Some results on U-cross Gram matrices by using K-frames

Abstract

U-cross Gram matrices are produced by frames and Riesz bases. In this paper, we represent bounded operators as matrix operators using K-frames. We study the invertibility matrices respect to K-frames. Moreover, we apply the concept of K-Riesz bases in Hilbert space ${\mathcal {H}}$ to the concept of matrix induced by U with respect to K-Riesz bases.

1 Introduction, notation and motivation

A unitary system is a set of unitary operators ${\mathcal {U}}$ acting on a Hilbert space ${\mathcal {H}}$ which contains the identity operator I of $B({\mathcal {H}})$. A Bessel generator for ${\mathcal {U}}$ is a vector $x\in {\mathcal {H}}$ with the property that ${\mathcal {U}}x :=\{Ux :U\in {\mathcal {U}}\}$ is Bessel sequence for ${\mathcal {H}}$. Many useful frames, which play an essential role in both theory and applications, can been considered as unitary systems, group-like unitary systems and atomic systems [16, 18]. K-frames were recently introduced by Gavruta to study atomic systems with respect to a bounded operator $K\in B({\mathcal {H}})$. It is a generalization of frame theory such that the lower bound is only satisfied for the elements in the range of K [17]. It is shown that an atomic system for K is a K-frame and vice versa. For this reason, K-frames are a useful mathematical tool to study the structure of unitary systems. Another purpose of this paper is to study Gram Matrices. The operator equation $Uf=v$ where $U\in B({\mathcal {H}})$ does not have a smooth solution (i.e. have all derivatives continuous) in general. It can be rewritten of the form

$$\begin{aligned} Ax=b \end{aligned}$$

(1.1)

where $A_{i,j}=\langle Ue_i,e_j\rangle$ and $\{e_i\}_{i\in I}$ is an orthonormal basis of H. To solve linear systems (1.1) variational method can be applied for example [25]. Recently, frames, Riesz bases and g-frames are applied to obtain (1.1) [3, 4, 12]. In this paper, we apply K-frames to get (1.1) as atomic decompositions of elements in the range of K which may not be closed.

Let ${\mathcal {H}}$ be a separable Hilbert space and K an operator from ${\mathcal {H}}$ to ${\mathcal {H}}$. A sequence $F:=\lbrace f_{i}\rbrace _{i \in I} \subseteq {\mathcal {H}}$ is called a K-frame for ${\mathcal {H}}$, if there exist constants $A, B > 0$ such that

$$\begin{aligned} A \Vert K^{*}f\Vert ^{2} \le \sum _{i\in I} \vert \langle f,f_{i}\rangle \vert ^{2} \le B \Vert f\Vert ^{2}, \quad (f\in {\mathcal {H}}). \end{aligned}$$

(1.2)

Clearly if $K=I_{{\mathcal {H}}}$, then F is an ordinary frame. The constants A and B in (1.2) are called lower and upper bounds of F,  respectively. We call F a A-tight K-frame if $A\Vert K^{*}f\Vert ^{2}=\sum _{i\in I}|\langle f,f_{i}\rangle |^{2}$ and a 1-tight K-frame as Parseval K-frame. A K-frame is called an exact K-frame, if by removing any element, the reminder sequence is not a K-frame.

Obviously, every K-frame is a Bessel sequence, hence similar to ordinary frames the synthesis operator can be defined as $T_{F}: l^{2}\rightarrow {\mathcal {H}}$; $T_{F}(\{ c_{i}\}_{i\in I}) = \sum _{i\in I} c_{i}f_{i}$. It is a bounded operator and its adjoint which is called the analysis operator given by $T_{F}^{*}(f)= \{ \langle f,f_{i}\rangle \}_{i\in I}$. Finally, the frame operator is given by $S_{F}: {\mathcal {H}} \rightarrow {\mathcal {H}}$; $S_{F}f = T_{F}T_{F}^{*}f = \sum _{i\in I}\langle f,f_{i}\rangle f_{i}$. Many properties of ordinary frames do not hold for K-frames, for example, the frame operator of a K-frame is not invertible in general. It is worthwhile to mention that if K has close range then $S_{F}$ from R(K) onto $S_{F}(R(K))$ is an invertible operator [24]. In particular,

$$\begin{aligned} B^{-1} \Vert f\Vert \le \Vert S_{F}^{-1}f\Vert \le A^{-1}\Vert K^{\dag }\Vert ^{2}\Vert f\Vert , \quad (f\in S_{F}(R(K))), \end{aligned}$$

(1.3)

where $K^{\dag }$ is the pseudo-inverse of K.

Let $\{ f_{i} \}_{i\in I}$ be a Bessel sequence. A Bessel sequence $\{ g_{i}\}_{i \in I}\subseteq {\mathcal {H}}$ is called a K-dual of $\{ f_{i} \}_{i\in I}$ if

$$\begin{aligned} Kf = \sum _{i\in I} \langle f,g_{i}\rangle \pi _{R(K)}f_{i}, \quad (f\in {\mathcal {H}}). \end{aligned}$$

(1.4)

In [17], it was shown that for every K-frame of ${\mathcal {H}}$ there exists at least a Bessel sequence $\{ g_{i}\}_{i \in I}$ which satisfies (1.4).

