[proofplan] We compute the commutator by applying it to an arbitrary vector $v\in\mathbb R^3$. The defining property of the hat map converts matrix multiplication into repeated cross products. A direct coordinate verification of the vector triple product identity then shows that the commutator action is exactly the action of $\widehat{a\times b}$. Since linear maps on $\mathbb R^3$ are equal when they agree on every vector, this proves the bracket identity; the [Lie algebra](/page/Lie%20Algebra) isomorphism follows from the linear bijectivity of the hat map. [/proofplan]

[step:Convert the matrix commutator into a triple cross product] Let $a,b\in\mathbb R^3$ be arbitrary, and let $v\in\mathbb R^3$ be arbitrary. The bracket on $\mathfrak{so}(3)$ is the commutator bracket, so \begin{align*} [\widehat a,\widehat b]v=(\widehat a\widehat b-\widehat b\widehat a)v. \end{align*} Using the defining property $\widehat x\,w=x\times w$ of the hat map, first with $x=b$ and then with $x=a$, gives \begin{align*} \widehat a\widehat b v=a\times(b\times v). \end{align*} Similarly, \begin{align*} \widehat b\widehat a v=b\times(a\times v). \end{align*} Therefore \begin{align*} [\widehat a,\widehat b]v=a\times(b\times v)-b\times(a\times v). \end{align*} [/step]

[step:Use the vector triple product identity to identify the resulting action]We use the vector triple product identity \begin{align*} x\times(y\times z)=y(x\cdot z)-z(x\cdot y) \end{align*} for $x,y,z\in\mathbb R^3$. Applying it twice gives \begin{align*} a\times(b\times v)=b(a\cdot v)-v(a\cdot b). \end{align*} Also, \begin{align*} b\times(a\times v)=a(b\cdot v)-v(a\cdot b). \end{align*} Subtracting these two identities yields \begin{align*} a\times(b\times v)-b\times(a\times v)=b(a\cdot v)-a(b\cdot v). \end{align*} The same vector triple product identity applied to $(a\times b)\times v$ gives \begin{align*} (a\times b)\times v=b(a\cdot v)-a(b\cdot v). \end{align*} Hence \begin{align*} [\widehat a,\widehat b]v=(a\times b)\times v. \end{align*}[/step]

[guided]The goal of this step is to recognize the expression obtained from the commutator as another hat-map action. Since $\widehat c$ acts on a vector $v$ by $c\times v$, we want to rewrite \begin{align*} a\times(b\times v)-b\times(a\times v) \end{align*} in the form $c\times v$ for a specific vector $c$. We use the vector triple product identity: for all $x,y,z\in\mathbb R^3$, \begin{align*} x\times(y\times z)=y(x\cdot z)-z(x\cdot y). \end{align*} With $x=a$, $y=b$, and $z=v$, this gives \begin{align*} a\times(b\times v)=b(a\cdot v)-v(a\cdot b). \end{align*} With $x=b$, $y=a$, and $z=v$, it gives \begin{align*} b\times(a\times v)=a(b\cdot v)-v(b\cdot a). \end{align*} Since the Euclidean dot product is symmetric, $b\cdot a=a\cdot b$. Therefore, subtracting the two displayed identities cancels the two terms involving $v(a\cdot b)$ and leaves \begin{align*} a\times(b\times v)-b\times(a\times v)=b(a\cdot v)-a(b\cdot v). \end{align*} Now apply the same triple product identity to $(a\times b)\times v$. This is the case $x=a\times b$, $y$ not used directly is better handled by the equivalent identity \begin{align*} (x\times y)\times z=y(x\cdot z)-x(y\cdot z). \end{align*} Taking $x=a$, $y=b$, and $z=v$, we obtain \begin{align*} (a\times b)\times v=b(a\cdot v)-a(b\cdot v). \end{align*} Combining this with the previous computation gives \begin{align*} a\times(b\times v)-b\times(a\times v)=(a\times b)\times v. \end{align*} Using the commutator computation from the preceding step, we conclude \begin{align*} [\widehat a,\widehat b]v=(a\times b)\times v. \end{align*}[/guided]

[step:Conclude equality of the matrices from equality on all vectors] By the defining property of the hat map applied to the vector $a\times b$, we have \begin{align*} \widehat{a\times b}\,v=(a\times b)\times v. \end{align*} The previous step proves that \begin{align*} [\widehat a,\widehat b]v=\widehat{a\times b}\,v \end{align*} for every $v\in\mathbb R^3$. Since two linear maps $\mathbb R^3\to\mathbb R^3$ that agree on every vector are equal, it follows that \begin{align*} [\widehat a,\widehat b]=\widehat{a\times b}. \end{align*} [/step]

[step:Identify the hat map as the Lie algebra isomorphism] The hat map $H:\mathbb R^3\to\mathfrak{so}(3)$, $a\mapsto\widehat a$, is linear and bijective by its standard coordinate description. The bracket identity just proved says precisely that for all $a,b\in\mathbb R^3$, \begin{align*} H(a\times b)=[H(a),H(b)]. \end{align*} Thus $H$ is a linear bijection preserving brackets. Therefore $H$ is an isomorphism from the Lie algebra $(\mathbb R^3,\times)$ to the Lie algebra $\mathfrak{so}(3)$. [/step]