$\det(I+\epsilon V)=1+\operatorname{trace}(V)\epsilon+O(\epsilon^2)$ 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.
 A: 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
$$\det (I+\epsilon V)=\prod_i (1+\epsilon \lambda_i) = 1+\epsilon  \sum_i \lambda_i +\mathcal O(\epsilon^2) = 1+\epsilon \operatorname{tr} V+\mathcal O(\epsilon^2).$$
A: 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.
A: 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.
A: 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
$$
\det(I+\epsilon V)=1+\epsilon\text{Tr}[V]
$$
A: 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.
