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Let $X = [x^i_j]$ be an $r \times r$ matrix of indeterminates and $f_k(X) \in \Bbb Z[x^i_j]$ be the coefficients of the characteristic polynomial $$\det(\lambda I + X) = \sum_{k=0}^r f_k(X)\lambda^{r-k}.$$

Given the above data, I am trying to get an idea about these polynomials $f_k(X)$. We were going over this in our lecture and our professor mentioned that each $f_k$ here is a polynomial of degree $k$. I find this hard to believe as if for example $X$ is the $2 \times 2$ matrix

$$ X=\begin{pmatrix}a&b\\ c&d\end{pmatrix} $$

then

$$ \begin{align*} \det(\lambda I + X)&=(\lambda+a)(\lambda+d)-bc \\ &= \lambda^2 + \lambda(a+d)+ad-bc\\ &= \lambda^2+\lambda\mathrm{tr}(X) + \det(X) \end{align*} $$

and $f_0(X)=1,f_1(X)=a+d,f_2(X)=ad-bc$, but none of these are polynomials with the respective degrees?

I might not have understood what $f_k(X) \in \Bbb Z[x^i_j]$ means here. What even is the ring $\Bbb Z[x^i_j]$ is it the polynomials with coefficients in $\Bbb Z$ in one variable $x^i_j$ for fixed $i$ and $j$ or something else?

Robert Israel
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Rene
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    In your example, $1$ is indeed a polynomial of degree $0$, $a+d$ is a polynomial of degree $1$ and $ad-bc$ is a polynomial of degree $2$, where by "degree" we mean total degree in the indeterminates. By $\mathbb Z[x^i_j]$ the prof means polynomials in the $r^2$ indeterminates $x^i_j$ with coefficients in $\mathbb Z$. – Robert Israel Jan 05 '24 at 17:47

1 Answers1

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The ring $\mathbb{Z}[x^i_j]$ here is the ring of polynomials in $n^2$ variables, which are the entries of the matrix.

We have $n^2$ variables $x^i_j$ for $i = 1, \dots, n$, $j = 1, \dots, n$. The entries of matrix $X$ are these variables, so in the $2 \times 2$ case we have $X = \begin{bmatrix} x^1_1 & x^1_2 \\ x^2_1 & x^2_2 \end{bmatrix}$. This matrix is sometimes called the generic $n \times n$ matrix.

The entries of $X - \lambda I$ are now polynomials in $x^i_j$ and $\lambda$. They are of the form $x^i_j$ for $i \neq j$ and $x^i_i - \lambda$, so they actually are homogeneous polynomials of deegree $1$ with integer coefficients. It follows that the determinant $\det(X - \lambda I) \in \mathbb{Z}[x^i_j,\lambda]$ is a homogeneous polynomial of degree $n$ in $x^i_j$ and $\lambda$, and if we write it by powers of $\lambda$ as $\det(X - \lambda I) = \sum_{i = 0}^n f_i(X) \lambda^{n - i}$, then $f_i \in \mathbb{Z}[x^i_j]$ are polynomials of degree $i$.

For $2 \times 2$ we have, as you write correctly with different variables $f_0(X) = 1$, $f_1(X) = x^1_1 + x^2_2$, and $f_2(X) = x^1_1 x^2_2 - x^1_2 x^2_1$. These are clearly of degree $0$, $1$ and $2$ in their variables.

Vladimir Lysikov
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  • Is there a general formula for $f_i(X)$ available from which it could be deduced that these indeed have total degree $i$? @vladimir-lysikov – Rene Jan 05 '24 at 18:05
  • Just use the Leibniz formula. – Martin Brandenburg Jan 05 '24 at 18:11
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    Yes — $f_i(X)$ is the sum of all principal minors of size $i \times i$ (principal minors are those where we choose the rows and the columns with the same indices). See for example this question: https://math.stackexchange.com/questions/28650/coefficients-of-characteristic-polynomial-of-a-matrix – Vladimir Lysikov Jan 05 '24 at 18:11