Let $T$ be a linear operator on a finite dimensional vector space $V$. If $f$ is the characteristic polynomial for $T$, then $f(T)=0$; in other words, the minimal polynomial divides the characteristic polynowlial for $T$.
Proof: Let $K=\{f(T)\in L(V)\mid f\in F[x]\}$ is an commutative ring with identity. Of course, $K$ is actually a commutative linear algebra with identity over the scalar field. Choose an ordered basis $\{\alpha_l,…,\alpha_n\}$ for $V$, and let $A$ be the matrix which represents $T$ in the given basis. Then $$ T(\alpha_i)=\sum_{j=1}^n A_{ji}\alpha_j,\ \forall 1\leq i\leq n$$ These equations may be written in the equivalent form $$\sum_{j=1}^n(\delta_{ij}T-A_{ji}I)\alpha_j=0,\ \forall 1\leq i\leq n$$ Let $B$ denote the element of $M_{n\times n}(K)$ with entries $B_{ij}=\delta_{ij}T-A_{ji}I$. When $n=2$ $$B=\begin{bmatrix} T-A_{11}I & -A_{21}I \\ -A_{12}I & T-A_{22}I\\ \end{bmatrix}$$ and $$\begin{align} \det (B) &= T^2-(A_{11}+A_{22})T+(A_{11}A_{22}-A_{12}A_{21})I \\ &= T^2-\text{tr}(A)T+\det (A)I\\ &= f(T) \end{align}$$ where $f$ is the characteristic polynomial. For $n\gt 2$, it is also clear that $\det (B)=f(T)$, since $f$ is the determinant of the matrix $xI-A$ whose entries are the polynomials $$(xI-A)_{ij}=\delta_{ij}x-A_{ji}.$$ We wish to show that $f(T)=0$. In order that $f(T)$ be the zero operator, it is necessary and sufficient that $(\det B)\alpha_k=0$ for $k=1,…,n$. By the definition of $B$, the vectors $\alpha_1,…,\alpha_n$ satisfy the equations $$\sum_{j=1}^n B_{ij}\alpha_j=0,\ \forall 1\leq i\leq n \tag{1}\label{1}$$ When $n=2$, it is suggestive to write $\eqref{1}$ in the form $$\begin{bmatrix} T-A_{11}I & -A_{21}I \\ -A_{12}I & T-A_{22}I\\ \end{bmatrix} \begin{bmatrix} \alpha_1\\ \alpha_2\\ \end{bmatrix}= \begin{bmatrix} 0\\ 0\\ \end{bmatrix}$$ In this case, the classical adjoint, adj $B$ is the matrix $$\bar{B}= \begin{bmatrix} T-A_{22}I & A_{21}I \\ A_{12}I & T-A_{11}I\\ \end{bmatrix} $$ and $\bar{B}B=\begin{bmatrix}\det (B)& 0\\ 0& \det (B)\\ \end{bmatrix} =\det (B)\cdot I_2$. Hence, we have $$(\det (B)\cdot I_2)\begin{bmatrix}\alpha_1\\ \alpha_2 \\ \end{bmatrix}=(\bar{B}B) \begin{bmatrix}\alpha_1\\ \alpha_2 \\ \end{bmatrix}=\bar{B}\left( B \begin{bmatrix}\alpha_1\\ \alpha_2 \\ \end{bmatrix} \right)= \begin{bmatrix}0\\ 0 \\ \end{bmatrix}$$ In the general case, let $\bar{B}=\text{adj} (B)$. Then by $\eqref{1}$ $$\sum_{j=1}^n\bar{B}_{ki}B_{ij}\alpha_j=0$$ for each pair $k$, $i$, and summing on $i$, we have $$0=\sum_{i=1}^n\sum_{j=1}^n\bar{B}_{ki}B_{ij}\alpha_j=\sum_{j=1}^n\left(\sum_{i=1}^n\bar{B}_{ki}B_{ij}\right)\alpha_j.$$ Now $\bar{B}B=\det(B)\cdot I_n$, so that $$\sum_{i=1}^n\bar{B}_{ki}B_{ij}=\delta_{kj}\det (B).$$ Therefore $$0= \sum_{j=1}^n \delta_{kj}\det (B)\alpha_j=(\det B)\alpha_k,\ \forall 1\leq k\leq n.$$
Question: (1) Since $\det (xI-A)=\det (xI-A)^t$, we have $f$ is determinant of matrix whose entries are the polynomials $[(xI-A)^t]_{ij}=(xI-A)_{ji}=\delta_{ij}x-A_{ji}$. How to “rigioursly” show $\det (B)=f(T)$?
(2) I don’t understand following step: Then by $\eqref{1}$ $\sum_{j=1}^n\bar{B}_{ki}B_{ij}\alpha_j=0$ for each pair $k$, $i$, and summing on $i$, we have $0=\sum_{i=1}^n\sum_{j=1}^n\bar{B}_{ki}B_{ij}\alpha_j=\sum_{j=1}^n\left(\sum_{i=1}^n\bar{B}_{ki}B_{ij}\right)\alpha_j.$
What if we do following instead, inspired by $n=2$ case. We know $\bar{B}B=\det (B)\cdot I_n$. So $$(\det (B)\cdot I_n)\begin{bmatrix}\alpha_1\\ \vdots \\ \alpha_n \\ \end{bmatrix}=(\bar{B}B) \begin{bmatrix}\alpha_1\\ \vdots \\ \alpha_n \\ \end{bmatrix}=\bar{B}( B \begin{bmatrix}\alpha_1\\\vdots \\ \alpha_n \\ \end{bmatrix} )=\bar{B}\begin{bmatrix} 0\\ \vdots \\ 0\\ \end{bmatrix}= \begin{bmatrix}0\\ \vdots \\ 0 \\ \end{bmatrix}.$$ Thus $$(\det (B)\cdot I_n)\begin{bmatrix}\alpha_1\\ \vdots \\ \alpha_n \\ \end{bmatrix}= \begin{bmatrix} \det (B)& & \\ &\ddots & \\ & & \det (B)\\ \end{bmatrix} \begin{bmatrix}\alpha_1\\ \vdots \\ \alpha_n \\ \end{bmatrix}= \begin{bmatrix}0\\ \vdots \\ 0 \\ \end{bmatrix}.$$ Hence $(\det B)\alpha_k=0$, $\forall 1\leq k\leq n$.