I'm reading about linear algebra and I came across with the following theorem where I have a problem convincing myself:
Theorem 2.1 $\,$ Every linear operator on a finite-dimensional complex vector space has an eigenvalue.
Proof:
To show that $T$ (our linear operator on $V$) has an
eigenvalue, fix any nonzero vector $v \in V$. The vectors $v, Tv,
T^2v,…, T^nv$ cannot be linearly independent, because $V$ has
dimension $n$ and we have $n + 1$ vectors. Thus there exist complex
numbers $a_0,…,a_n$, not all $0$, such that$$a_0v + a_1Tv + ··· + a_nT^nv = 0.$$ Make the $a$’s the coefficients
of a polynomial, which can be written in factored form as $$a_0 + a_1z
+ ··· + a_nz^n = c(z − r_1)\cdots(z − r_m),$$ where $c$ is a non-zero complex number, each $r_j$ is complex, and the equation holds for all
complex $z$. We then have$${\color{red}{ 0=(a_0I + a_1T + ··· + a_nT^n)v= c(T − r_1I)\cdots(T −
r_mI)v}},$$ which means that $T − r_j$ I is not injective for at least
one $j$. In other words, $T$ has an eigenvalue.$\;\blacksquare$
I'm having trouble with the factorized form of the matrix polynomial (in red). I understood that the factorization holds for a polynomial by the fundamental theorem of algebra but why does this also hold for matrices?
In other words, why is the the part I highlighted true? Does such an factorization always exist?
Could I have some help to see this? Thank you =)
P.S. here is my reference (page 3).
UPDATE:
Someone else also has asked the same question before it seems.
Best Answer
What matters is that the matrices involved, namely powers of $T$, commute with each other. With that in mind, the legitimacy of the factorisation should be clear: just think about expanding the brackets using the associative and distributive properties of matrix multiplication. A more sophisticated argument can be obtained by viewing the equation in terms of an action of the polynomial ring $\mathbb{C}[x]$, in which factorisation is more familiar.