Coordinate-free proof of the identity for two times cross product operator

Question

Consider an oriented three-dimensional Euclidean space $V$. Let $[a]$ be the operator of cross product by the vector $a$: $[a] b = a \times b$. It is easy to check that $$[a]^2 = a \otimes \flat a - \langle a, a \rangle \mathrm{id},$$ where $\flat: V \to V^*: a \mapsto \langle a, \_ \rangle$ is the "index lowering", and in the first term the canonical isomorphism $V \otimes V^* \cong \mathrm{End}(V)$ is assumed. Indeed, a direct computation shows that in the positive oriented orthonormal basis the matrix of $[a]^2$ is $$ \begin{pmatrix} -a_2^2 - a_3^2 & a_1 a_2 & a_1 a_3 \\ a_1 a_2 & -a_1^2 - a_3^2 & a_2 a_3 \\ a_1 a_3 & a_2 a_3 & -a_1^2 - a_2^2 \end{pmatrix} = \begin{pmatrix} a_1 a_1 & a_1 a_2 & a_1 a_3 \\ a_1 a_2 & a_2 a_2 & a_2 a_3 \\ a_1 a_3 & a_2 a_3 & a_3 a_3 \end{pmatrix} - (a_1^2 + a_2^2 + a_3^2) E. $$ And this is just the matrix of the operator above. I'm curious is there a coordinate free way to prove that identity?