What's the associated matrix of this linear operator? Let $V$ be a complex vector space of dimension $n$ with a scalar product, and let $u$ be an unitary vector in $V$. Let $H_u: V \to V$ be defined as
$$H_u(v) = v - 2 \langle v,u \rangle u$$
for all $v \in V$. I need to find the characteristic polynomial of this linear operator, but the only way to find it that I know of is using the associated matrix of the operator.
I don't know how to find this matrix because I don't know how to deal with the scalar product. Is there some other way to find the characteristic polynomial? If not, how can I find the associated matrix of this linear operator?
Thanks in advance.
 A: I am assuming that $\langle v,u \rangle = v^T \overline{u}$, otherwise $H_u$ would be conjugate linear.
$H_u v = v-2 v^T \overline{u} u = v-2 \overline{u}^T v u = v- 2u \overline{u}^T v = (I-2u u^*)v$, so $H_u = I-2u u^*$. The characteristic polynomial is $\chi_{H_u}(s) = \det( sI-H_u) = \det ( (s-1)I +2 u u^*) $.
Let $U$ be an orthogonal matrix such that $U e_1 = u$ (such a unitary matrix exists because $u$ is a unit vector), then 
$\chi_{H_u}(s) = \det ((s-1)I +2 Ue_1 e_1 U^*) = \det ((s-1)I +2 e_1 e_1)$. Since $(s-1)I +2 e_1 e_1$ is diagonal, the determinant is sinfully easy to compute, so we have $\chi_{H_u}(s) = (s-1)^{n-1}(s+1)$. 
Alternatively, you could find $u_2,...,u_n$ so that $u,u_2,...,u_n$ form an orthonormal basis. Then we have $H_u u = -u$, $H_u u_k = u_k$ for $k=2,...,n$. Hence the eigenvalues are $-1$ with multiplicity $1$ and $+1$ with multiplicity $n-1$. Since $\chi_{H_u}(s) = \Pi_{k=1}^n (s-\lambda_k)$, where $\lambda_k$ are the eigenvalues of $H_u$, we have the desired result.
Finally, you could try using the matrix determinant lemma (see related Sherman-Morrison formula) which states $\det(A+u v^*) = (1+v^* A^{-1}u) \det A$.
If $s \neq 1$, we have $\det ( (s-1)I +2 u u^*) = (1+\frac{u^* u}{s-1}) \det ((s-1)I) = \frac{s+1}{s-1} (s-1)^n = (s-1)^{n-1}(s+1)$. Continuity shows this is true for all $s$, hence we have the desired result.
A: It is important to have a geometric understanding of projections. Suppose $P$ is a linear map on a vector space $V$ such that $P^2=P$. Then $P$ acts as the identity on its image, and furthermore $V=\ker P\oplus\operatorname{img}P$. The idea here is that every vector can be decomposed into two parts, one annihilated by $P$ and the other fixed by $P$. Suppose $v\in V$. Then $P(v)$ is the image-component of the vector $v$. The other component adds to $P(v)$ to make $v$, so the other component can be found using subtraction as $v-P(v)$. Clearly $v=(v-P(v))+P(v)$ and $v-P(v)\in\ker P$ and $P(v)$ is in the image of $P$. This is exactly like writing vectors in "$x$ and $y$" coordinates.
Suppose we want to project orthogonally onto a one-dimensional subspace $\Bbb Ru$ (I will work with real spaces for the geometric insight). For convenience make $u$ have norm $1$. Then the magnitude of the $u$-component of $v$ is $\|v\|\sin\theta=\langle v,u\rangle$ by vector geometry. The direction of the $v$-component is simply $v$. Thus the $v$-component is $\langle v,u\rangle u$ and the orthogonal component is $v-\langle v,u\rangle u$. Notice what happens if we subtract $\langle v,u\rangle u$ from $v$ twice: we now have the same $u$-component as $v$ had originally, but pointed in the opposite direction. This is a reflection across the hyperplane $u^{\perp}$.
$\hskip 1.5in$ 
Thus we have: $V=\langle u\rangle\oplus \langle u\rangle^\perp$. On $\langle u\rangle$, $H_u$ reflects across $0$ so what is its eigenvalue? On $\langle u\rangle^\perp$, $P$ acts as the identity, so what are its eigenvalues there? What are the dimensions of $\langle u\rangle$ and $\langle u\rangle^\perp$?
The above argument works over $\Bbb C$ too as-is, there just isn't a $2$D picture showing it there.
A: This is the abstract kind of way of defining linear operators.  You should understand that a matrix is just a collection of coefficients describing the action of an abstract linear operator according to some basis.
That is, you can find the components of the matrix $(H_u)_{ij}$ simply by choosing some basis vectors $e_k, k \in [1, 2, \ldots, n]$ and plugging them into the expression
$$\langle H_u(e_i), e_j \rangle = \langle e_i, e_j \rangle - 2 \langle e_i, u \rangle \langle u, e_j \rangle$$
for each $i, j \in [1, 2, \ldots, n]$.
Alternatively, you should realize that the first term is just the expression of the identity map, so you only need to exert effort on the second term.

