# determinant of a sum

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?

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what do you mean by first, second and third order? Are you looking for an expansion of $\det(I+tM)$ as a function of $t$? –  user20266 May 7 '12 at 11:02
@Thomas, Here's my interpretation: $\det(I+M)$ is a polynomial in the coefficients of $M$, and as such can be decomposed into homogeneous parts according to degree. Truncating this polynomial below or up to a certain degree defines the determinant up to the given order. So in dimension two, $\det=\color{Green}1+\color{Blue}{a+d}+\color{Purple}{ad-bc}$ highlights the differences between orders $0,1,2$. –  anon May 7 '12 at 11:28
We're having trouble figuring out what you mean. Maybe you could include a link to what you found on the internet? –  Gerry Myerson May 7 '12 at 12:33

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.
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. Even though OP has yet to clarify I still think this answer is more likely than not useful for his/her query. –  anon May 7 '12 at 13:27