Let $F=\{f_{i}\}_{i\in I}$ be a K-frame. The Bessel sequence $\{ K^{*}S_{F}^{-1}\pi _{S_{F}(R(K))}f_{i} \}_{i\in I}$ can be considered as the canonical K-dual of F [1]. For simplicity, the canonical K-dual is denoted by $F^{\ddag }=\{f_{i}^{\ddag }\}_{i\in I}$. In the sequel, we show that for each $f\in {\mathcal {H}}$, the sequence $\{\langle f,f_{i}^{\ddag }\rangle \}_{i\in I}$ has minimal $\ell ^{2}$-norm among all sequences representating Kf.

The next proposition is important in K-frame theory.

Proposition 1.1

[14] Let $L_{1}\in B({\mathcal {H}}_{1},{\mathcal {H}})$ and $L_{2}\in B({\mathcal {H}}_{2},{\mathcal {H}})$ be two bounded operators. The following statements are equivalent:

• (1) $R(L_{1})\subseteq R(L_{2})$.

• (2) $L_{1}L_{1}^{*}\le \lambda ^{2}L_{2}L_{2}^{*}$ for some $\lambda \ge 0$.

• (3) there exists a bounded operator $X\in B({\mathcal {H}}_{1},{\mathcal {H}}_{2})$ so that $L_{1}=L_{2}X$

In this paper, we establish the notion of K-Riesz bases and show that, similar to ordinary frames, a K-Riesz basis has a unique K-dual. Also, try to state an operator as a matrix operator induced by K-frames and K-Riesz bases. More precisely, every K-frame is a Bessel sequence, and therefore we can induce matrix representations (3.1) for operators by K-frames. The inverse of such matrices are computed if they are exist. Moreover, we investigate sufficient conditions such that a matrix operator induced by K-frames is invertible. For more similar information see [5].

2 K-Riesz bases

In this section, we present K-Riesz sequences in ${\mathcal {H}}$ and investigate their properties. Also, we state K-Riesz bases and give some characterizations of this concept such as we prove that they are a unique K-dual. Throughout this paper we suppose K is a bounded operator with closed range.

Definition 2.1

A family $F:=\left\{ f_{i}\right\} _{i\in I}$ is called a K-Riesz sequence for ${\mathcal {H}}$ if there exists an injective bounded operator $U:{\mathcal {H}}\rightarrow {\mathcal {H}}$ such that $\left\{ \pi _{R(K)}f_{i}\right\} _{i\in I}=\left\{ Ue_{i}\right\} _{i\in I}$, where $\{e_{i}\}_{i\in I}$ is an orthonormal basis for ${\mathcal {H}}$. In addition, if F is a K-frame, then $\{f_{i}\}_{i\in I}$ is called a K-Riesz basis.

The next theorem, which used frequently throughout the paper, gives an equivalent condition for K-Riesz sequences.

Theorem 2.2

For a K-frame $F=\{f_{i}\}_{i\in I}$ in ${\mathcal {H}}$, the following are equivalent:

• (1) $\{f_{i}\}_{i\in I}$ is a K-Riesz basis for ${\mathcal {H}}$.

• (2) There exist constants $A,B>0$ such that for every finite scalar sequence $\{c_{i}\}_{i\in I}$,

  $$\begin{aligned} A\sum _{i\in I}|c_{i}|^{2}\le \left\| \sum _{i\in I}c_{i}\pi _{R(K)}f_{i}\right\| ^{2}\le B\sum _{i\in I}|c_{i}|^{2}. \end{aligned}$$

  (2.1)

Proof

$(1)\Rightarrow (2)$ Let $\{f_{i}\}_{i\in I}$ be a K-Riesz sequence. Then there exists an injective bounded operator $U:{\mathcal {H}}\rightarrow {\mathcal {H}}$ such that $Ue_{i}=\pi _{R(K)}f_{i}.$ Moreover, applying the lower K-frame condition and Theorem 1.1 we have

$$\begin{aligned} R(U)= & {} \pi _{R(K)}\pi _{R(T_F)}{\mathcal {H}}\\= & {} \pi _{R(T_F)}\pi _{R(K)}{\mathcal {H}}\\= & {} \pi _{R(K)}{\mathcal {H}}=R(K). \end{aligned}$$

In particular, R(U) is closed, and so U has a bounded left inverse denoted by L. Hence,

$$\begin{aligned} \left\| \sum _{i\in I}c_{i}\pi _{R(K)}f_{i}\right\| ^{2}=\left\| \sum _{i\in I}c_{i}Ue_{i}\right\| ^{2} \le \left\| U\right\| ^{2}\sum _{i\in I}|c_{i}|^{2} \end{aligned}$$

and

$$\begin{aligned} \sum _{i\in I}|c_{i}|^{2}=\left\| LU\sum _{i\in I}c_{i}e_{i}\right\| ^{2}\le \left\| L\right\| ^{2}\left\| \sum _{i\in I}c_{i}\pi _{R(K)}f_{i}\right\| ^{2}, \end{aligned}$$

for every finite scalar sequence $\{c_{i}\}_{i\in I}$.

