Trace and Determinant Relationship: Proving the Relation with O(x^2)

[alle.fabbri]

Hi,
can anyone give me an hint to proof the relation

det(I + x A) = 1 + x tr A + O(x^2)

thank you.

 

[Ben Niehoff]

Write out the general form of (I+xA). The determinant must be some polynomial in x, right? So, carefully pick out the constant- and linear-order terms of this polynomial by choosing the appropriate factors in the determinant.

 

[alle.fabbri]

Write out the general form of (I+xA). The determinant must be some polynomial in x, right? So, carefully pick out the constant- and linear-order terms of this polynomial by choosing the appropriate factors in the determinant.

ok that's the idea behind...but what about some explicit calculations? Are them possible? I thought i can use the general determinant expansion in terms of the total antisymmetric tensor which, for a nxn matrix A must be something like

<br /> det A = \frac{1}{n!} \sum_{i_1, i_2, ... , i_n = 1}^{n} \epsilon_{i_1\;i_2\;...\;i_n} A_{1\;i_1} A_{2\;i_2} A_{3\;i_3} ... A_{n\;i_n}<br />

the expression of the fact that the determinant is the average of all possible products of the elements picked one per column keeping them in different rows, thanks to totally antisymmetric tensor.

Any idea to be a bit formal?

 

[Ben Niehoff]

There should be no 1/n! in there. The determinant is the anti-symmetrized sum of all the products; not the average.

If you want to do the proof formally, the simplest way would be to diagonalize A. Or in general, bring A to upper-triangular form by a similarity transformation. Note that the identity matrix is invariant under similarity transformations. The determinant and trace are also invariant.