Tricky question on induction and characteristic polynomials I am to prove via induction that for any $n \times n$ matrix $A$, the characteristic polynomial of $A$ has


*

*degree $n$;

*$(-1)^n$ as the coefficient of the $\lambda ^n$ terms;

*$(-1)^{n-1}\cdot \text{Trace}(A)$ as the coefficient of $\lambda ^{n-1}$.
I have shown the first two, but can't seem to prove the third.  Could somebody give me a hint?
 A: Eigenvalue arguments tend to work well when the base field $\Bbb F$, with $A \in M_{n \times n}(\Bbb F)$, is algebraically closed, so that all roots of the characteristic polynomial are guaranteed to exist in $\Bbb F$.  Here's an inductive demonstration not based on eigenvalues and eigenvectors which works over any field $\Bbb F$:
First, the result is clearly true if $n = 2$, for if
$A = \begin{bmatrix} a_{11} & a_{22} \\ a_{21} & a_{22} \end{bmatrix}, \tag{1}$
then
$A - \lambda I = \begin{bmatrix} a_{11} - \lambda & a_{22} \\ a_{21} & a_{22} - \lambda\end{bmatrix}, \tag{2}$
so the characteristic polynomial $p_A(\lambda)$ is
$p_A(\lambda) = \det(\begin{bmatrix} a_{11} - \lambda & a_{22} \\ a_{21} & a_{22} - \lambda\end{bmatrix})$
$= \lambda^2 - (a_{11} + a_{22}) \lambda + (a_{11}a_{22} - a_{12}a_{21}) = \lambda^2 - \text{Tr}(A)\lambda + \det(A); \tag{3}$
the coefficient of $\lambda^2$ is $1 = (-1)^2$ and that of $\lambda = \lambda^1$ is $(-1)^1\text{Tr}(A) = (-1)^{2 - 1}\text{Tr}(A)$, and $\deg p_A(\lambda) = 2$ in this the base case for the induction.  Now assume that $\deg p_A(\lambda) = k$, the coefficient of $\lambda^k$ is $(-1)^k$, and the coefficient of  $\lambda^{k - 1}$ in $p_A(\lambda)$ is $(-1)^{k - 1}\text{Tr}(A)$ for all $A \in M_{k \times k}(\Bbb F)$, and consider some $A \in M_{(k + 1) \times (k + 1)}(\Bbb F)$; its characteristic polynomial is
$p_A(\lambda) = \det(A - \lambda I)$
$= \det(\begin{bmatrix} a_{11} - \lambda & a_{12} & \ldots & a_{1 \; k} & a_{1 \; (k + 1)} \\ a_{21} & a_{22} - \lambda & \ldots & a_{2 \; k} & a_{2 \; (k + 1)} \\
a_{31} & a_{32} & a_{33} - \lambda & \ldots & a_{3 \; (k + 1)} \\
\ldots \\ a_{(k + 1) \; 1} & a_{(k + 1) \; 2} & \ldots & a_{(k + 1) \; k} & a_{(k + 1) \; (k + 1)} - \lambda \end{bmatrix}). \tag{4}$
When we evaluate the determinant in (4) by expanding in minors along the first row, we see that
$p_A(\lambda) = (a_{11} - \lambda) \det(\begin{bmatrix} a_{22} - \lambda & \ldots & a_{2 \; k} & a_{2 \; (k + 1)} \\
a_{32} & a_{33} - \lambda & \ldots & a_{3 \; (k + 1)} \\
\ldots \\ a_{(k + 1) \; 2} & \ldots & a_{(k + 1) \; k} & a_{(k + 1) \; (k + 1)} - \lambda \end{bmatrix}) + \Theta(\lambda), \tag{5}$
where $\Theta(\lambda)$ is a polynomial of degree at most $k - 1$ in $\lambda$, since deleting the row and column containing $a_{1 \; j}$, $2 \le j \le k + 1$ leaves a matrix with at most $k - 1$ entries of the form $a_{ii} - \lambda$.  Furthermore, our inductive hypothesis implies that
$\det(\begin{bmatrix} a_{22} - \lambda & \ldots & a_{2 \; k} & a_{2 \; (k + 1)} \\
a_{32} & a_{33} - \lambda & \ldots & a_{3 \; (k + 1)} \\
\ldots \\ a_{(k + 1) \; 2} & \ldots & a_{(k + 1) \; k} & a_{(k + 1) \; (k + 1)} - \lambda \end{bmatrix})$
$= (-1)^k \lambda^k + (-1)^{k - 1} \text{Tr} (A_{11}) \lambda^{k - 1} +\theta(\lambda), \tag{6}$
where $A_{11}$ is the matrix obtained from $A$ by deleting the first row and column, and $\theta(\lambda)$ is a polynomial of degree at most $k - 2$ in $\lambda$.  Thus the $\lambda^{k + 1}$ and $\lambda^k$ terms of $p_A(\lambda)$ are given by the product
$(a_{11} - \lambda)((-1)^k \lambda^k + (-1)^{k - 1} \text{Tr} (A_{11}) \lambda^{k - 1})$
$= (-1)^{k + 1}\lambda^{k + 1} + a_{11}(-1)^k \lambda^k + (-1)^k \text{Tr} (A_{11})\lambda^k + a_{11}(-1)^{k - 1} \text{Tr} (A_{11}) \lambda^{k - 1}$
$= (-1)^{k + 1}\lambda^{k + 1} + (-1)^k \text{Tr} (A)\lambda^k + a_{11}(-1)^{k - 1} \text{Tr} (A_{11}) \lambda^{k - 1} \tag{7};$
we see from (7) that not only is the coefficient of $\lambda^k$ $(-1)^k \text{Tr} (A)$ and , but that the coefficient of $\lambda^{k + 1}$ is $(-1)^{k + 1}$ and also that the degree of $p_A(\lambda)$ is $k + 1$, thus inductively establishing all three of the bullet points in the question.  And all this over an arbitrary field $\Bbb F$!  QED.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!!
A: Assuming that we have $n$ eigenvalues $c_1,c_2,...,c_n$, we can express the characteristic polynomial as
$\sum_{k=0}^na_k\lambda^k=\prod_{k=1}^n(\lambda-c_k)$ (Eq. $2$)
where $a_k$ is the coefficient of $\lambda^k$
The trace of a matrix $A$ is the sum of its eigenvalues, hence
$Trace(A)=\sum_{k=1}^nc_k$
If you were to expand the RHS of (Eq. $2$), what do you think the coefficient of $\lambda^{n-1}$ will be in terms of the eigenvalues, given that the coefficient of $\lambda^n$  is $(-1)^n$. Hint: Vieta's formula
