This is used in one of the many proofs for the Cayley-Hamilton theorem. My professor noted that this should be proved. However, the proof of this fact is rather straightforward, no? Is the proof I made underneath correct? I am mainly asking this question because, since I find the proof to be very straightforward, I find it odd that my professor wouldn't have added it to his notes and am therefore afraid I might be missing something.
Let $k$ be a commutative algebra with one, and let $P = (p_{ij})$ be an $n\times n$ matrix whose entries are polynomials $p_{ij}\in k[X]$.
For each $\lambda\in k$ we can form the matrix $$ P(\lambda) = \bigl(p_{ij}(\lambda)\bigr)\in M_n(k). $$ Prove that $$ \det\bigl(P(\lambda)\bigr) = \bigl(\det P\bigr)(\lambda) \qquad\text{for all }\lambda\in k. $$
Using the Leibniz formula for the determinant of the matrix $P$ with polynomial entries, $$ \det P = \sum_{\sigma\in S_n} \mathrm{sgn}(\sigma)\, p_{1,\sigma(1)}\, p_{2,\sigma(2)}\cdots p_{n,\sigma(n)} \in k[X]. $$
Evaluating at $\lambda\in k$ and using properties of product and sums of the evaluation of a polynomial (like evaluation of summation is summation of evaluations and evaluation of products is product of evaluation; here is where we are doing the "proof" I reckon) gives $$ (\det P)(\lambda) = \sum_{\sigma\in S_n} \mathrm{sgn}(\sigma)\, p_{1,\sigma(1)}(\lambda)\, p_{2,\sigma(2)}(\lambda)\cdots p_{n,\sigma(n)}(\lambda). \tag{1} $$
Now use the Leibniz formula again for the numerical matrix $$ P(\lambda)=\bigl(p_{ij}(\lambda)\bigr). $$
This yields $$ \det\bigl(P(\lambda)\bigr) = \sum_{\sigma\in S_n} \mathrm{sgn}(\sigma)\, p_{1,\sigma(1)}(\lambda)\, p_{2,\sigma(2)}(\lambda)\cdots p_{n,\sigma(n)}(\lambda). \tag{2} $$
Comparing the right–hand sides of (1) and (2), they coincide term by term. Hence $$ \det\bigl(P(\lambda)\bigr) = (\det P)(\lambda). $$
Is this "proof" (if it can even be called a proof) correct?
