Let ${(\phi_n(.),\psi_n(.))}_{n\geq 1}$ be a Riesz basis in $H^1_0(0,1) \times L^2(0,1)$. My question is the following: If we multiply the basis by the matrix $e^{Mx}$ where $M$ is a constant matrix of dimension $2$, is this still a Riesz basis? I can prove that the operator $T: H^1_0(0,1) \times L^2(0,1) \rightarrow H^1_0(0,1) \times L^2(0,1)$ taking $(f,g)\rightarrow e^{Mx}(f,g)$ is bounded from above, but not from below. Is this statement true?.

Thank you in advance.

Let ${(\phi_n(.),\psi_n(.))}_{n\geq 1}$ be a Riesz basis in $H^1_0(0,1) \times L^2(0,1)$. My question is the following: If we multiply the basis by the matrix $e^{Mx}$, $x \in (0,1)$ where $M$ is a constant matrix of dimension $2$, is this still a Riesz basis? I can prove that the operator $T: H^1_0(0,1) \times L^2(0,1) \rightarrow H^1_0(0,1) \times L^2(0,1)$ taking $(f,g)\rightarrow e^{Mx}(f,g)$ is bounded from above, but not from below. Is this statement true?.

Thank you in advance.

Let ${(\phi_n(.),\psi_n(.))}_{n\geq 1}$ be a Riesz basis in $H^1_0(0,1) \times L^2(0,1)$. My question is the following: If we multiply the basis by the matrix $e^{Mx}$ where $M$ is a constant matrix of dimension $2$, is this still a Riesz basis? I can prove that the operator $T: H^1_0(0,1) \times L^2(0,1) \rightarrow H^1_0(0,1) \times L^2(0,1)$ taking $(f,g)\rightarrow e^{Mx}(f,g)$ is bounded from above, but not from below. Is this statement true?

Let ${(\phi_n(.),\psi_n(.))}_{n\geq 1}$ be a Riesz basis in $H^1_0(0,1) \times L^2(0,1)$. My question is the following: If we multiply the basis by the matrix $e^{Mx}$ where $M$ is a constant matrix of dimension $2$, is this still a Riesz basis? I can prove that the operator $T: H^1_0(0,1) \times L^2(0,1) \rightarrow H^1_0(0,1) \times L^2(0,1)$ taking $(f,g)\rightarrow e^{Mx}(f,g)$ is bounded from above, but not from below. Is this statement true?.

Thank you in advance.

Let ${(\phi_n(.),\psi_n(.))}_{n\geq 1}$ be a Riesz basis in $H^1_0(0,1) \times L^2(0,1)$. My question is the following: If we multiply the basis by the matrix $e^{Mx}$ where $M$ is a constant matrix of dimension $2$. Is this still a Riesz basis?. I can prove that the operator $T: H^1_0(0,1) \times L^2(0,1) \rightarrow H^1_0(0,1) \times L^2(0,1)$ taking $(f,g)\rightarrow e^{Mx}(f,g)$ is bounded from above, but not from below. Is this statement if true?.

Thank you.

Let ${(\phi_n(.),\psi_n(.))}_{n\geq 1}$ be a Riesz basis in $H^1_0(0,1) \times L^2(0,1)$. My question is the following: If we multiply the basis by the matrix $e^{Mx}$ where $M$ is a constant matrix of dimension $2$, is this still a Riesz basis? I can prove that the operator $T: H^1_0(0,1) \times L^2(0,1) \rightarrow H^1_0(0,1) \times L^2(0,1)$ taking $(f,g)\rightarrow e^{Mx}(f,g)$ is bounded from above, but not from below. Is this statement true?