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Let $A=(A_{ij})$ be a square matrix of order $n$. Verify that the determinant of the matrix

$\left( \begin{array}{ccc} a_{11}+x & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22}+x & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn}+x \end{array} \right)$,

can be represented as the polynomial $x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_{n-1}x+a_n$, where each coefficient $a_k$ is the sum of the minors of order $k$ of the matrix $A$.

I tried to use the definition of determinant by cofactor expansion but it's very long, I was wondering if there's a shorter way to show this.

Twnk
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2 Answers2

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Write each column$~j$ of $A+xI_n$ as a sum of the column$~j$ of$~A$ and $x$ times column $j$ of$~I_n$. Now apply multi-linearity of the determinant with respect to the columns for each of the columns, to obtain a sum of $2^n$ determinants (each column was a sum of $2$ terms, and doubled the number of terms obtained for the whole determinant). The sum can be indexed by the $2^n$ subsets of $\{1,2,\ldots,\}$, namely the subset of the columns for which the term involving $x$ was chosen. One obtains $$ \det(A+xI_n) = \sum_{S\subseteq\{1,2,\ldots,\}}x^{|S|}\det M(S,A) $$ where $M(S,A)$ denotes the matrix obtained from$~A$ by replacing the columns whose index appears in$~S$ by the corresponding column of$~I_n$.

Now fixing $S$, the determinant of $M(S,A)$ can be successively developed by the columns selected by$~S$, those that are taken from$~I_n$, which development involves only a single nonzero term each time. What remains is the determinant of the matrix obtained from $A$ by removing the rows and the columns whose index lies in$~S$. This is a principal minor of order $n-|S|$. Thus after collecting the terms with the same power of$~x$, the coefficient$~a_k$ of $x^{n-k}$ is the sum of all principal minors of order$~k$ of$~A$. With that correction, the statement given has been proved.

Marc van Leeuwen
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Simply note that a determinant is computed as the sum and product of finitely many polynomial terms, which results in a polynomial. You can show that with an induction argument, although I don't really think that's necessary.

Eurakarte
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  • I think the induction is necessary if you want to argue that the degree of the polynomial is $n$. – EuYu Sep 02 '13 at 06:34
  • This is not what the OP asked for. That the determinant is a polynomial is obvious - but it's not obvious that each $a_k$ is the sum of the degree $k$ minors. – Fredrik Meyer Sep 02 '13 at 08:52