Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A$ be an $n\times n$ non-singular matrix with real entries. How can I prove the following equation? Any references would be helpful. $$ \det(A) = \frac 1{n!} \left| \begin{array}{cccccc}\operatorname{tr}(A) & 1 & 0 & \cdots & \cdots & 0 \\ \operatorname{tr}(A^2) & \operatorname{tr}(A) & 2 & 0 & \cdots & 0 \\ \operatorname{tr}(A^3) & \operatorname{tr}(A^2) & \operatorname{tr}(A) & 3 & & \vdots \\ \vdots & & & & & n-1 \\ \operatorname{tr}(A^n) & \operatorname{tr}(A^{n-1}) & \operatorname{tr}(A^{n-2}) & \cdots & \cdots & \operatorname{tr}(A) \end{array}\right|$$

share|cite|improve this question
@DietrichBurde yes. – user117432 Jan 1 '14 at 14:32
why is this helpful for you? What is the background for this? What have you tried on your part? do you immediately see this for a $2\times 2$ matrix? – Praphulla Koushik Jan 1 '14 at 14:34
This looks like a result of the Leverrier-Faddejev algorithm. At the same time, the matrix is part of the system of Newton identities relating the power sums of the eigenvalues to the coefficients of the characteristic polynomial. – LutzL Jan 1 '14 at 14:34
up vote 11 down vote accepted

The Newton identities relating power sums of eigenvalues $s_k=tr(A^k)$ to the coefficients of the characteristic polynomial $\chi_A(t)=t^n+c_1t^{n-1}+\dots+c_{n-1}t+c_n$ with $c_n=(-1)^n\det(A)$ read as

\begin{align} s_1 &= -c_1,\\ s_2 &= -c_1 s_1 - 2 c_2,\\ s_3 &= -c_1 s_2 - c_2 s_1 - 3 c_3,\\ s_4 &= -c_1 s_3 - c_2 s_2 - c_3 s_1 - 4 c_4, \\ & {} \ \ \vdots\\ s_n &= -c_1 s_{n-1}-\dots-c_{n-1} s_1 - n c_n \end{align}


$$-\begin{bmatrix} s_1\\s_2\\s_3\\\vdots\\s_{n-1}\\s_n \end{bmatrix} = \begin{bmatrix} 1&0&0&\dots&0&0\\ s_1&2&0&\dots&0&0\\ s_2&s_1&3&&0&0\\ \vdots&\vdots&&&&\vdots\\ s_{n-2}&s_{n-3}&s_{n-4}&\dots&n-1&0\\ s_{n-1}&s_{n-2}&s_{n-3}&\dots&s_1&n \end{bmatrix} \begin{bmatrix} c_1\\c_2\\c_3\\\vdots\\c_{n-1}\\c_n \end{bmatrix}$$

Now apply Cramers rule to the computation of $c_n$ to obtain the stated formula.

Note that the solution of this triangular system constitutes the computational core of the Leverrier-Faddejev algorithm for the (mostly) division free computation of the characteristic polynomial of a matrix.

share|cite|improve this answer

Determinants and traces are invariant and everything is polynomial, so it is enough to check it for diagonal matrices.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.