I have a linear algebra question that is a proof and I am unsure how to approach this problem.
For any vector $s$, show that $$(s\times)^2 = ss^T-s^TsI$$ where the skew operator $s\times$ is defined by $$s\times\triangleq\begin{bmatrix}0&-c_s&b_s\\c_s&0&-a_s\\-b_s&a_s&0\end{bmatrix}$$
So you want to show
$$\begin{bmatrix} 0 & -c & b \\ c & 0 & -a \\ -b & a & 0\end{bmatrix}^2=\begin{bmatrix}a \\ b \\ c\end{bmatrix}\begin{bmatrix} a & b & c\end{bmatrix}-(a^2+b^2+c^2)\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}.$$
What's wrong with simplifying both sides and seeing that they're the same?
You should already understand the following:
$$\begin{bmatrix}a \\ b \\ c\end{bmatrix}\begin{bmatrix} a & b & c\end{bmatrix}=\begin{bmatrix}a^2 & ab & ac \\ ba & b^2 & bc \\ ca & cb & c^2\end{bmatrix}$$
$$(a^2+b^2+c^2)\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}a^2+b^2+c^2 & 0 & 0 \\ 0 & a^2+b^2+c^2 & 0 \\ 0 & 0 & a^2+b^2+c^2\end{bmatrix}$$
This is simply how matrix multiplication and scalar multiplication work.