This formula can be obtained during a proof of the Cayley-Hamilton theorem, as is indicated on its Wikipedia article. The essence of the argument is that Euclidean division by a monic polynomial (on the left, say), can be performed in the polynomial ring over any (unitary) ring, not necessarily commutative; this follows directly from consideration of what Euclidean division does, or by a simple inductive argument.

Since I care about polynomials being monic, I'l define the characteristic polynomial of a matrix $A$ to be $\chi_A=\det(I_nX-A)=\sum_{i=0}^nc_iX^i$ where $c_n=1$ (and $c_0=\det(-A)$), so the result to prove becomes $\mathrm{adj}(-A)=c_1I_n+c_2A+\cdots+c_{n-1}A^{n-2}+A^{n-1}=\sum_{i=1}^nc_iA^{i-1}$

Consider the noncommutative ring $M=\mathrm{Mat}_n(R)$, and using Euclidean division in $M[X]$ (in which $R[X]$ is embedded by mapping $r$ to $rI_n$) divide $\chi_A$ on the left by $X-A$. Since we know that $(X-A)\mathrm{adj}(X-A)=\det(X-A)=\chi_A$, uniqueness of quotient and remainder in Euclidean division implies they will have to be $\mathrm{adj}(X-A)$ and $0$, respectively. Writing the quotient $\mathrm{adj}(X-A)=\sum_{i=0}^{n-1}B_iX^i$, its coefficients $B_i\in M$ are determined in the division successively as $B_{n-1}=c_n=1$ and $B_{i-1}=c_i+AB_i$ for $i=n-1,\ldots,1$ (these are just the intermediate values while computing the evaluation of $\chi_A$ at$~X=A$ using the Horner scheme), which expands to $B_{i-1}=c_iA^0+c_{i+1}A^1+\cdots+c_nA^{n-i}$. In particular the constant coefficient of the quotient $\mathrm{adj}(X-A)$ equals $B_0=\sum_{i=1}^nc_iA^{i-1}$, but this is also $\mathrm{adj}(-A)$ (by substituting $X=0$).

To retrieve the Cayley-Hamilton theorem from the formula found, multiply on the left or right by $A$ and move the left hand side to the right.

This formula can be obtained during a proof of the Cayley-Hamilton theorem, as is indicated on its Wikipedia article. The essence of the argument is that Euclidean division by a monic polynomial (on the left, say), can be performed in the polynomial ring over any (unitary) ring, not necessarily commutative; this follows directly from consideration of what Euclidean division does, or by a simple inductive argument.

Since I care about polynomials being monic, I'l define the characteristic polynomial of a matrix $A$ to be $\chi_A=\det(I_nX-A)=\sum_{i=0}^nc_iX^i$ where $c_n=1$ (and $c_0=\det(-A)$), so the result to prove becomes $\mathrm{adj}(-A)=c_1I_n+c_2A+\cdots+c_{n-1}A^{n-2}+A^{n-1}=\sum_{i=1}^nc_iA^{i-1}$

Consider the noncommutative ring $M=\mathrm{Mat}_n(R)$, and using Euclidean division in $M[X]$ (in which $R[X]$ is embedded by mapping $r$ to $rI_n$) divide $\chi_A$ on the left by $X-A$. Since we know that $(X-A)\mathrm{adj}(X-A)=\det(X-A)=\chi_A$, uniqueness of quotient and remainder in Euclidean division implies they will have to be $\mathrm{adj}(X-A)$ and $0$, respectively. Writing the quotient $\mathrm{adj}(X-A)=\sum_{i=0}^{n-1}B_iX^i$, its coefficients $B_i\in M$ are determined in the division successively as $B_{n-1}=c_n=1$ and $B_{i-1}=c_i+AB_i$ for $i=n-1,\ldots,1$ (these are just the intermediate values while computing the evaluation of $\chi_A$ at$~X=A$ using the Horner scheme), which expands to $B_{i-1}=c_iA^0+c_{i+1}A^1+\cdots+c_nA^{n-i}$. In particular the constant coefficient of the quotient $\mathrm{adj}(X-A)$ equals $B_0=\sum_{i=1}^nc_iA^{i-1}$, but this is also $\mathrm{adj}(-A)$ (by substituting $X=0$).

