1
$\begingroup$

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.

$\endgroup$
6
  • 1
    $\begingroup$ How do you define characteristic polynomial without determinants? $\endgroup$
    – José Carlos Santos
    Dec 12, 2018 at 7:25
  • 1
    $\begingroup$ @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. $\endgroup$
    – Masacroso
    Dec 12, 2018 at 7:52
  • 1
    $\begingroup$ 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. $\endgroup$
    – it's a hire car baby
    Dec 12, 2018 at 10:58
  • $\begingroup$ Also, your title should be a question with a question mark at the end. $\endgroup$
    – it's a hire car baby
    Dec 12, 2018 at 11:01
  • 1
    $\begingroup$ Thanks @user334732 I've updated my question accordingly. $\endgroup$
    – Chayan Ghosh
    Dec 12, 2018 at 14:55

1 Answer 1

Reset to default
1
$\begingroup$

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 $T$.

$\endgroup$
2
  • 1
    $\begingroup$ So you are showing that p(T) is 0 and deg(p) = dim V. Does that mean p is the characteristic polynomial? I believe all it says is that p is some multiple of the minimal polynomial. The polynomial p with degree dimV and p(T) = 0 is not unique and the characteristic polynomial is just one such polynomial. Am I mistaken? $\endgroup$
    – Chayan Ghosh
    Dec 12, 2018 at 8:23
  • $\begingroup$ I think that the proof is right now $\endgroup$
    – Masacroso
    Dec 12, 2018 at 8:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .