# How to represent characteristic polynomial in terms of those of invariant direct sum subspaces?

Suppose $$V$$ is a complex vector space and $$V_1,...,V_m$$ are nonzero subspaces of $$V$$ such that $$V = V_1 \oplus ... \oplus V_m$$. Suppose $$T \in \mathcal{L}(V)$$ and each $$V_j$$ is invariant under $$T$$. For each $$j$$, let $$p_j$$ denote the characteristic polynomial of $$T|_{V_j}$$. Prove that the characteristic polynomial of $$T$$ equals $$p_1...p_m$$.

The above question is from Linear Algebra Done Right 3rd ed. Chapter 8C problem 20.

I found and understood the proof from here. However since the book above I am (self) studying doesn't introduce determinant till the last chapter, I would be very grateful if someone could help me with a determinant free proof.

Note:
In the book the characteristic polynomial is defined as the product $$(z−\lambda_1)^{m_1}⋯(z−\lambda_n)^{m_n}$$ for some $$T \in \mathcal{L}(V)$$ where $$V$$ is a finite-dimensional complex vector space, where $$\lambda_k$$ are the distinct eigenvalues of $$T$$ and $$m_k$$ it algebraic multiplicities.

• How do you define characteristic polynomial without determinants? Dec 12, 2018 at 7:25
• @José in the book the characteristic polynomial is defined as the product $$(z-\lambda_1)^{m_1}(z-\lambda_2)^{m_2}\cdots(z-\lambda_n)^{m_n}$$ for some $T\in\mathcal L(V)$ where $V$ is a finite-dimensional complex vector space, where $\lambda_k$ are the distinct eigenvalues of $T$ and $m_k$ it multiplicities. Determinants are not used in the book except in the last chapter. And for real-valued vector spaces the characteristic polynomial is defined via complexification. Dec 12, 2018 at 7:52
• Try this: math.meta.stackexchange.com/questions/5020 people love it when you do that as the maths is (to some degree) searchable and the image cannot vanish later. This might be why you've received close votes. Dec 12, 2018 at 10:58
• Also, your title should be a question with a question mark at the end. Dec 12, 2018 at 11:01
• Thanks @user334732 I've updated my question accordingly. Dec 12, 2018 at 14:55

From the decomposition $$V=\bigoplus_{j=1}^m V_j\tag1$$ define a basis $$B_j$$ for each $$V_j$$. Then $$B:=\bigcup_{j=1}^m B_j$$ is a basis of $$V$$.
Because $$p_j$$ is the characteristic polynomial of $$T|_{V_j}$$ and $$V$$ is complex then $$\deg p_j=\dim V_j$$, thus $$\deg (\prod_{j=1}^m p_j)=\dim V$$.
Now if $$p_j(T)v=0$$ then clearly $$p(T)v=0$$, where we define $$p:=\prod_{j=1}^m p_j$$. Then is easy to see that for each $$v_k\in B$$ we have that $$p(T)v_k=0$$, hence $$p(T)=0$$.
Now suppose that $$p$$ is not the characteristic polynomial of $$T$$. Then it must be a multiple of the minimal polynomial, and because $$\deg p=\dim V$$ this would imply that there is some factor $$(z-\lambda)^r$$ in $$p$$ where $$r$$ is bigger than the multiplicity of $$\lambda$$, but this would imply that there are $$T|_{V_j}$$ and $$T|_{V_k}$$ that share some generalized eigenvector of $$\lambda$$. But this is impossible because $$T$$ is invariant in each $$V_k$$, what means that if $$v\in V_k$$ then $$(w I-T)^nv\in V_k$$ for all $$n\in\Bbb N$$ and any chosen $$w\in\Bbb C$$.
Hence $$p$$ is the characteristic polynomial of $$T$$.