The Determinant of an Operator Equals the Product of Determinants of its Restrictions

Motivated by the answer of Fei Li, I think the following is the main key point to prove this theorem.

$$\begin{align} V &= {\Large\oplus}_{i=1}^{m}G(\lambda_i,T) ={\Large\oplus}_{i=1}^{m}\left[{\Large\oplus}_{j=1}^{M}G(\lambda_i,T|_{V_j})\right] ={\Large\oplus}_{j=1}^{M}\left[{\Large\oplus}_{i=1}^{m}G(\lambda_i,T|_{V_j})\right] \\ &={\Large\oplus}_{j=1}^{M}V_j \end{align} \tag{1}$$

where we have used the commutativity and associativity of direct sums and the following important identity

$$G(\lambda_i,T)={\Large\oplus}_{j=1}^{M}G(\lambda_i,T|_{V_j}) \tag{2}$$

In this way, we don't need too much complicated notation too. I don't give a complete proof of $(2)$ but I will mention the things one needs to prove it. Theorems $1$ through $4$ are mentioned to conclude theorem $5$.

Theorem 1. If $T \in \mathcal{L}(V)$ and $\lambda \in \Bbb{F}$ and $V_j$ is a subspace of $V$ invariant under $T$ then $$G(\lambda,T|_{V_j}) = \{v_j \in V_j : \exists k, \, (T|_{V_j}-\lambda I|_{V_j})^{k}v_j=0\}$$

Theorem 2. If $T \in \mathcal{L}(V)$ and $\lambda \in \Bbb{F}$ and $V_j$ is a subspace of $V$ invariant under $T$ then $V_j$ is also invariant under $\lambda T$ and $$\lambda (T |_{V_j})=(\lambda T) |_{V_j}$$

Theorem 3. If $T_p \in \mathcal{L}(V), \, p=1,2,\cdots,n$ and $V_j$ is a subspace of $V$ invariant under each $T_p, \, p=1,2,\cdots,n$ then $V_j$ is also invariant under $\sum_{p=1}^{n}T_p$ and $$\sum_{p=1}^{n}T_p|_{V_j}=(\sum_{p=1}^{n}T_p)|_{V_j}$$

Theorem 4. If $T_p \in \mathcal{L}(V), \, p=1,2,\cdots,n$ and $V_j$ is a subspace of $V$ invariant under each $T_p, \, p=1,2,\cdots,n$ then $V_j$ is also invariant under ${\LARGE\circ}_{p=1}^{n}T_p$ and $${\LARGE\circ}_{p=1}^{n}T_p|_{V_j}=({\LARGE\circ}_{p=1}^{n}T_p)|_{V_j}$$

Theorem 5. If $T \in \mathcal{L}(V)$ and $\lambda \in \Bbb{F}$ and $V_j$ is a subspace of $V$ invariant under $T$ then $V_j$ is also invariant under $(T-\lambda I)^{k}$ and $$G(\lambda,T|_{V_j}) = \{v_j \in V_j : \exists k, \, (T-\lambda I)^{k}|_{V_j}v_j=0\}$$

Theorem 6. If $T \in \mathcal{L}(V)$ and $\lambda \in \Bbb{F}$ and $V={\Large\oplus}_{j=1}^{M}V_j$ with each $V_j,\,j=1,\cdots,M$ invariant under $T$ then $\sum_{j=1}^{M}G(\lambda,T|_{V_j})$ is a direct sum.

Now, knowing the previous theorems, it is easy to prove what we wanted

Theorem 7. If $T \in \mathcal{L}(V)$ and $\lambda \in \Bbb{F}$ and $V={\Large\oplus}_{j=1}^{M}V_j$ with each $V_j,\,j=1,\cdots,M$ invariant under $T$ then $$G(\lambda,T)={\Large\oplus}_{j=1}^{M}G(\lambda,T|_{V_j})$$

Sketch of Proof

According to the proved theorem, for $V_1$ and $T|_{V_1}\in\mathcal{L}(V_1)$, we have

$$V_1=G(\lambda_1^1,T|_{V_1})\oplus\cdots\oplus G(\lambda_{n_1}^1,T|_{V_1}),\tag{1}$$

where the "$1$" in the upper-right of $\lambda$ indicates $V_1$. Thus $\lambda_1^1,\ldots,\lambda_{n_1}^1$ are distinct eigenvalues of $T|_{V_1}$. According to definition $6$, $$\det(T|_{V_1})=(\lambda_1^1)^{d_1^1}\cdots(\lambda_{n_1}^1)^{d_{n_1}^1},\tag{2}$$ where $d_1^1,\ldots,d_{n_1}^1$ are multiplicities of the corresponding eigenvalues.

Similarly, for $V_2$ and $T|_{V_2}\in\mathcal{L}(V_2)$, we have

$$V_2=G(\lambda_1^2,T|_{V_2})\oplus\cdots\oplus G(\lambda_{n_2}^2,T|_{V_2}).\tag{3}$$

$\lambda_1^2,\ldots,\lambda_{n_2}^2$ are distinct eigenvalues of $T|_{V_2}$. According to definition $6$, $$\det(T|_{V_2})=(\lambda_1^2)^{d_1^2}\cdots(\lambda_{n_2}^2)^{d_{n_2}^2},\tag{4}$$ where $d_1^2,\ldots,d_{n_2}^2$ are again multiplicities of the corresponding eigenvalues.

$\vdots$

Continuing this way, we have

$$V_M=G(\lambda_1^M,T|_{V_M})\oplus\cdots\oplus G(\lambda_{n_M}^M,T|_{V_M}).\tag{5}$$

$\lambda_1^M,\ldots,\lambda_{n_M}^M$ are distinct eigenvalues of $T|_{V_M}$, and $$\det(T|_{V_M})=(\lambda_1^M)^{d_1^M}\cdots(\lambda_{n_M}^M)^{d_{n_M}^M},\tag{6}$$ where $d_1^M,\ldots,d_{n_M}^M$ are multiplicities of the corresponding eigenvalues.

Now,

$\begin{matrix}(\det T|_{V_1})(\det T|_{V_2})\cdots(\det T|_{V_M})\\ =\left[(\lambda_1^1)^{d_1^1}\cdots(\lambda_{n_1}^1)^{d_{n_1}^1}\right]\left[(\lambda_1^2)^{d_1^2}\cdots(\lambda_{n_2}^2)^{d_{n_2}^2}\right]\cdots \left[(\lambda_1^M)^{d_1^M}\cdots(\lambda_{n_M}^M)^{d_{n_M}^M}\right] .\end{matrix}\tag{7}$

Since $V$ has eigenvalues $\{\lambda_1,\ldots,\lambda_m\}$, the above eigenvalues are all from the set $\{\lambda_1,\ldots,\lambda_m\}$. So we can pick up all $\lambda_1$ among $(7)$ and group them together, obtaining $\lambda^{d_1}$. Note that $d_1$ is precisely the multiplicity of $\lambda_1$ in $V$, no more, no less. Similarly, after picking out $\lambda_2,\ldots,\lambda_m$ and group them together, the above equation is equal to $(\lambda_1)^{d_1}(\lambda_2)^{d_2}\cdots(\lambda_m)^{d_m}$, which is $\det T$.$\blacksquare$