The adjugate of a matrix $A$ is defined as $$ (\mathrm{adj}(A))_{ij} = (-1)^{i+j}M_{ji}(A) $$ where $M_{ji}(A)$ is the determinant of the matrix $A$ after row $j$ and column $i$ have been removed. It is well-known that $$ A\,\mathrm{adj}(A) = \mathrm{adj}(A)A = \det(A) I $$ Therefore, if $A$ is invertible, we get the famous formula $A^{-1} = \det(A)^{-1}\mathrm{adj}(A)$. There is also a well-known expansion of $A$ using the Cayley-Hamilton theorem, that looks like this: $$ p_0 I + p_1 A + p_2 A^2 + \dots + p_n A^n = 0 $$ where the $p_i$ are the coefficients of the characteristic polynomial of $A$. I have seen it written in many places (wikipedia, planetmath, textbooks) that the following also holds: $$ \mathrm{adj}(A) = -(p_1 I + p_2 A + \dots + p_n A^{n-1}) $$ where the $p_i$ are the same as in the previous equation. It is easy to prove this formula in the case where $A$ is invertible by applying the identities given above together with the fact that $p_0 = \det(A)$. However, I think the adjugate expansion still holds when $A$ is not invertible... how do I prove it for such a case?
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1The function $A \mapsto \mathrm{adj}(A)$ is continuous, and the functions $A \mapsto p_k(A)$ (the coefficients of the characteristic polynomial) are continuous. The set of invertible matrices is dense, hence a continuity argument shows that the formula must hold for singular matrices. – copper.hat Aug 28 '13 at 22:38
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While this argument works for real matrices, I would like a proof that holds for matrices whose entries belong to a general commutative ring. This argument would fail if applied for example to matrices of integers. – Laurent Lessard Aug 28 '13 at 22:48
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1I think the fastest way to prove this is by noting that $p_k$ is the sum of all $(n-k)$-minors of $A$. – walcher Aug 28 '13 at 22:48
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Beware! you have $p_0=\det(A)$ only if your definition of characteristic polynomial is $P_A(x)=\det(A-xI)$, but in this case your characteristic polynomial is not monic. If you define instead $P_A(x)=\det(xI-A)$, then your characteristic polynomial is monic, with constant coefficient equal to $\det(-A)=(-1)^n\det(A)$. – Matemáticos Chibchas Aug 28 '13 at 22:58
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@walcher, I am aware of the formula you make reference to, but I fail to see how it can help me with the proof. – Laurent Lessard Aug 28 '13 at 23:06
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@MatemáticosChibchas, yes you are correct. $p_0 = \pm \det(A)$ depending on which definition of the characteristic polynomial you are using. – Laurent Lessard Aug 28 '13 at 23:07
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2See this excellent expository paper from Keith Conrad, where it is shown how to prove general identities by specializing to real or complex numbers. – Matemáticos Chibchas Aug 28 '13 at 23:11
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1You can find an elementary proof here. – Pedro Aug 28 '13 at 23:23
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2Answered in MO: Expressing adj(A) as a polynomial in A? – user1551 Aug 28 '13 at 23:25
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Thank you all for the responses. The Conrad paper was very interesting. Thanks @MatemáticosChibchas. – Laurent Lessard Aug 28 '13 at 23:32
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aha! seems I asked this question some years ago in MO. Thank you @user1551. – Laurent Lessard Aug 28 '13 at 23:33
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For sake of having an answer: the OP had asked this question on MO before and got a number of good answers there. See Expressing adj(A) as a polynomial in A?