$(2)\Rightarrow (1)$ Given

$$\begin{aligned} U:{\mathcal {H}}\rightarrow {\overline{span}}\left\{ \pi _{R(K)}f_{i}\right\} _{i\in I}, (e_{i}\mapsto \pi _{R(K)}f_{i}). \end{aligned}$$

Then (2.1) yields

$$\begin{aligned} A\sum _{i\in I}|c_{i}|^{2}= & {} \left\| \sum _{i\in I}c_{i}\pi _{R(K)}f_{i}\right\| ^{2}\\= & {} \left\| U\sum _{i\in I}c_{i}e_{i}\right\| ^{2}\le B\sum _{i\in I}|c_{i}|^{2}. \end{aligned}$$

So, U is bounded and injective. $\square$

The next corollary gives equivalent conditions for a Bessel sequence being a K-Riesz basis.

Corollary 2.3

Let $F=\{f_i\}_{i\in I}$ be a Bessel sequence in ${\mathcal {H}}$. The following are equivalent:

• (1) F is a K-Riesz basis.

• (2) F is a K-frame and

  $$\begin{aligned} A\sum _{i\in I}|c_{i}|^{2}\le \left\| \sum _{i\in I}c_{i}\pi _{R(K)}f_{i}\right\| ^{2}\le B\sum _{i\in I}|c_{i}|^{2}. \end{aligned}$$

• (3) $\pi _{R(K)}T_F$ is invertible from $\ell ^2$ onto R(K).

An overcomplete or redundant K-frame is a K-frame $\{f_{i}\}_{i\in I}$ such that $\{f_{i}\}_{i\in I}$ is not a K-Riesz basis. In other word, a $K^{*}$-frame $\{f_{i}\}_{i\in I}$ is redundant, if there exist coefficients $\{c_{i}\}_{i\in I}\in \ell ^{2}\setminus \{0\}$ for which $\sum _{i\in I}c_{i}\pi _{R(K)}f_{i}=0$. In fact, a K-frame $\{f_{i}\}_{i\in I}$ is a K-Riesz basis if the elements of $\{\pi _{R(K)}f_{i}\}_{i\in I}$ are independent.

Proposition 2.4

Let $\{f_{i}\}_{i\in I}$ be a Bessel sequence in ${\mathcal {H}}$. The following are equivalent:

• (1) $\{f_{i}\}_{i\in I}$ is K-Riesz sequence for ${\mathcal {H}}$.

• (2) $\{\pi _{R(K)}f_{i}\}_{i\in I}$ is a Riesz sequence.

• (3) $\{\pi _{R(K)}f_{i}\}_{i\in I}$ is $\omega -$independent.

Moreover, let $\{f_{i}\}_{i\in I}$ be a K-frame. Then $\{f_{i}\}_{i\in I}$ is a K-Riesz basis if and only if $\{\pi _{R(K)}f_{i}\}_{i\in I}$ is $\omega -$independent.

The relationship between K-Riesz bases and exact $K^{*}$-frames is discussed on the following proposition, see Theorem 3.3.2 of [11] for the ordinary case.

Proposition 2.5

Let $F=\{f_{i}\}_{i\in I}$ be a K-frame in ${\mathcal {H}}$. The following are equivalent.

• (1) F is a K-Riesz basis.

• (2) F has a unique K-dual in ${\mathcal {H}}$.

Proof

$(1)\Rightarrow (2)$ Assume that F is a K-Riesz basis of the form of $\{Ue_{i}\}_{i\in I},$ where $U\in B({\mathcal {H}})$ is injective. If $\{g_{i}\}_{i\in I}$ and $\{h_{i}\}_{i\in I}$ are K-dual of F, then

$$\begin{aligned} U\sum _{i\in I}\left\langle f,g_{i}\right\rangle e_{i}= & {} \sum _{i\in I}\left\langle f,g_{i}\right\rangle Ue_{i}\\= & {} \sum _{i\in I}\left\langle f,g_{i}\right\rangle \pi _{R(K)}f_{i}\\= & {} Kf\\= & {} \sum _{i\in I}\left\langle f,h_{i}\right\rangle \pi _{R(K)}f_{i}\\= & {} \sum _{i\in I}\left\langle f,h_{i}\right\rangle Ue_{i} =U\sum _{i\in I}\left\langle f,h_{i}\right\rangle e_{i}, \end{aligned}$$

for every $f\in {\mathcal {H}}$. The injectivity U induces that $\{g_{i}\}_{i\in I}=\{h_{i}\}_{i\in I}$.

