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Let $A \in M(\mathbb{R}, n)$ and $C_A(\lambda)$ its characteristic polynomial. Let

$$ \Gamma_A(\lambda) := (-1)^{n+1}\frac{C_A(\lambda) - C_A(0)}{\lambda} $$

Then, $\text{adj}(A) = \Gamma_A(A)$ which is a form of the Cramer's rule $\text{adj}(A)A = \det(A) I$ using the Cayley-Hamilton theorem.

There is an identitiy that

$$ \text{tr} ( \text{adj} (A) ) = (-1)^{n+1} \Gamma_A(0)$$

Question: how is this identity derived?

Background: I was studying Lemma 1.4 of this book on page 86, claim 6.

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    $\begingroup$ By the method of universal identities, or by the denseness of diagonalisable matrices over $\mathbb C$, you may assume that $A$ is diagonalisable and in turn you may assume that it is diagonal. The identity then becomes trivial. $\endgroup$ – user1551 Dec 20 '16 at 11:56
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Here is, I think, a possible answer.

From the Jacobi's identity, it follows that

$$ - \frac{d}{d \lambda} C_A(\lambda) = \frac{d}{d \lambda} \det ( \lambda I - A) = \text{tr} \left( \text{adj} ( \lambda I - A ) \frac{d}{d \lambda} ( \lambda I - A) \right) = \text{tr} ( \text{adj} ( \lambda I - A ) ) $$

Therefore,

$$ \frac{d}{d \lambda} C_A(\lambda) \Bigg|_{\lambda=0} = \text{tr} ( \text{adj} ( A ) )$$

Observe that

$$ \frac{d}{d \lambda} C_A(\lambda) \Bigg|_{\lambda=0} = (-1)^{n+1} \lim_{\lambda \rightarrow 0} \frac{C_A(\lambda) - C_A(0)}{\lambda} = (-1)^{n+1} \lim_{\lambda \rightarrow 0} \Gamma_A(\lambda) = (-1)^{n+1} \Gamma_A(0)$$

Therefore,

$$ (-1)^{n+1} \Gamma_A(0) = \text{tr} ( \text{adj} ( A ) ) $$

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This result is true for matrices over any field, we can give a simple algebraic proof:

The trace of $\text{adj}(A)$ is the sum of the principal minors of $A$, which corresponds, modulo the sign $(-1)^{n-1}$, with the coefficient $c_1$ of $X$ in $C_A(X)$. Now $\Gamma_A(X)$ has independent term $c_1$ (by construction), so $\Gamma_A(0)=c_1$ and $$\text{trace(adj}(A))=(-1)^{n-1}\Gamma_A(0).$$

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