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I want to find an $R \in \mathbb{R}^{2 \times 2}$ such that the following equality is satisfied: $$gRg^T = I_{3 \times 3},$$

where $I_{3 \times 3}$ is the identity matrix of dimension 3 and $g \in \mathbb{R}^{3 \times 2}$ and $g$ has a full (column) rank of 2.

My approach was to solve this, using the Pseudo inverse, for which I obtain $$R = g^{\dagger} (g^{T})^{\dagger}.$$

However, this doesn’t work because for the matrix $g$ only a left-pseudo inverse exists and not a right one which of course doesn’t satisfy the equality, i.e., $$g g^{\dagger} (g^{T})^{\dagger} g^T \neq I_{3 \times 3}.$$

So, is there a way to solve for $R$?

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    $\begingroup$ the rank of your $g R g^T$ is, at most, 2. $\endgroup$ Commented Mar 17 at 3:26

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