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I am currently working with Andrew McInerney book "First Steos in Differential Geometry: Riemannian, Contact, Symplectic". There is a problem in chapter 2 (Linear Algebra Essentials) where I could solve the (a) part of the question but I struggled with the (b). Therefore I hope someone can give a hint or solution to the (b). The problem is stated as follows:

Let $\alpha \in (\mathbb{R}^3)^*$ be given by $\alpha(x,y,z) = 4y+z$.

  • (a) Find the $\ker \alpha$.
  • (b) Find all linear tranformations $\psi : \mathbb{R}^3 \to \mathbb{R}^3$ with the property that $\psi^*\alpha=\alpha$.
ShireHobbit145
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    Could you tell us what you've tried so far? That helps providing you with a better answer. – HSN Apr 04 '23 at 11:50
  • Yes, in particular how you express $\psi^*\alpha.$ – Anne Bauval Apr 04 '23 at 12:13
  • If you like to solve this matricially, this can help. – Anne Bauval Apr 04 '23 at 12:19
  • I tried with $(\psi^* \alpha)(v) = \alpha(\psi(v))$, if $\alpha(x,y,z)= 4y+z$, then I concluded that $(\psi^* \alpha)(v) = \alpha(\psi(v))= 4\psi_{2}(v) + \psi_{3}(v)$. Assuming the standard dual basis then $4\psi_{2}(0,1,0) + \psi_{3}(0,1,0)= 4$ and $4\psi_{2}(0,0,1) + \psi_{3}(0,0,1)= 1$. And I dont know where the next step is or if the begging of my solutions is wrongly reasoned. – ShireHobbit145 Apr 04 '23 at 12:28
  • The space of linear transformations $\Bbb R^3\to\Bbb R^3$ is $9$-dimensional, and you have only three equations to determine $\psi$. So you expect a (linear) $6$-dimensional space of such linear transformations. – Ted Shifrin Apr 04 '23 at 16:31
  • In what sense do you want to find the linear transformations in (b)? Parametrize that set? Find a basis for ${\psi - 1}$? – ronno Apr 05 '23 at 14:36

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