$(2)\Rightarrow (1)$ Assume that F has a unique K-dual in ${\mathcal {H}}$. On the contrary, suppose that F is not a K-Riesz basis. Using Proposition 2.4 follows that $\{\pi _{R(K)f_i}\}$ is not a Riesz basis, or equivalently, $\pi _{R(K)}T_F$ is not injective. Choose $0\ne \{c_i\}_{i\in I}\in l^2$ such that

$$\begin{aligned} \pi _{R(K)} T_F\{c_i\}_{i\in I}=0. \end{aligned}$$

(2.2)

Defining the sequence $\{g_i\}_{i\in I}$ in ${\mathcal {H}}$ weakly by

$$\begin{aligned} \langle f,g_i\rangle =\langle f,f_{i}^{\ddag }\rangle + c_i, \quad (i\in I, f\in {\mathcal {H}}). \end{aligned}$$

(2.3)

Then $\{g_i\}_{i\in I}$ is a Bessel sequence. Moreover, applying (2.2) and (2.3) we obtain

$$\begin{aligned} \sum _{i\in I}\langle f,g_i\rangle \pi _{R(K)}f_i= & {} \sum _{i\in I}\left( \langle f,f_{i}^{\ddag }\rangle + c_i\right) \pi _{R(K)}f_i\\= & {} \sum _{i\in I}\langle f,f_{i}^{\ddag }\rangle \pi _{R(K)}f_i+\pi _{R(K)} T_F\{c_i\}_{i\in I}=Kf \end{aligned}$$

Hence, $\{g_i\}_{i\in I}$ is a K-dual of F and so $g_i=f_i^{\ddag }$, for all $i\in I$ by the assumption. This easily follows that $\{c_i\}=0$ which is impossible. $\square$

The question that may involve is that the relationship between Riesz bases and K-Riesz bases.

Corollary 2.6

Let $\{f_{i}\}_{i\in I}$ be a Bessel sequence in ${\mathcal {H}}$. If $\{f_{i}\}_{i\in I}$ is a Riesz basis, then $\{Kf_{i}\}_{i\in I}$ is a K-Riesz basis in ${\mathcal {H}}$. If $\{Kf_{i}\}_{i\in I}$ is a Riesz basis, then $\{f_{i}\}_{i\in I}$ is a K-Riesz sequence in ${\mathcal {H}}$.

3 U-Gram matrix with respect to K-frames

The standard matrix description of an operator U, using an orthonormal basis $\{e_{i}\}_{i\in I}$, is the matrix M defined by

$$\begin{aligned} (Mc)_{j}=\sum _{k}M_{jk}c_{k},\qquad (c=\{c_{k}\}_{k}\in \ell ^{2}), \end{aligned}$$

where $M_{jk}=\left\langle Ue_{k},e_{j}\right\rangle$. The same can be constructed with frames and their duals. More precisely, assume that $\Phi =\{\phi _{i}\}_{i\in I}$ and $\Psi =\{\psi _{i}\}_{i\in I}$ are a pair of dual frames and $Uf=v$ is an operator equation, then

$$\begin{aligned} \sum _{j}\langle U\psi _j,\phi _i\rangle \langle f,\phi _j\rangle= & {} \left\langle \sum _{j}\langle f,\phi _j\rangle \psi _j, U^*\phi _i\right\rangle \\= & {} \left\langle Uf, \phi _i\right\rangle =\left\langle v, \phi _i\right\rangle . \end{aligned}$$

Thus, the operator equation can be reduced to the linear system

$$\begin{aligned} \left( \left\langle U\psi _j,\phi _i\right\rangle \right) _{i,j}\left( \langle f,\phi _j\rangle \right) = \left( \left\langle v, \phi _i\right\rangle \right) . \end{aligned}$$

We say that $\left( \left\langle U\psi _j,\phi _i\right\rangle \right) _{i,j}$ is the matrix representation of U by using dual pairs $\Phi$ and $\Psi$. In [22], it is shown that operators can be described as the form of matrices by using fusion frames.

In this section, we represent an operator in $B({\mathcal {H}})$ as the form of a matrix in the base of K-frames. Also, we investigate its inverse if there exists.

Definition 3.1

Let $\Psi =\{\psi _{i}\}_{i\in I}$ be a Bessel sequence in ${\mathcal {H}}_{1}$ and $\Phi =\{\phi _{i}\}_{i\in I}$ a Bessel sequence in ${\mathcal {H}}_{2}$. For $U\in B({\mathcal {H}}_{1},{\mathcal {H}}_{2})$, the matrix induced by operator U with respect to the Bessel sequences $\Phi =\{\phi _{i}\}_{i\in I}$ and $\Psi =\{\psi _{i}\}_{i\in I}$, denoted by ${\mathbf {G}}_{U,\Phi ,\Psi }$, is given by

$$\begin{aligned} \left( {\mathbf {G}}_{U,\Phi ,\Psi }\right) _{i,j}=\left\langle U\psi _{j},\phi _{i}\right\rangle ,\qquad \left( i,j\in I\right) , \end{aligned}$$

(3.1)

for more details see [4]. It is straightforward to see that

$$\begin{aligned} {\mathbf {G}}_{U,\Phi ,\Psi }=T_{\Phi }^{*}UT_{\Psi }. \end{aligned}$$

