Bocher’s formula

A colleague of mine mentioned to me today the Bocher’s formula for computing the coefficients of the characteristic polynomial of a matrix. It seems that this formula does not appear too often in textbooks or literature. I’ll just write down the formula and the idea of a simple proof here.

Let the characteristic polynomial of a matrix A be


Then the coefficients can be computed by

To prove the formula, note that the coefficient a_j is the summation of all possible products of j eigenvalues, i.e.,

\displaystyle{a_j=(-1)^j\sum_{\{t_1\cdots t_j\}\in C_n^j}\lambda_{t_1}\cdots\lambda_{t_j},}

where C_n^j denotes the j-combination of numbers from 1 to n, and the trace of A^i is the sum of the jth power of the eigenvalues, i.e.,


In addition, we have

\displaystyle{\left(\sum_{\{t_1\cdots t_j\}\in C_n^j}\lambda_{t_1}\cdots\lambda_{t_j}\right)\left(\sum_{t_{j+1}=1}^n\lambda_{t_{j+1}}^i\right)=\sum_{\{t_1\cdots t_{j+1}\}\in C_n^{j+1}}\lambda_{t_1}\cdots\lambda_{t_j}\lambda_{t_{j+1}}^i+\sum_{\{t_1\cdots t_j\}\in C_n^j}\lambda_{t_1}\cdots\lambda_{t_{j-1}}\lambda_{t_j}^{i+1}.}

The above indicates that the first part of (-1)^ja_jtr(A^i) cancels the second part of (-1)^{j-1}a_{j-1}tr(A^{i+1}), whereas the second part of (-1)^ja_jtr(A^i) cancels the first part of (-1)^{j+1}a_{j+1}tr(A^{i-1}). The rest of proof becomes obvious now.


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