[proofplan] We compare the two characteristic polynomials from their defining determinant formula. The coefficient inclusion $k[t]\hookrightarrow K[t]$ sends the matrix $tI_n-A$ over $k[t]$ to the same formal matrix $tI_n-A$ over $K[t]$. Expanding the determinant by the permutation formula shows directly that determinants are compatible with this coefficient inclusion, so the two characteristic polynomials agree after extending coefficients. [/proofplan]

[step:Distinguish the two characteristic polynomials] Let $\iota:k\hookrightarrow K$ denote the inclusion of fields. Define the induced polynomial map \begin{align*} \iota_*:k[t]\to K[t] \end{align*} by applying $\iota$ to each coefficient. Since $\iota$ is injective, we regard $k[t]$ as a subring of $K[t]$ through $\iota_*$. Let $I_{n,k}\in M_n(k)$ be the identity matrix over $k$, and let $I_{n,K}\in M_n(K)$ be the identity matrix over $K$. By the definition of the [characteristic polynomial](/page/Characteristic%20Polynomial) over the coefficient field, we have \begin{align*} \chi_{A,k}(t)=\det_{k[t]}(tI_{n,k}-A). \end{align*} Here $tI_{n,k}-A$ is regarded as a matrix with entries in $k[t]$. Similarly, after viewing $A$ as a matrix with entries in $K$, we have \begin{align*} \chi_{A,K}(t)=\det_{K[t]}(tI_{n,K}-A). \end{align*} [/step]

[step:Show the determinant commutes with the coefficient inclusion]Let $B=(b_{ij})\in M_n(k[t])$ be any matrix. We claim that \begin{align*} \iota_*(\det_{k[t]}B)=\det_{K[t]}(\iota_*(B)), \end{align*} where $\iota_*(B)\in M_n(K[t])$ denotes the matrix whose $(i,j)$-entry is $\iota_*(b_{ij})$. Using the permutation formula for the determinant, and writing $S_n$ for the symmetric group on $\{1,\dots,n\}$, we have \begin{align*} \det_{k[t]}B=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n b_{i,\sigma(i)}. \end{align*} Since $\iota_*:k[t]\to K[t]$ is a ring homomorphism, it preserves finite sums, finite products, and the elements $1$ and $-1$. Applying $\iota_*$ to the determinant expansion gives \begin{align*} \iota_*(\det_{k[t]}B)=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n \iota_*(b_{i,\sigma(i)}). \end{align*} The right-hand side is exactly the permutation formula for $\det_{K[t]}(\iota_*(B))$. Therefore \begin{align*} \iota_*(\det_{k[t]}B)=\det_{K[t]}(\iota_*(B)). \end{align*}[/step]

[guided]The only algebraic point that must be checked is that extending coefficients does not change a determinant except by applying the same coefficient map to the final polynomial. Let $B=(b_{ij})\in M_n(k[t])$ be an arbitrary matrix. We define $\iota_*(B)\in M_n(K[t])$ by applying the coefficientwise inclusion $\iota_*:k[t]\to K[t]$ to each entry of $B$. The determinant of $B$ over $k[t]$ is defined by the permutation formula \begin{align*} \det_{k[t]}B=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n b_{i,\sigma(i)}. \end{align*} Here $S_n$ is the group of all permutations of $\{1,\dots,n\}$, and $\operatorname{sgn}(\sigma)\in\{1,-1\}$ is the sign of the permutation $\sigma$. Now apply $\iota_*$ to this formula. Because $\iota_*:k[t]\to K[t]$ is a ring homomorphism, it preserves addition and multiplication. It also sends $1$ to $1$ and $-1$ to $-1$, so the sign factor is unchanged. Hence \begin{align*} \iota_*(\det_{k[t]}B)=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n \iota_*(b_{i,\sigma(i)}). \end{align*} But this last expression is precisely the determinant expansion of the matrix $\iota_*(B)$ over the ring $K[t]$. Therefore \begin{align*} \iota_*(\det_{k[t]}B)=\det_{K[t]}(\iota_*(B)). \end{align*} This proves determinant compatibility for the coefficient inclusion $k[t]\hookrightarrow K[t]$.[/guided]

[step:Apply determinant compatibility to $tI_n-A$] Set \begin{align*} B:=tI_{n,k}-A\in M_n(k[t]). \end{align*} Applying $\iota_*$ entrywise sends $B$ to \begin{align*} \iota_*(B)=tI_{n,K}-A\in M_n(K[t]), \end{align*} because $\iota_*(t)=t$ and each entry of $A$ is sent from $k$ to the same entry viewed in $K$. Using the determinant compatibility proved above with this matrix $B$, we obtain \begin{align*} \iota_*(\chi_{A,k}(t))=\iota_*(\det_{k[t]}(tI_{n,k}-A)). \end{align*} Thus \begin{align*} \iota_*(\chi_{A,k}(t))=\det_{K[t]}(tI_{n,K}-A). \end{align*} By the definition of $\chi_{A,K}(t)$, the right-hand side is $\chi_{A,K}(t)$. Hence \begin{align*} \chi_{A,K}(t)=\iota_*(\chi_{A,k}(t)). \end{align*} This is the desired compatibility of the characteristic polynomial with [field extension](/page/Field%20Extension). [/step]