# Injectivity of a function with matrix

I'm trying to prove injectivity of $$g_a:$$ $$\begin{array}[t]{lrcl} \mathcal{S}_d^{++}(\mathbb{R}) & \longrightarrow & \mathcal{S}_d(\mathbb{R}) \\ \Gamma & \longmapsto & a\Gamma - \Gamma^{-1} \end{array}$$ for $$a > 0$$ (domain is symetric definite positive matrices and codomain symetric matrices).

Let $$\Gamma \in \mathcal{S}_d^{++}(\mathbb{R})$$ and $$\Sigma \in \mathcal{S}_d^{++}(\mathbb{R})$$ be such that $$g_a(\Gamma) = g_a(\Sigma)$$.

Thus, $$a(\Gamma - \Sigma) = \Gamma^{-1} - \Sigma^{-1}$$ and, with $$\Lambda := \sqrt{\Sigma}^{-1}\Gamma \sqrt{\Sigma}^{-1}$$,

$$a(\sqrt{\Sigma}\Lambda \sqrt{\Sigma} - \sqrt{\Sigma}\sqrt{\Sigma}) = \sqrt{\Sigma}^{-1}\Lambda^{-1}\sqrt{\Sigma}^{-1} - \sqrt{\Sigma}^{-1}\sqrt{\Sigma}^{-1}$$ (since $$\sqrt{\Sigma}^{-1}\sqrt{\Sigma}^{-1} = (\sqrt{\Sigma}\sqrt{\Sigma})^{-1}$$)

so $$a\sqrt{\Sigma}(\Lambda - I_d) \sqrt{\Sigma} = \sqrt{\Sigma}^{-1}(\Lambda^{-1} - I_d)\sqrt{\Sigma}^{-1}$$

or $$a\, \Sigma(\Lambda - I_d) \Sigma = \Lambda^{-1}- I_d$$.

We then notice $$\Lambda = \sqrt{\Sigma}^{-1}\Gamma \sqrt{\Sigma}^{-1}$$ is still definite positive so $$\Lambda = PDP^{-1}$$ for $$P \in \text{GL}_d(\mathbb{R})$$ and $$D = \text{diag}(\lambda_1, \dots, \lambda_d)$$ where $$\forall i \in \{1, \dots, d\}$$, $$\lambda_i > 0$$.

Thus, $$a\,\Sigma(PDP^{-1} - I_d)\Sigma = PD^{-1}P^{-1} - I_d$$

so $$a\,\Sigma P(D - I_d) P^{-1}\Sigma = P(D^{-1} - I_d)P^{-1}$$

and $$a\, P^{-1}\Sigma P(D - I_d)P^{-1}\Sigma P = D^{-1}- I_d$$

and finally $$a\,Q\,\text{diag}(\lambda_1 \,-\, 1, \dots, \lambda_d \,- \,1)\,Q = \text{diag}(\lambda_1^{-1} \,-\, 1, \dots, \lambda_d^{-1}\, - \,1) \text{ with } Q := P^{-1}\Sigma P$$.

Now, if it was $$a\,Q\,\text{diag}(\lambda_1 \,-\, 1, \dots, \lambda_d \,- \,1)\,Q^{-1} = \text{diag}(\lambda_1^{-1} \,-\, 1, \dots, \lambda_d^{-1}\, - \,1)$$

I could conclude by equating the eigenvalues but it's not: how can I finish? I want to prove that all $$\lambda_i = 1$$ so that $$\Lambda = I_d$$ and $$\Sigma = \Gamma$$.