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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
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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

1 Answer 1

up vote 5 down vote accepted

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.

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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

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