Why is $a \times b$ the null vector of the anti-symmetric matrix $M = ab^{\mathrm T} - ba^{\mathrm T}$?

Question

Let $a$ and $b$ be two vectors of size $3 \times 1$. The matrix $M = ab^{\mathrm T} - ba^{\mathrm T}$ is $3 \times 3$ anti-symmetric (easy to prove), and also $a \times b$ is its null vector. I have proven this, but only by substituting $a$ as $a = (a_1, a_2, a_3)^{\mathrm T}$ and $b$ as $b = (b_1, b_2, b_3)^{\mathrm T}$ and doing the calculations.

Is there any other (shorter) proof or any identity/theorem I am missing?

Also, let's say we didn't know the null vector was $a \times b$. How could one try and approach the problem of finding the null vector of $M$ from "scratch"?

Answer 1

For a vector $\vec a=[a_1,a_2,a_3]^{T}$, define $[\vec a]_{\times}=\left[\begin{matrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0\end{matrix}\right]$. Then we have some properties like: $[\vec a]_{\times}\vec a=\vec 0$, $\:\:\vec a\times\vec b=[\vec a]_{\times}\vec b$, $\;\;[\vec a\times\vec b]_{\times}=[\vec a]_{\times}[\vec b]_{\times}-[\vec b]_{\times}[\vec a]_{\times}$. See: https://en.wikipedia.org/wiki/Skew-symmetric_matrix

Now, I think here, $\vec a\vec b^{T}-\vec b\vec a^{T}=[\vec a\times\vec b]_{\times}$. Therefore, $(\vec a\vec b^{T}-\vec b\vec a^{T})(\vec a\times\vec b)=\vec 0$.

There is a connection with lie algebras. But, I don't know about them.

Answer 2

Note that $x^t y$ is just the inner product $x.y$ (at least, identifying $\mathbb{R}$ with the space of $1\times 1$ real matrices). Since $a\times b$ is orthogonal to $a$ and $b$, we have $M(a\times b) = a(b.(a\times b)) - b(a.(a\times b)) = 0$.