Let $A$ be an $n\times n$ matrix. Then its characteristic polynomial is
$$\phi_A(x) = \det(xI – A) = x^n – (\operatorname{tr} A)x^{n-1} + \cdots + (-1)^{n-1}(\operatorname{tr}(\operatorname{adj}A))x + (-1)^n\det A,$$
where $\operatorname{adj}A$ denotes the adjugate matrix of $A$.
My question: Why are the coefficients of $\phi_A$ those shown above?
I can see why $(-1)^n\det A$ is the constant term, since
$$\det(A) = \phi_A(0) = (0-\mu_1)\cdots(0-\mu_n) = (-1)^n\mu_1\cdots\mu_n,$$
where $\mu_i$ are the roots ($i=1,\dots,n$). I also see why the coefficient of $x^{n-1}$ is minus the trace, since by Laplace expansion (along the first row):
\begin{align*}
&\det(xI-A)\\[10pt]
&= \begin{vmatrix}
x-a_{11} & -a_{12} & \cdots & -a_{1n}\\
-a_{21} & x-a_{22} & \cdots & -a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
-a_{n1} & -a_{n2} & \cdots & x-a_{nn}
\end{vmatrix}\\[10pt]
&= (x-a_{11})\begin{vmatrix}
x-a_{22} & \cdots & -a_{2n}\\
\vdots & \ddots & \vdots\\
-a_{n2} & \cdots & x-a_{nn}
\end{vmatrix} + \underbrace{a_{12} \begin{vmatrix}
-a_{21} & \cdots & -a_{2n}\\
\vdots & \ddots & \vdots\\
-a_{n1} & \cdots & x-a_{nn}
\end{vmatrix} + \cdots}_{\text{at most polynomial of degree $n-2$}}\\[10pt]
&=(x-a_{11})\bigg((x-a_{22})\begin{vmatrix}
x-a_{33} & \cdots & -a_{3n}\\
\vdots & \ddots & \vdots\\
-a_{n3} & \cdots & x-a_{nn}
\end{vmatrix}
+\underbrace{\underbrace{a_{23}\begin{vmatrix}
-a_{32} & \cdots & -a_{3n}\\
\vdots & \ddots & \vdots\\
-a_{n2} & \cdots & x-a_{nn}
\end{vmatrix}+\cdots}_{\text{at most polynomial of degree $n-3$}}
\bigg) + \cdots}_{\text{at most polynomial of degree $n-2$}}\\
&\kern10pt\vdots\\[10pt]
&=(x-a_{11})(x-a_{22})\cdots(x-a_{nn}) + \underbrace{\cdots}_{\text{at most polynomial of degree $n-2$}}\\[10pt]
&= x^n – (a_{11}+a_{22}+\cdots + a_{nn})x^{n-1} + \cdots\\[10pt]
&= x^n – (\operatorname{tr}A)x^{n-1} + \cdots
\end{align*}
(Let me know if there is a more elegant way to obtain this).
But what I cannot seem to prove is that the coefficient of $x$ is $ (-1)^{n-1}\operatorname{tr}(\operatorname{adj}A)$. Attempting to work with the Laplace expansion (as I did to obtain the trace above) is leading me nowhere.
Best Answer
Suppose first $B$ is an $n \times n$ matrix with $\det(B) = 1$. Then $\text{adj}(B) = B^{-1}$, and
$$ \phi_{B^{-1}}(x) = \det(x I - B^{-1}) = \det(x B - I) = (-x)^n \det(x^{-1} I - B) = (-x)^n \phi_B(1/x)$$
The coefficient of $x^{1}$ in $\phi_B(x)$ is the coefficient of $x^{-1}$ in $\phi_B(1/x)$, i.e. $(-1)^n$ times the coefficient of $x^{n-1}$ in $(-x)^n \phi_B(1/x)$, but you know the coefficient of $x^{n-1}$ in $\phi_{B^{-1}}(x)$ is $\text{tr}(B^{-1})$. That is, the coefficient of $x^1$ in $\phi_B(x)$ is $(-1)^n \text{tr}(B^{-1}) = (-1)^n \text{tr}(\text{adj}\; B)$.
Now let $A = t B$, where $t \ne 0$. Then $\phi_A(x) = \det(x I - t B) = t^n \det((x/t)I - A) = t^n \phi_B(x/t)$. The coefficient of $x^1$ in $\phi_A(x)$ is then $t^{n-1}$ times the coefficient of $x^1$ in $\phi_B(x)$. But also $\text{adj}\; A = t^{n-1} \text{adj}\; B$. So we again obtain that the coefficient of $x^1$ in $\phi_A(x)$ is $(-1)^n \text{tr}(\text{adj}\; A)$.
Every nonsingular matrix $A = \det(A)^{1/n} B$ where $\det(B) = 1$, so the formula for the coefficient holds for every nonsingular matrix. (Note that in the case where $\det(A)$ is negative and $n$ is even, we must use complex numbers here, but that's not a problem)
Now the coefficient of $x^1$ in $\phi_A(x)$ and $(-1)^n \text{tr}(\text{adj}\; A)$ are both polynomials in the coefficients of $A$.
Thus the equation must hold for all $n \times n$ matrices, not just the nonsingular ones.