(3.2)

Because of the operator representation (3.2), we call ${\mathbf {G}}_{U,\Phi ,\Psi }$ the U-cross Gram matrix of $\Phi$ and $\Psi$, respectively. In other word, ${\mathbf {G}}_{U,\Phi ,\Psi }$ is a bounded operator on $\ell ^{2}$ with $\left\| {\mathbf {G}}_{U,\Phi ,\Psi }\right\| \le \sqrt{B_{\Phi }B_{\Psi }}\Vert U\Vert$ and $\left( {\mathbf {G}}_{U,\Phi ,\Psi }\right) ^{*}={\mathbf {G}}_{U^{*},\Psi ,\Phi }$. If ${\mathcal {H}}_{1}={\mathcal {H}}_{2}$ and $U=I_{{\mathcal {H}}_{1}}$ it is called the cross Gram matrix and denoted by ${\mathbf {G}}_{\Phi ,\Psi }$. We use ${\mathbf {G}}_{\Psi }$ for ${\mathbf {G}}_{\Psi ,\Psi }$; the so called the Gram matrix [11].

An operator $U\in B({\mathcal {H}})$ has a K-right inverse (K-left inverse) if there exists an operator ${\mathcal {R}}\in B({\mathcal {H}})$ (resp. ${\mathcal {L}}\in B({\mathcal {H}})$), so that

$$\begin{aligned} U{\mathcal {R}}=K,\qquad (resp. \quad {\mathcal {L}}U=K), \end{aligned}$$

for $K\in B({\mathcal {H}})$. If ${\mathcal {R}}={\mathcal {L}}$, then ${\mathcal {R}}$ is the K-inverse of U.

Example 3.2

Let $K\in B({\mathcal {H}})$ and $\Phi =\{\phi _{i}\}_{i\in I}$ be a K-frame in ${\mathcal {H}}$. Then

• (1) ${\mathbf {G}}_{S_{\Phi }\left( K^{\dag }\right) ^{*},\Phi ,\Phi ^{\ddag }} ={\mathbf {G}}_{\Phi }$, when K is a closed range operator in $B({\mathcal {H}})$ and $\Phi \subseteq S_{\Phi }(R(K))$. Indeed, since $R(S_{\Phi }^{-1})\subseteq R(K)$ on $S_{\Phi }(R(K))$ and $KK^{\dag }=I\mid _{R(K)}$ we have

  $$\begin{aligned} {\mathbf {G}}_{S_{\Phi }\left( K^{\dag }\right) ^{*},\Phi ,\Phi ^{\ddag }}= & {} T_{\Phi }^{*}S_{\Phi }\left( K^{\dag }\right) ^{*} T_{\Phi ^{\ddag }}\\= & {} T_{\Phi }^{*}T_{S_{\Phi }\left( K^{\dag }\right) ^{*}K^{*}S_{\Phi }^{-1}\pi _{S_{\Phi }(R(K))}\Phi }\\= & {} T_{\Phi }^{*}T_{\Phi }={\mathbf {G}}_{\Phi }. \end{aligned}$$

• (2) If $R(U)\subseteq R(K)$ and ${\mathbf {G}}_{\left( S_{\Phi }^{-1}\right) ^{*}U\pi _{R(K)},\Phi ,\Phi }=I_{\ell ^{2}}$, then U is K-right invertible. According to the fact that $\Phi ^{\ddag }$ is a K-dual of $\Phi$, we have

  $$\begin{aligned} K= & {} {\pi _{R(K)}}T_{\Phi }T_{\Phi ^{\ddag }}^{*}\\= & {} {\pi _{R(K)}}T_{\Phi }{\mathbf {G}}_{\left( S_{\Phi }^{-1}\right) ^{*}U\pi _{R(K)},\Phi ,\Phi } T_{\Phi ^{\ddag }}^{*}\\= & {} {\pi _{R(K)}}T_{\Phi }T_{\Phi }^{*}\left( S_{\Phi }^{-1}\right) ^{*}U\pi _{R(K)}T_{\Phi } T_{\Phi ^{\ddag }}^{*}\\= & {} {\pi _{R(K)}}S_{\Phi }\left( S_{\Phi }^{-1}\right) ^{*}UK\\= & {} {\pi _{R(K)}}UK=UK. \end{aligned}$$

In the following we state a sufficient condition such that ${\mathbf {G}}_{U,\Phi ,\Phi ^{\ddag }}=I_{\ell ^2}.$

Theorem 3.3

Let $U,K\in B({\mathcal {H}})$ and $R(U)\subseteq R(K)$. If $\Phi$ is a K-frame such that ${\mathbf {G}}_{U,\Phi ,\Phi ^{\ddag }}=I_{\ell ^2}$, then $UK^*$ is a biorthogonal projection on R(K). The converse is true if $\Phi$ is a K-Riesz basis.

