$$

Characteristic polynomial

Problem

Let

$$A=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& 2& 4& 8\\ 1& 3& 9& 27\\ 1& 4& 16& 64\end{array}\right].$$

Find the characteristic polynomial

$p_A(x)$ of

$A$.

Thought

There are several ways to compute

$p_A(x)=det(A-xI)$, including the Laplace expansion and the permutation expansion. They are doable, but it would be easier to compute

$p_A(x)$ by the formula

$$p_A(x)=(-x{)}^n+s_1(-x{)}^{n-1}+\cdots +s_n,$$

where

$$s_k=\sum _{\begin{array}{c}\alpha \subseteq [n]\\ |\alpha |=k\end{array}}det(A[\alpha ]).$$

Sample answer

For

$k=1$, the set

$\alpha$ can be

$\{1\}$,

$\{2\}$,

$\{3\}$,

$\{4\}$. The corresponding matrices

$A[\alpha ]$ are

$\left[\begin{array}{c}1\end{array}\right]$,

$\left[\begin{array}{c}2\end{array}\right]$,

$\left[\begin{array}{c}9\end{array}\right]$,

$\left[\begin{array}{c}64\end{array}\right]$ with their determinant

$1$,

$2$,

$9$,

$64$, respectively. Therefore,

$$s_1=\mathrm{tr}(A)=1+2+9+64=76.$$

For

$k=2$, the set

$\alpha$ can be

$$\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\}.$$

The corresponding matrices

$A[\alpha ]$ are

$$\left[\begin{array}{cc}1& 1\\ 1& 2\end{array}\right],\left[\begin{array}{cc}1& 1\\ 1& 9\end{array}\right],\left[\begin{array}{cc}1& 1\\ 1& 64\end{array}\right],\left[\begin{array}{cc}2& 4\\ 3& 9\end{array}\right],\left[\begin{array}{cc}2& 8\\ 4& 64\end{array}\right],\left[\begin{array}{cc}9& 27\\ 16& 64\end{array}\right]$$

with their determinant

$1$,

$8$,

$63$,

$6$,

$96$,

$144$, respectively. Therefore,

$$s_2=1+8+63+6+96+144=318.$$

For

$k=3$, the set

$\alpha$ can be

$$\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\}.$$

The corresponding matrices

$A[\alpha ]$ are

$$\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 4\\ 1& 3& 9\end{array}\right],\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 8\\ 1& 4& 64\end{array}\right],\left[\begin{array}{ccc}1& 1& 1\\ 1& 9& 27\\ 1& 16& 64\end{array}\right],\left[\begin{array}{ccc}2& 4& 8\\ 3& 9& 27\\ 4& 16& 64\end{array}\right]$$

with their determinant

$2$,

$42$,

$114$,

$48$, respectively. Therefore,

$$s_3=2+42+114+48=206.$$

For

$k=4$, the only

$\alpha$ is

$\{1,2,3,4\}$ and

$A[\alpha ]=A$. Since

$A$ is a Vandermonde matrix, one may easily calculate its determinant

$$s_4=det(A)=12.$$

In summary, the characteristic polynomial is

$$\begin{array}{rl}{p}_A(x)& =(-x{)}^4+s_1(-x{)}^3+s_2(-x{)}^2+s_3(-x)+s_4\\ & =x^4-76x^3+318x^2-206x+12.\end{array}$$

Note

With Sage, you may run the code below or simply click here to find the answer.

A = matrix([
    [1,1,1,1],
    [1,2,4,8],
    [1,3,9,27],
    [1,4,16,64]
])
n = A.dimensions()[0]

show((-1)**n * A.charpoly())

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