I'm reading the proof of Cayley-Hamilton theorem and I got stuck on the proof.

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 polynomial for $T$.

(proof) Let $K$ be the commutative ring with identity consisting of all polynomials in $T$. Choose an ordered basis $\{\alpha_1, \ldots, \alpha_n\}$ for $V$, and let $A$ be the matrix which represents $T$ in the given basis. Then $$T\alpha_i = \sum_{j=1}^{n} A_{ji} \alpha_j, \quad 1\leq i \leq n.$$ These equations may be written in the equivalent form $$\sum_{j=1}^{n} (\delta_{ij} T - A_{ji}I) \alpha_j = 0, \quad 1\leq i\leq n$$ Let $B$ denote the element of $K^{n \times n}$ with entries $$B_{ij} = \delta_{ij}T - A_{ji}I.$$ When $n=2$, $\det B =f(T)$ makes sense. For the case $n>2$, it is also clear that $$\det B = f(T)$$ since $f$ is the determinant of matrix $xI - A$ whose entries are the polynomials $$(xI - A)_{ij}= \delta_{ij}x - A_{ji}$$

I got stuck on the last part of the above proof.

I think it should be $(xI - A)_{ij}= \delta_{ij}x - A_{ij}$ not $(xI - A)_{ij}= \delta_{ij}x - A_{ji}$, and also I don't understand why this verifies $\det B = f(T)$.


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