That said, there are ways of finding determinants (and thus, characteristic polynomials) without finding the matrix representation.  These typically rely on exterior algebra or something similar.  The key idea to use here is something called a wedge product.  Define an extension of the map over wedge products to be
$$H_u(v \wedge w) \equiv H_u(v) \wedge H_u(w)$$
Objects formed by wedge products are called blades.  The wedge product of two vectors is a $2$-blade (vectors are defined to be 1-blades).  Once you form an $n$-blade of the space, you can find the determinant.  The $n$-blade is an "eigen"blade of all operators, and the determinant is its eigenvalue.
Great, how can you find the determinant that way?  Well, here you can factor an $n$-blade into two parts.  Let $E$ be an $n$-blade on the space.  We can write $E$ as
$$E = A_{n-1} \wedge u$$
where $A_{n-1}$ is an $n-1$-blade totally orthogonal to $u$ (all vectors lying in the subspace $A$ represents are orthogonal to $u$).
Crucially, this means you can write the action of $H_u$ on $E$ as
$$H_u(E) = H_u(A_{n-1}) \wedge H_u(u)$$
The first factor is easy: $H_u(A_{n-1}) = A_{n-1}$.  Why?  Because $A_{n-1}$ is formed from $n-1$ vectors that are all orthogonal to $u$ and so the second term that would be $-2 \langle v, u \rangle u$ is zero.  So all that's left is the identity.
The second factor is just $u - 2 \langle u, u\rangle u = -u$, so we get
$$H_u(E) =(\det H_u) E =  A_{n-1} \wedge (-u) = -E$$
Therefore, the determinant is $-1$.

Exterior algebra (or its superior cousin of clifford algebra) is intensely useful for building up linear maps to act on objects higher than vectors, without having to get mired in tensor index gymnastics or a proliferation of maps that only produce scalars without any geometric interpretation.
As has been said, this is merely a reflection map (anyone familiar with 3d geometry recognizes it instantly), and a lot of the logic here works equally well for real vector spaces as opposed to complex.  Keeping a geometric interpretation on what you're doing helps immensely.  $E$, the $n$-blade, can be said to represent an oriented $n$-volume, and the determinant just tells us if that volume increases or decreases when acted upon by the operator, and the change in sign tells us that that volume has a reversed orientation (think left-hand vs. right-hand rule).

Edit: you were interested in characteristic polynomials in particular.  This is slightly more complicated because the operator doesn't split as cleanly.  Using clifford (or geometric) algebra, however, this is something that has a known solution.
The key to the solution lies with "characteristic multivectors".  Examples of these are the trace and determinant, which also show up as coefficients in the characteristic polynomial, but there are many others, and traditional linear algebra doesn't usually tell us about them.
Write the characteristic polynomial as $\sum_{i=0}^n c_i (-\lambda)^{n-i}$.  The coefficient $c_0$ is always 1.  The coefficient $c_1$ is the trace, but we write it in geometric algebra and calculus as
$$c_1 = \text{Tr} \, H_u = \partial_{a_1} * H_u(a_1)$$
This means "take the vector derivative with respect to a vector $a_1$ and find the scalar part".  You could say this is the divergence.
Further coefficients all take the form
$$c_i = \partial_{a_i} * H_u(a_i)$$
which means "take a derivative with respect to the $i$-blade $a_i$, and return the scalar part".  I'll illustrate for $i=2$.  $H_u$ acting on 2-blades is
$$H_u(v \wedge w) = v \wedge w - 2 (v \cdot u) (u \wedge w) - 2 (v \wedge u)(u \cdot w)$$
Take a derivative with respect to $\partial_v$ and then $\partial_w$:
$$\partial_v \cdot H_u(v \wedge w)  = (n-1)w - 2 u\cdot (u \wedge w) - 2(n-1) u(u \cdot w)$$
and then
$$\begin{align*}\partial_w \cdot (\partial_v \cdot H_u[v\wedge w]) &= (n)(n-1) - 2n + 2 - 2(n-1) \\ &= n^2 -n - 2n + 2 - 2n + 2 \\ &= n^2 - 5n + 4 \\ &= (n-1)(n-4)\end{align*}$$
By no means is this a simple process.  However, the process can be made somewhat easier by recognizing that
$$H_u(v \wedge w) = v \wedge w - 2 u \wedge (u \cdot [v \wedge w])$$
so that for any $k$-blade $A_k$, we have
$$H_u(A_k) = A_k - 2 u \wedge (u \cdot A_k)$$
I'm confident that all the necessary coefficients can be worked out from this equation once one is familiar with some helpful calculus identities, but I don't know how the right identities for differentiating with respect to blades work.
A: $$
\left\langle v',H_u(v)\right\rangle = \left\langle v',v\right\rangle - 2 \langle v,u \rangle \left\langle v',u\right\rangle
$$
A: I am a little surprised how complicated most of these answers look! I'll assume (cf. Hurkyl's comment below the question) that $H_u$ is meant to be complex-linear in the argument $v$. 
If you want just the characteristic polynomial of $H_u$, then note that $H_u(u) = -u$, and that $H_u(v) = v$ whenever $\langle u, v\rangle = 0$; such $v$ form the hyperplane orthogonal to $u$, a subspace of dimension $n-1$. Therefore, if $u_1, u_2, \ldots, u_{n-1}$ is any basis of the hyperplane, we have that the matrix representation of $H_u$ on the basis $u, u_1, \ldots, u_{n-1}$ of $V$ is a diagonal matrix with $-1$ as the first diagonal entry, and $1$'s for the rest. Hence the characteristic polynomial $\det(t I_n - H_u)$ is $(t+1)(t-1)^{n-1}$. 
