How to prove this property of outer product? I encountered the following equation about vector outer product:
$$nn^T = [n]_{\times}^2 + I$$
where $n = [n_1, n_2, n_3]^T$ is a unit column vector, $I$ is the identity matrix, and $[n]_{\times}$ is a skew matrix of the form:
$$
[n]_{\times} =
\begin{bmatrix}
 0   & -n_3 & n_2 \\ 
 n_3 & 0    & -n_1\\ 
 -n_2& n_1  & 0
\end{bmatrix} 
$$
I want to know if there is a clever method to prove this equation, not by laborious matrix computation. The reason for my wondering is that this matrix $[n]_{\times}$ is a special matrix and used extensively in rotation transformation:
$n \times v = [n]_{\times}\cdot v$. 
Any advice?
 A: One method would be to see what each matrix does, geometrically.
The left side, $nn^T$, is easy: for any vector $x$, we find
$$
(nn^T)x = n(n^Tx) = (n \cdot x)n
$$
where the dot denotes the dot-product. In other words, $nn^T$ takes a vector $x$ as its input and produces the projection of $x$ onto $n$ as its output.  Now, for this next part, we'll need something that I've always known as the "bac-cab" formula, which says that
$$
a \times (b \times c) = b(a \cdot c) - c(a \cdot b)
$$
(hence its name).  So, we find that
$$
[n]_{\times}^2x = n\times (n \times x)=n(n \cdot x) - x(n \cdot n) = (n \cdot x)n - x
$$
Hey, don't we recognize that first term at the end there?  All together, we have
$$
([n]_\times^2 + I)x = [n]_\times^2 x + x = (n \cdot x)n - x + x = (n \cdot x)n = nn^Tx
$$
Since our two matrices do the same thing, they must be the same matrix.
A: Some easy ways to prove the identity (in order of increasing difficulty):


*

*Calculate $[v]_\times^2$ directly.

*Verify that equality holds when both sides are applied to $n$ or any unit vector $v\perp n$.

*Prove that $n\times (n\times x)=(n\cdot x)n-x$. (The bac-cab formula is an overkill; as two of the vectors here are equal to the unit vector $n$, you may prove the equality directly.)

*Since $[n]_\times$ is $3\times3$ real skew-symmetric, its complex eigenvalues must be 0 and $\pm ib$ for some $b$. However, $[n]_\times$ is an isometry on $n^\perp$. Hence $|b|=1$ and the eigenvalues of $A=[n]_\times^2$ are $0,-1,-1$. It follows that $A$ has a one-dimensional nullspace $L$ and $A$ is a reflection on $L^\perp$. Clearly, $L$ is the linear span of $n$. Hence $A=nn^T-I$.

A: A direct computation is obtainable using Einstein notation. In terms of Einstein notation, the relevant operations are
\begin{align*}
 u\cdot v &\equiv u_i v_i & \text{(dot product)} \\
    (u\otimes v)_{ij} &\equiv u_iv_j & \text{(tensor product)} \\ 
    (AB)_{ik} &\equiv A_{ij}B_{jk} & \text{(matrix product)} \\ 
    ([u]_\times)_{ij} &\equiv u_k\epsilon_{ijk} & \text{(your defined operation)} \\
    (I)_{ij} &\equiv \delta_{ij} & \text{(identity)}
\end{align*}
To use this notation, there is an implicit summation over any repeated indices. Note that I have not bothered to distinguish between raised and lowered indices due to the implicit assumption of purely Euclidean space ($\mathbb R^3$) as the underlying field. The terms $\delta_{ij}$ and $\epsilon_{ijk}$ are the Kronecker delta and the antisymmetric Levi-Civita symbol respectively.
Now, trudging through the calculation, the matrix elements of the LHS are simply
$$(n\otimes n)_{ik} =n_in_k$$
The matrix elements of the RHS are given by
\begin{align*}
\left([n]^2_\times + I\right)_{ik} &= ([n]_\times)_{ij}([n]_x)_{jk} + \delta_{ik} \\
&= (n_l\epsilon_{ijl})(n_{m}\epsilon_{jkm}) + \delta_{ik}\\
&= -n_ln_m\epsilon_{jil}\epsilon_{jkm} + \delta_{ik} \\
&= -n_ln_m(\delta_{ik}\delta_{lm} - \delta_{im}\delta_{kl}) + \delta_{ik} \\
&= -n_{l}n_{l}\delta_{ik} + n_kn_i + \delta_{ik} \\
&= n_in_k
\end{align*}
In the last step, we have used the fact that $n$ is a unit vector which means it satisfies $n_ln_l = n_1 n_1 + n_2n_2 + n_3n_3 = 1$. The identity for the product of Levi-Civita symbols with one matching index can be found on the Wikipedia page.
