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How to show that $$\det(I+\epsilon V)=1+\operatorname{trace}(V)\epsilon+O(\epsilon^2)$$ for any $n\times n$ real matrix $V$?

This is used a lot in the theory Lie groups, but I never saw a proof of it.

Peter
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    Use the usual Leibniz formula for the determinant (with the sum over all permutations...). – PhoemueX Jan 22 '15 at 19:35
  • For reference, the extended question of finding $\det(A + \epsilon V)$ can be answered by assuming $A$ invertible, calculating $\det(A)\det(I + \epsilon A^{-1}V)$ by methods found here, and using continuity of the determinant together with the density of invertible matrices to prove the formula for general $A$. – user157227 Jan 22 '15 at 22:56

5 Answers5

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The determinat of a matrix is the product of its eigenvalues. On the other hand, the eigenvalues of the matrix $$I+\epsilon V$$ have the form $1+\epsilon \lambda_i$, where $\lambda_i$ is the eigenvalue of $V$. The multiplicity of eigenvalues is also preserved, hence

$$\det (I+\epsilon V)=\prod_i (1+\epsilon \lambda_i) = 1+\epsilon \sum_i \lambda_i +\mathcal O(\epsilon^2) = 1+\epsilon \operatorname{tr} V+\mathcal O(\epsilon^2).$$

TZakrevskiy
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Hint : You can see this easily by triangulating $V$. (which is always possible in $M_n(\mathbb{C})$)

You would get that $\det(I+\epsilon V) = \textstyle \prod_{i=1}^{n} (1+\epsilon \lambda_i)$ where $\lambda_i$ are the eigenvalues of V.

The conclusion follows immediately.

Jack D'Aurizio
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We have $\det(I+\epsilon V)=\epsilon^n\det\left(\frac1\epsilon I-(-V)\right)$. Since the characteristic polynomial of $-V$ is monic and the coefficient of its $(n-1)$-th power term in is $\operatorname{tr}(V)$, the result follows.

user1551
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I want to supplement the above answers in a way how we can derive the desired result from a more general formula.

Lets suppose that $V$ be diagonalisable by $U$. Then

$$ \det(I+\epsilon V)=\det(U(I+\epsilon V)U^{-1})=\det(I+\epsilon \lambda)= \\\prod_i(1+\epsilon \lambda_i) $$ Now take the logarithm of this expression:

$$ \log[\det(I+\epsilon V)]=\text{Tr}\{\log[I+\epsilon V]\} $$

It follows that

$$ \det(I+\epsilon V)=e^{\text{Tr}\{\log[I+\epsilon V]\}} $$

Now make use of the Taylorexpansions (this is all well justified if $\epsilon$ is small)

$\log(1+x)\approx x$ and $e^x\approx 1+x$

to get $$ \det(I+\epsilon V)=1+\epsilon\text{Tr}[V] $$

tired
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Let $e_1, e_2, \ldots, e_n$ be the canonical basis vectors. Then exterior algebra tells us that, for any linear operator $A$ on the vector space,

$$A(e_1) \wedge A(e_2) \wedge \ldots \wedge A(e_n) = (\det A) (e_1 \wedge e_2 \ldots \wedge e_n)$$

For $A = I + \epsilon V$, the zeroth order term in $\epsilon$ is clearly $\det I$.

The first-order term in $\epsilon$ has the form

$$\epsilon [V(e_1) \wedge I(e_2) \wedge \ldots \wedge I(e_n) + I(e_1) \wedge V(e_2) \wedge \ldots \wedge I(e_n) + \ldots ]$$

This reduces to $\epsilon [V^{11} + V^{22} + \ldots + V^{nn}] (e_1 \wedge e_2 \wedge \ldots \wedge e_n)$, and of course the term in square brackets is the trace.

Muphrid
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