In the proof presented in my textbook, it utilises the equation
$$(A-\lambda I) \operatorname{Adj}(A - \lambda I) = \det(A-\lambda I) \, I$$
And it stated that,
$$P(A-\lambda I) = \det(A-\lambda I)= a_0 + a_1\lambda + ... + a_n\lambda^n$$
and that
$$Q(A-\lambda I) = \operatorname{Adj}((A-\lambda I) = b_0 + b_1\lambda + ... + b_k\lambda^k$$
I have no problem understanding that $P(A-\lambda I)$ is the characteristic polynomial for matrix $A$. However, I have problem understanding 2 things,
Why $\operatorname{Adj}(A-\lambda I)$ can be written into a polynomial?
What does it mean to then multiply the above polynomial to matrix $(A-\lambda I)$?
Many thanks!
I hope it's not too much trouble for me to talk in terms of endomorphisms rather than square matrices.
If we want to evaluate the polynomial $\text{det}(\phi - t I)$ at a matrix, it is natural to view it as an element of $\text{End}_{\mathbb{F}} (V) [t]$. The factorization $\text{det}(t I - \phi ) = \text{adj}(tI - \phi ) (tI - \phi)$ makes sense if we allow ourselves to think of $\text{adj}(tI - \phi)$ as a polynomial with endomorphisms as entries.
After we think of these as elements in $\text{End}_{\mathbb{F}} (V)[t]$, the product $\text{adj}(A - \lambda I) ( A - \lambda I)$ works just like any polynomial ring over a noncommutative ring:
$$\sum_{i = 1}^n \phi_i t^i \sum_{j = 1}^m \psi_j t^j = \sum_{i = 1}^{n + m} \sum_{j + k = i, 1 \leq j \leq n, 1 \leq k \leq m} \phi_j \circ \psi_k t^{i}$$
In fact, this language is convenient for the proof: we have an isomorphism $$\text{End}_{\mathbb{F}} (V) [t] \cong \text{End}_{\mathbb{F}} (V \otimes_{\mathbb{F}} \mathbb{F}[t])$$ In $\text{End}_{\mathbb{F}} (V \otimes_{\mathbb{F}} \mathbb{F}[t])$, we have a factorization $$ \text{det}(\mu_t - \phi) 1_V = \text{adj}(\mu_t - \phi) (\mu_t - \phi) $$ Where $\mu_t$ is multiplication by $t$. Under the isomorphism, this becomes a factorization $$ p(t) = f(t)(t - \phi) $$ where $p(t) \in \text{End}_{\mathbb{F}} (V)[t]$ is the characteristic polynomial.