Proof

Using the duality formula we have

$$\begin{aligned} T_{\Phi ^{\ddag }}T_{\Phi }^*\pi _{R(K)}=K^*. \end{aligned}$$

Then

$$\begin{aligned} S_{\Phi }\pi _{R(K)}= & {} T_{\Phi }T_{\Phi }^*\pi _{R(K)}\\= & {} T_{\Phi }{\mathbf {G}}_{U,\Phi ,\Phi ^{\ddag }}T_{\Phi }^*\pi _{R(K)}\\= & {} T_{\Phi }T_{\Phi }^*UT_{\Phi ^{\ddag }}T_{\Phi }^*\pi _{R(K)}=S_{\Phi }UK^*. \end{aligned}$$

Applying $R(U)\subseteq R(K)$ implies that $S_{\Phi }$ is an invertible on R(U) and then $UK^*=\pi _{R(K)}$. Conversely, it is easy to see that

$$\begin{aligned} {\mathbf {G}}_{U,\Phi ,\Phi ^{\ddag }}T_{\Phi }^*\pi _{R(K)}= & {} T_{\Phi }^*UT_{\Phi ^{\ddag }}T_{\Phi }^*\pi _{R(K)}\\= & {} T_{\Phi }^*UK^*=T_{\Phi }^*\pi _{R(K)}. \end{aligned}$$

So, ${\mathbf {G}}_{U,\Phi ,\Phi ^{\ddag }}=I_{\ell ^2}$ if $T_{\Phi }^*\pi _{R(K)}$ is invertible and so by Proposition 2.3 if $\Phi$ is a K-Riesz basis. $\square$

It is worthwhile to mention that from the one sided invertibility of Gram matrix induced by $K\in B({\mathcal {H}})$ with respect to Bessel sequences $\Psi$ and $\Phi$, respectively, it follows that the Bessel sequences $K\Psi$ and $K^*\Phi$ are K-Riesz sequence and $K^*$-Riesz sequence, respectively.

Theorem 3.4

Let $\Psi =\{\psi _{i}\}_{i\in I}$ be a Bessel sequence in ${\mathcal {H}}$.

• (1) If ${\mathbf {G}}_{K,\Phi ,\Psi }$ has a left inverse, then $K\Psi$ is a K-Riesz sequence in ${\mathcal {H}}$.

• (2) If ${\mathbf {G}}_{K,\Phi ,\Psi }$ has a right inverse, then $K^*\Phi$ is a $K^{*}$-Riesz sequence in ${\mathcal {H}}$.

Proof

Let ${\mathcal {L}}$ be a left inverse of ${\mathbf {G}}_{K,\Phi ,\Psi }$. Then

$$\begin{aligned} I_{\ell ^{2}}= & {} {\mathcal {L}}{\mathbf {G}}_{K,\Phi ,\Psi }\\= & {} {\mathcal {L}}T_{\Phi }^{*}KT_{\Psi }={\mathcal {L}}T_{\Phi }^{*}T_{K\Psi }. \end{aligned}$$

The above computations show that $T_{K\Psi }$ has a left inverse and so $T_{K\Psi }$ is an injective operator. Hence, by applying Lemma 2.4.1 of [11] there exists $A>0$ such that

$$\begin{aligned} A\sum _{i\in I}|c_{i}|^{2}\le & {} \left\| \sum _{i\in I}c_{i}K\psi _{i}\right\| ^{2}= \left\| \sum _{i\in I}c_{i}\pi _{R(K)}K\psi _{i}\right\| ^{2}. \end{aligned}$$

Therefore, $K\Psi$ is a K-Riesz sequence in ${\mathcal {H}}$ by Theorem 2.2.

The proof of (2) is similar. $\square$

Now, we study several K-duals of a K-frame by its Gram matrix.

Theorem 3.5

Let $\Phi =\{\phi _{i}\}_{i\in I}$ be a K-frame in ${\mathcal {H}}$. Then ${\mathbf {G}}_{\Phi }=I_{\ell ^{2}}$ on $R(T_{\Phi }^{*}K)$ if and only if $S_{\Phi }=I_{{\mathcal {H}}}$ on R(K). In this case $\phi _{i}^{\ddag }=K^{*}\pi _{R(K)}\phi _{i}$, for all $i\in I$.

Proof

We first claim that ${\mathbf {G}}_{\Phi }T_{\Phi }^{*}K=T_{\Phi }^{*}K$ if and only if $S_{\Phi }K=K$. Assume that ${\mathbf {G}}_{\Phi }T_{\Phi }^{*}K=T_{\Phi }^{*}K$. Then

$$\begin{aligned} S_{\Phi }S_{\Phi }K= & {} T_{\Phi }T_{\Phi }^{*}T_{\Phi }T_{\Phi }^{*}K\\= & {} T_{\Phi }{\mathbf {G}}_{\Phi }T_{\Phi }^{*}K\\= & {} T_{\Phi }T_{\Phi }^{*}K=S_{\Phi }K. \end{aligned}$$

