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In the textbook that I am reading, the following is how a polynomial is applied to a linear map.

Definition: For a polynomial $p(z)=a_0+a_1z+\dots+a_nz^n$ and a linear map $T\in L(V, V)$, we define the polynomial $p(T)$ to be the polynomial $p(T)=a_0I+a_1T+\dots+a_nT^n$.

My question is: Can results regarding a polynomial $p(z)$ be interpreted to be the same as for the polynomial $p(T)$? For example, suppose that a polynomial $p(z)=a_0+a_1z+\dots+a_nz^n$ over $\Bbb{C}$ has a factorisation $$p(z)=c(z-\lambda_1)\cdots(z-\lambda_m)$$

Does the polynomial $p(T)$ also have a factorisation that mimics this factorisation? Something like $$p(T)=c(T-\lambda_1I)\cdots(T-\lambda_mI)$$

I should also add that polynomials were defined to be functions $p : F\to F$ where $F$ denotes the field of real or complex numbers.

Seeker
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  • https://math.stackexchange.com/questions/1165539/factoring-a-matrix-polynomial Does this help? – jcneek Sep 14 '22 at 22:01
  • @jcneek It doesn't answer my question. Even the answer to that question says that they don't know how to answer the second question, which is my example in this question. Can a factorisation of a polynomial be considered as a factorisation for the polynomial of the linear map. – Seeker Sep 14 '22 at 22:05
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    https://math.stackexchange.com/questions/1882069/why-does-polynomial-factorization-generalize-to-matrices – jcneek Sep 14 '22 at 22:10
  • @jcneek Thanks, didn't think about multiplying the factored form. – Seeker Sep 14 '22 at 22:18

2 Answers2

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If the operators $T\in L(V,V)$ are linear, then yes, the factorization is valid. That's because you can just multiply out $c(T-\lambda_1 I)\cdots(T-\lambda_m I)$ and you get the same polynomial $p(T)$ as before based on how $c$ and the $\lambda_i$ combine to form the coefficients $a_k$.

As for the definition: You say that $p:F\rightarrow F$ is how the polynomial is defined. What we are doing when we are writing $p(T)$ is that we are saying "Well, we know what that polynomial is based on $F$. Let us extend it to other kinds of objects by defining $p(T)$ as a polynomial on operators". It is, strictly speaking, not the same function as the original polynomial because it has a different domain. Rather, given a polynomial $p=p_F$, we have defined a different function $p_{L(V,V)}$ by using the same polynomial coefficients as $p_F$ has. We just find it convenient to use the same symbol, even though it really is a different function.

Wolfgang Bangerth
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Let $(V,F,+,\cdot)$ be a vector space over $F$. $(V,F,+,\cdot,\times)$ is linear algebra over $F$ if $\times:V\times V\to V$ have following properties:

$(1)$ $\alpha \times (\beta \times \gamma)$ $=(\alpha \times \beta)\times \gamma$, for all $\alpha, \beta, \gamma\in V$

$(2)$ $\alpha \times (\beta+\gamma)$ $=\alpha \times \beta + \alpha \times \gamma$ and $(\alpha +\beta)\times \gamma$ $=\alpha \times \gamma +\beta \times \gamma$, for all $\alpha, \beta, \gamma\in V$

(3) $c\cdot (\alpha \times \beta)$ $=(c\cdot \alpha)\times \beta$ $=\alpha \times (c\cdot \beta)$, for all $c\in F$.

If $\exists 1_V\in V$ such that $1_V \times \alpha$ $=\alpha \times 1_V$ $=\alpha$, for all $\alpha \in V$, then we say $(V,F,+,\cdot, \times)$ is linear algebra with identity over $F$. If $\alpha \times \beta$ $=\beta \times \alpha$, for all $\alpha ,\beta \in V$, then we say $V$ is commutative.

Definition: Let $V$ be a linear algebra with identity over $F$. Suppose $\alpha ^0=1_V$, $\forall \alpha\in V$. Define $F[x]\times V\to V$ such that $(p,\alpha)$ $=p(\alpha)$ $=\sum_{i=0}^na_i \cdot \alpha^i\in V$.

It’s easy to check $(L(V,V),F,+,\cdot ,\circ)$ is linear algebra with identity over $F$. By above definition $p(T)\in L(V,V)$. I don’t think it is accurate to say $p(T)$ is a “polynomial”.

user264745
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