplicity of $\lambda$ in $m_L(x)$, we can factor the minimal polynomial as

$$m_L(x) = (x-\lambda)^{\ell}\cdot c(x)$$ where $c(x)$ is a polynomial with $c(\lambda)\ne0$. In particular we have

$$\gcd\left((x-\lambda)^m,m_L(x)\right) = (x-\lambda)^{\ell}$$ The extended Euclidean algorithm gives polynomials $a(x)$ and $b(x)$ that satisfy

$$a(x)\cdot (x-\lambda)^m+b(x)\cdot m_L(x) = (x-\lambda)^{\ell}$$ If we substitute $L$ in all polynomials of this equality, we find, thanks to the fact that $m_L(L)=0$,

$$a(L)\, (L-\lambda\,I_V)^m = (L-\lambda\,I_V)^{\ell}$$

such that from $\vec{w}\in \ker{(L-\lambda\, I_V)^m}$ follows $$(L-\lambda\,I_V)^{\ell}(\vec{w}) = a(L)\left( (L-\lambda\,I_V)^m(\vec{w}) \right) = a(L)(\vec{0}) = \vec{0}$$ showing that $\vec{w}$ belongs to $\ker{(L-\lambda\, I_V)^{\ell}}$, such that $\ker{(L-\lambda\, I_V)^{\ell}}$ coincides with ${E_\lambda^*}$. This proves the first statement.

Since $k\ge\ell$ the equality $E_\lambda^* =\ker{(L-\lambda\, I_V)^k}$ in the second statement follows directly from the first. We will also show that $\dim{E_\lambda^*}= k$. Reasoning for $k$ and $p_L(x)$ along the same lines as above for $\ell$ and $m_L(x)$ we find a polynomial $d(x)$ with $d(\lambda)\ne0$, in such a way that $$p_L(x) = (x-\lambda)^k\cdot d(x)$$ According to Invariant direct sum we have the following direct sum decomposition of $V$ into subspaces that are invariant under $L$: $$V = \ker{(L-\lambda\,I_V)^k}\oplus \ker{d(L)}$$ We have already seen that the first summand is equal to $E_\lambda^*$. According to the theorem determinants of some special matrices the characteristic polynomial $p_L(x)$ is the product of the characteristic polynomials of $L$ restricted to each summand. The characteristic polynomial of $L$ restricted to $E_\lambda^*$ is of the form $(x-\lambda)^t$, where $t=\dim{E_\lambda^*}$, because the minimal polynomial is a divisor of $(x-\lambda)^k$. On the other hand $x-\lambda$ is no divisor of the characteristic polynomial of $L$ restricted to $\ker{d(L)}$, since the corresponding minimal polynomial divides $d(x)$ and $d(\lambda)\ne0$. We conclude that $k=t$, such that $k=\dim{E_\lambda^*}$.

Concerning the third statement: it is obvious that $e_i\le e_{i+1}$ since $\ker{(L-\lambda\,I_V)^i}$ is enclosed in $\ker{(L-\lambda\,I_V)^{i+1}}$. To prove that $e_i\lt e_{i+1}$ if $i\lt\ell$, we assume that $e_i= e_{i+1}$. We claim that from this follows that $e_i=e_{i+m}$ for each natural number $m$. For $m=1$ this follows from that assumption. We use full induction to prove this for all $m$. For that use $m\gt 1$ and assume that $e_i=e_{i+m-1}$ (this is the induction hypothesis). If $\vec{v}$ lies in $\ker{(L-\lambda\,I_V)^{i+m}}$, then $$(L-\lambda\,I_V)^{i+m-1}\left((L-\lambda\,I_V)\vec{v}\right)=(L-\lambda\,I_V)^{i+m}(\vec{v})=\vec{0}$$ so that $\left(L-\lambda\,I_V\right)\vec{v}$ lies in $\ker{(L-\lambda\,I_V)^{i+m-1}}$. Because $e_i=e_{i+m-1}$ and $\ker{(L-\lambda\,I_V)^i}$ is included in $\ker{(L-\lambda\,I_V)^{i+m-1}}$, we have $\ker{(L-\lambda\,I_V)^i}=\ker{(L-\lambda\,I_V)^{i+m-1}}$. This means that $\left(L-\lambda\,I_V\right)\vec{v}$ lies in $\ker{\left(L-\lambda\,I_V\right)^i}$, so that $\vec{v}$ lies in $\ker{\left(L-\lambda\,I_V\right)^{i+1}}$. But $\ker{\left(L-\lambda\,I_V\right)^{i+1}}=\ker{\left(L-\lambda\,I_V\right)^i}$, hence $\vec{v}$ even lies in $\ker{\left(L-\lambda\,I_V\right)^i}$. Thus, we have deduced that $\ker{(L-\lambda\,I_V)^{i+m}}=\ker{\left(L-\lambda\,I_V\right)^i}$, hence, $e_{i+m}=e_i$.

From the statement we have just proven it follows that if $e_i=e_{i+1}$, also $e_m=e_i$ for all $m\gt i$, so $e_i=e_\ell=\dim{E^*_\lambda}$. Since $\ell$ is the smallest natural number that satisfies $e_\ell=\dim{E^*_\lambda}$, we see that $e_i=e_{i+1}$ only holds for $i\geq\ell$ so that $e_1\lt e_2\lt \cdots\lt e_{\ell-1}\lt e_\ell$, which concludes the proof of the theorem.