To retrieve the Cayley-Hamilton theorem from the formula found, multiply on the left or right by $A$ and move the left hand side to the right.

This formula can be obtained during a proof of the Cayley-Hamilton theorem, as is indicated on its Wikipedia article. The essence of the argument is that Euclidean division by a monic polynomial (on the left, say), can be performed in the polynomial ring over any (unitary) ring, not necessarily commutative; this follows directly from consideration of what Euclidean division does, or by a simple inductive argument.

Since I care about polynomials being monic, I'l define the characteristic polynomial of a matrix $A$ to be $\chi_A=\det(I_nX-A)=\sum_{i=0}^nc_iX^i$ where $c_n=1$ (and $c_0=\det(-A)$), so the result to prove becomes $\mathrm{adj}(-A)=c_1I_n+c_2A+\cdots+c_{n-1}A^{n-2}+A^{n-1}=\sum_{i=1}^nc_iA^{i-1}$

Consider the noncommutative ring $M=\mathrm{Mat}_n(R)$, and using Euclidean division in $M[X]$ (in which $R[X]$ is embedded by mapping $r$ to $rI_n$) divide $\chi_A$ on the left by $X-A$; the quotient and remainder will be $\mathrm{adj}(X-A)$ and $0$ (by uniqueness). Writing the quotient $\mathrm{adj}(X-A)=\sum_{i=0}^{n-1}B_iX^i$, its coefficients $B_i\in M$ are determined in the division successively as $B_{n-1}=c_n=1$ and $B_{i-1}=c_i+AB_i$ for $i=n-1,\ldots,1$, which expands to $B_{i-1}=c_iA^0+c_{i+1}A^1+\cdots+c_nA^{n-i}$. In particular the constant coefficient of the quotient is $B_0=\sum_{i=1}^nc_iA^{i-1}$, but this is also $\mathrm{adj}(-A)$ (by substituting $X=0$ into $\mathrm{adj}(X-A)$).

To retrieve the Cayley-Hamilton theorem from the formula found, multiply on the left or right by $A$ and move the left hand side to the right.

This formula can be obtained during a proof of the Cayley-Hamilton theorem, as is indicated on its Wikipedia article. The essence of the argument is that Euclidean division by a monic polynomial (on the left, say), can be performed in the polynomial ring over any (unitary) ring, not necessarily commutative; this follows directly from consideration of what Euclidean division does, or by a simple inductive argument.

Since I care about polynomials being monic, I'l define the characteristic polynomial of a matrix $A$ to be $\chi_A=\det(I_nX-A)=\sum_{i=0}^nc_iX^i$ where $c_n=1$ (and $c_0=\det(-A)$), so the result to prove becomes $\mathrm{adj}(-A)=c_1I_n+c_2A+\cdots+c_{n-1}A^{n-2}+A^{n-1}=\sum_{i=1}^nc_iA^{i-1}$

Consider the noncommutative ring $M=\mathrm{Mat}_n(R)$, and using Euclidean division in $M[X]$ (in which $R[X]$ is embedded by mapping $r$ to $rI_n$) divide $\chi_A$ on the left by $X-A$. Since we know that $(X-A)\mathrm{adj}(X-A)=\det(X-A)=\chi_A$, uniqueness of quotient and remainder in Euclidean division implies they will have to be $\mathrm{adj}(X-A)$ and $0$, respectively. Writing the quotient $\mathrm{adj}(X-A)=\sum_{i=0}^{n-1}B_iX^i$, its coefficients $B_i\in M$ are determined in the division successively as $B_{n-1}=c_n=1$ and $B_{i-1}=c_i+AB_i$ for $i=n-1,\ldots,1$ (these are just the intermediate values while computing the evaluation of $\chi_A$ at$~X=A$ using the Horner scheme), which expands to $B_{i-1}=c_iA^0+c_{i+1}A^1+\cdots+c_nA^{n-i}$. In particular the constant coefficient of the quotient $\mathrm{adj}(X-A)$ equals $B_0=\sum_{i=1}^nc_iA^{i-1}$, but this is also $\mathrm{adj}(-A)$ (by substituting $X=0$).

To retrieve the Cayley-Hamilton theorem from the formula found, multiply on the left or right by $A$ and move the left hand side to the right.