The invertibility $S_{\Phi }$ on R(K) implies that $S_{\Phi }K=K$. For the reverse,

$$\begin{aligned} {\mathbf {G}}_{\Phi }T_{\Phi }^{*}K=T_{\Phi }^{*}T_{\Phi }T_{\Phi }^{*}K=T_{\Phi }^{*} S_{\Phi }K=T_{\Phi }^{*}K. \end{aligned}$$

Moreover,

$$\begin{aligned} \phi _{i}^{\ddag }= & {} K^{*}S_{\Phi }^{-1}\pi _{S_{\Phi }(R(K))}\phi _{i}\\= & {} K^{*}S_{\Phi }S_{\Phi }^{-1}\pi _{S_{\Phi }R(K)}\phi _{i}\\= & {} K^{*}\pi _{S_{\Phi }R(K)} \phi _{i}=K^{*}\pi _{R(K)} \phi _{i}, \end{aligned}$$

for all $i\in I$. It completes the proof. $\square$

We are going to construct several K-duals for some Bessel sequences in which their Gram matrices are invertible from the left or from the right.

Theorem 3.6

Let $\Phi =\{\phi _{i}\}_{i\in I}$ and $\Psi =\{\psi _{i}\}_{i\in I}$ be a K-frame and a Bessel sequence in ${\mathcal {H}}$, respectively. Also, $U\in B({\mathcal {H}})$ and ${\mathbf {G}}_{U,\Phi ,\Psi }$ has a right inverse as ${\mathcal {R}}$. Then

• (1) $\left\{ S_{\Phi }^{-1}\pi _{S_{\Phi }(R(K))}\phi _{i}\right\} _{i\in I}$ is a K-frame. Moreover, $\left\{ K^{*}T_{\Phi }{\mathcal {R}}^{*}T_{\Psi }^{*}U^{*}\phi _{i}\right\} _{i\in I}$ is a K-dual of $\left\{ S_{\Phi }^{-1}\pi _{S_{\Phi }(R(K))}\phi _{i}\right\} _{i\in I}$.

• (2) for every K-dual $\Phi ^{d}=\left\{ \phi _{i}^{d}\right\} _{i\in I}$, the $K^{*}$-frame $\left\{ T_{\Phi ^{d}}{\mathcal {R}}^{*}T_{\Psi }^{*}U^{*}\phi _{i}\right\} _{i\in I}$ is a K-dual of $\Phi$.

Proof

Notice that $S_{\Phi }^{-1}S_{\Phi }=I$ on R(K). So, for all $f\in {\mathcal {H}}$, we obtain

(1)

$$\begin{aligned} Kf= & {} S_{\Phi }^{-1}S_{\Phi }Kf\\= & {} S_{\Phi }^{-1}T_{\Phi }{\mathbf {G}}_{U,\Phi ,\Psi }{\mathcal {R}}T_{\Phi }^{*}Kf\\= & {} S_{\Phi }^{-1}T_{\Phi }T_{\Phi }^{*}UT_{\Psi }{\mathcal {R}}T_{\Phi }^{*}Kf\\= & {} \sum _{i\in I}\left\langle f,K^{*}T_{\Phi }{\mathcal {R}}^{*} T_{\Psi }^{*}U^{*}\phi _{i}\right\rangle \pi _{R(K)} S_{\Phi }^{-1}\pi _{S_{\Phi }(R(K))}\phi _{i}. \end{aligned}$$

The existence of K-dual for $\left\{ S_{\Phi }^{-1}\pi _{S_{\Phi }(R(K))}\phi _{i}\right\} _{i\in I}$ proves the considered sequence is a K-frame.

(2) Using (1.4) we obtain

$$\begin{aligned} Kf= & {} \pi _{R(K)}T_{\Phi }T_{\Phi ^{d}}f\\= & {} \pi _{R(K)}T_{\Phi }{\mathbf {G}}_{U,\Phi ,\Psi }{\mathcal {R}}T_{\Phi ^{d}}f\\= & {} \pi _{R(K)}T_{\Phi }T_{\Phi }^{*}UT_{\Psi }{\mathcal {R}}T_{\Phi ^{d}}f\\= & {} \sum _{i\in I}\left\langle f,T_{\Phi ^{d}}{\mathcal {R}}^{*} T_{\Psi }^{*}U^{*}\phi _{i}\right\rangle \pi _{R(K)}\phi _{i}. \end{aligned}$$

$\square$

In the following, we present pairs of K-duals by means of the one sided invertibility of Gram matrices.

Corollary 3.7

Let $\Phi =\{\phi _{i}\}_{i\in I}$ be a Bessel sequence and $\Psi =\{\psi _{i}\}_{i\in I}$ a K-frame in ${\mathcal {H}}$. Also, let $U\in B({\mathcal {H}})$, and ${\mathcal {L}}$ be a left inverse of ${\mathbf {G}}_{U,\Phi ,\Psi }$. Then

• (1) $\left\{ K^{*}\psi _{i}\right\} _{i\in I}$ is a K-dual of $\left\{ S_{\Psi }^{-1}\pi _{S_{\Psi }(R(K))}T_{\Psi }{\mathcal {L}} T_{\Phi }^{*}U\psi _{i}\right\} _{i\in I}$.

