1
$\begingroup$

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 $\Sigma = \Gamma$.

$\endgroup$

0

You must log in to answer this question.

Browse other questions tagged .