How to prove this property of outer product?

Question

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?

Answer 1

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.

Answer 2

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 $A=nn^T-I$.