# Proving a well-known formula regarding adj(A)

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?

• The 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
• I 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
• See 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
• You can find an elementary proof here. – Pedro Tamaroff Aug 28 '13 at 23:23
• Answered in MO: Expressing adj(A) as a polynomial in A? – user1551 Aug 28 '13 at 23:25