# 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?

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!!!!

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