1
$\begingroup$

I need a formula for the determinant of the sum of two matrices: $\det(\mathbb{I}+M)$. On the internet I found it for the first order but i need it at second or even third order. Where can I find the proof?

  • 2
    We're having trouble figuring out what you mean. Maybe you could include a link to what you found on the internet?2012-05-07

1 Answers 1

5

Note that $\det(I+M)$ is just the characteristic polynomial $\det(M - \lambda I)$ evaluated at $\lambda = -1$. There is a well known relationship between the elementary symmetric polynomials in the eigenvalues of $M$ and the coefficients of the characteristic polynomial of $M$, which result in equations like yours when evaluated at $\lambda = 1$. E.g. in anon's example you have $1 + \mbox{trace} M + det M$. In general the coefficient of $\lambda^{(n-1)}$ is (up to sign) the trace of $M$ (the sum of the eigenvalues), the constant coefficient the determinant of $M$ (again up to sign), which is the product of the eigenvalues. Maybe this is what you are looking for? See, e.g., this and this wikipedia entry.

  • 0
    Note also that if we want these in terms of coefficients of the matrices (via $\mathrm{tr}(A^k)$'s specifically), because we regard the eigenvalues as unavailable for whatever reason, we can recursively use those Newton-Girard formulas, [as I discussed here](http://math.stackexchange.com/questions/137951/symmetric-and-exterior-power-of-representation/138794#138794). Even though OP has yet to clarify I still think this answer is more likely than not useful for his/her query.2012-05-07