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

Consider the following definitions and proved theorems

Definitions

$1$. The field $\Bbb{F}$ is $\Bbb{R}$ or $\Bbb{C}$.

$2$. $V$ is a vector space over $\Bbb{F}$.

$3$. $\mathcal{L}(V)$ is the vector space of all operators on $V$ (that is, linear maps $V \to V$) and $T \in \mathcal{L}(V)$.

$4$. $T|_{U}$ is the restriction of $T$ to the invariant subspace $U$.

$5$. $G(\lambda_i,T)$ is the generalized eigen-space of $T$ corresponding to eigen-value $\lambda_i$. The multiplicity of $\lambda_i$ is defined to be $d_i=\dim G(\lambda_i,T)$.

$6$. If $\Bbb{F}=\Bbb{C}$ then $\lambda_1,\cdots,\lambda_m$ are distinct eigen-values of $T$. If $\Bbb{F}=\Bbb{R}$ then $\lambda_1,\cdots,\lambda_m$ are distinct eigen-values of the complexification of $T$, denoted by $T_{\Bbb{C}}$. Then $\det T = \lambda_1^{d_1}\cdots\lambda_m^{d_m}$ where each $d_i$ is the multiplicity of $\lambda_i$.

Proved Theorems

$1$. Suppose $V$ is a vector space over $\Bbb{C}$ and $T \in \mathcal{L}(V)$. Let $\lambda_1,\cdots,\lambda_m$ be distinct eigen values of $T$. Then $V=G(\lambda_1,T) \oplus \cdots \oplus G(\lambda_m,T)$ where each $G(\lambda_i,T)$ is invariant under $T$ . Furthermore, $\dim V = d_1+\cdots+d_m$ where $d_i$ is the multiplicity of $\lambda_i$.

Now, I want to prove the following theorem just using the above tools. Any hint or help is appreciated.

Question

If $T \in \mathcal{L}(V)$ and $V=V_1 \oplus \cdots \oplus V_M$ with each $V_j,\,j=1,\cdots,M$ invariant under $T$ then $\det T = \det T|_{V_1} \cdots \det T|_{V_M}$.

This problem happened in example $10.28$ of the book Linear Algebra Done Right. There was just one sentence for the proof of this in the text.

Because the dimensions of generalized eigen-spaces in $V_j$ add up to $\dim V$.

but I don't understand it!

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

• (+1) Your proof is more elegant than mine. Sep 4, 2016 at 8:04
• However, I'd like to see how you proof the part about multiplicities. Sep 4, 2016 at 8:04
• @FeiLi: Thanks. Do you mean, how to prove $(2)$? :) Because, taking $\dim$ from both sides of $(2)$, states the thing about multiplicities! :) Sep 4, 2016 at 9:44
• @FeiLi: Now, do you agree with me that Sheldon is a little LAZY in writing things down? :D Sep 4, 2016 at 9:48
• Indeed, this can be seen as a gap in the book. Sep 4, 2016 at 10:04

### 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 $\det T$.$\blacksquare$

• @H.R. You are welcome…... Sep 3, 2016 at 13:25
• I am having problem to understand the first equality in $(9)$. Can you add some more explanation on that? :) I think we should prove that $\lambda_1^1,\ldots,\lambda_{n_1}^1,\lambda_1^2,\ldots,\lambda_{n_M}^2,\cdots,\lambda_1^M,\ldots,\lambda_{n_M}^M$ are all the distinct eigen-values of $T$. Also, I think we should prove something about the multiplicities. Sep 3, 2016 at 14:51
• For proving $G(\lambda_j^i,T|_{V_i})=G(\lambda_j^i,T)$, we need more argument than just $V_i$ being invariant under $T$, I think. :) Sep 3, 2016 at 15:00
• I feel something is wrong with your proof as you go from $(7)$ to $(8)$. Take a look at my answer below. :) Sep 3, 2016 at 16:19
• Indeed, the above eigenvalues are not necessarily distinct. I've corrected my error. Sep 4, 2016 at 8:03