• (2) $\left\{ K^{*}S_{\Psi }^{-1}\pi _{S_{\Psi }(R(K))}\psi _{i}\right\} _{i\in I}$ is a K-dual of $\left\{ T_{\Psi }{\mathcal {L}}T_{\Phi }^{*}U\psi _{i}\right\} _{i\in I}$.

• (3) $\Psi ^{d}=\{\psi _{i}^{d}\}_{i\in I}$ is a K-dual of $\left\{ T_{\Psi }{\mathcal {L}}T_{\Phi }^{*}U\psi _{i}\right\} _{i\in I}$, where $\Psi ^{d}$ is a K-dual of $\Psi$.

Proof

(1) Let ${\mathcal {L}}\in B({\mathcal {H}})$ be a left inverse of ${\mathbf {G}}_{U, \Phi ,\Psi }$. It is clear to see that $\left\{ K^{*}\psi _{i}\right\} _{i\in I}$ and $\left\{ S_{\Psi }^{-1}\pi _{S_{\Psi }(R(K))}T_{\Psi }{\mathcal {L}} T_{\Phi }^{*}U\psi _{i}\right\} _{i\in I}$ are Bessel sequences in ${\mathcal {H}}$. Moreover,

$$\begin{aligned} Kf= & {} S_{\Psi }^{-1}S_{\Psi }Kf\\= & {} \pi _{R(K)}S_{\Psi }^{-1}S_{\Psi }Kf\\= & {} \pi _{R(K)}S_{\Psi }^{-1}\pi _{S_{\Psi }R(K)}T_{\Psi }T_{\Psi }^*Kf\\= & {} \pi _{R(K)}S_{\Psi }^{-1}\pi _{S_{\Psi }R(K)}T_{\Psi }{\mathcal {L}}{\mathbf {G}}_{U,\Phi ,\Psi }T_{\Psi }^*Kf\\= & {} \sum _{i\in I}\left\langle f,K^*\psi _i\right\rangle \pi _{R(K)}S_{\Psi }^{-1}\pi _{S_{\Psi }R(K)} T_{\Psi }{\mathcal {L}}T_{\phi }^*U\psi _i, \end{aligned}$$

for all $f\in {\mathcal {H}}$. The rest is similar. $\square$

In the following we present a K-dual for a K-frame by some its K-duals.

Proposition 3.8

Assume that $\Psi =\{\psi _{i}\}_{i\in I}$ is a K-dual of a K-frame $\Phi =\{\phi _{i}\}_{i\in I}$. If $U\in B({\mathcal {H}})$ such that ${\mathbf {G}}_{U,\Phi ,\Psi }=I_{\ell ^{2}}$, then $S_{\Psi }^{*}U^{*}\Phi$ is a K-dual of $\Phi$. In particular, if $\Psi$ is the canonical K-dual of $\Phi$ and $KU^*=I_{{\mathcal {H}}}$, then $S_{\Psi }^{*}U^{*}\Phi =\Psi$.

Proof

For all $f\in {\mathcal {H}}$ we have

$$\begin{aligned} Kf= & {} \pi _{R(K)}T_{\Phi }T_{\Psi }^{*}f\\= & {} \pi _{R(K)}T_{\Phi }{\mathbf {G}}_{U,\Phi ,\Psi }T_{\Psi }^{*}f\\= & {} \pi _{R(K)}T_{\Phi }T_{\Phi }^{*}UT_{\Psi }T_{\Psi }^{*}f\\= & {} \pi _{R(K)}S_{\Phi }US_{\Psi }f\\= & {} \sum _{i\in I}\left\langle f,S_{\Psi }^{*}U^{*}\phi _{i}\right\rangle \pi _{R(K)}\phi _{i}. \end{aligned}$$

In particular, if $\Psi$ is the canonical K-dual of $\Phi$, then we have

$$\begin{aligned} S_{\Psi }f= & {} \sum _{i\in I}\left\langle f , K^{*}S_{\Phi }^{-1}\pi _{S_{\Phi }R(K)}\phi _{i}\right\rangle K^{*}S_{\Phi }^{-1}\pi _{S_{\Phi }R(K)}\phi _{i}\\= & {} K^{*}S_{\Phi }^{-1}\pi _{S_{\Phi }R(K)}S_{\Phi }(S_{\Phi }^{-1})^{*}Kf. \end{aligned}$$

Now, by using $KU^{*}=I_{{\mathcal {H}}}$ we obtain

$$\begin{aligned} S_{\Psi }^{*}U^*\Phi= & {} K^{*}S_{\Phi }^{-1}\pi _{S_{\Phi }R(K)}S_{\Phi }^{*}(S_{\Phi }^{-1})^{*}KU^*\Phi \\= & {} K^{*}S_{\Phi }^{-1}\pi _{S_{\Phi }R(K)}\Phi =\Psi . \end{aligned}$$

$\square$