Eigenvalues of $T(v) = (a\times v)\times b$

Question

Fix two elements $ a,b$ of $\mathbb{R}^3$ with $ a\cdot b\neq 0$. Let $T : \mathbb{R}^3\to \mathbb{R}^3$ be linear transformation given by $$T(v) = (a\times v)\times b$$ where $\times$ is crossproduct. What is the eigenvalue of T?

It is clear that $0$ is an eigenvalue and $a$ is its corresponding eigenvector. Is there any other eigenvalue?

Similar problem that I can solve is $T_1$ defined by $T_1(v) = a\times v$ with only $0$ as its eigenvalue, but the same argument don't work so far. Am I missing something? Any hint?

Answer 1

Recall the vector triple product $$ (\mathbf a\times\mathbf v)\times\mathbf b=(\mathbf a\cdot\mathbf b)\mathbf v-(\mathbf b\cdot\mathbf v)\mathbf a $$ so if $\mathbf{b}\cdot\mathbf{v}=0$, we have an eigenvector with eigenvalue $\mathbf{a}\cdot\mathbf{b}\neq 0$.

Answer 2

First, observe that you have a triple product. By the following formula: $$(a\times b)\times c=-(c\cdot b)a+(c\cdot a)b$$ we have $T(v)=-(b\cdot v)a+(b\cdot a)v$.

Now let $(\lambda,v)$ such that $T(v)=\lambda v$. Then

$$(b\cdot v)a=(b\cdot a-\lambda)v$$

Case $1$: $b\cdot v=0$ and $\lambda=b\cdot a$. Actually in this case you get $b^{\perp}$ as the eigenspace of the eigenvalue $b\cdot a$. Its dimension is $2$.

Case $2$: $b\cdot v\neq 0$. In this case you can write $v=\mu a$. It yields $\lambda =0$. Indeed, $T(a)=0$. Hence you get $a$ as the eigenspace of the eigenvalue $0$. Its dimension is $1$.

Putting this together, we conclude that there exists a basis of $\mathbb{R}^3$ in which $T$ is the diagonal matrix $\text{diag}(b\cdot a,b\cdot a,0)$, i.e. $T$ is the projection into the plane $b^{\perp}$ in the direction of $a$.