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Let $\alpha_1^{(s)},\dots, \alpha_n^{(s)} \in \mathbb{C}, \quad s=1, \ldots, r.$

Question:

Are the following conditions equivalent:

1. There are unitary operators $U_1, \ldots, U_n\in B(H)$ such that $\alpha_1^{(s)} U_1 +\dots +\alpha_n^{(s)} U_n =0, \quad s=1, \ldots, r.$

2. There are $k$ and unitary operators $V_1, \ldots, V_n \in M_{k}(\mathbb{C})$ such that $\alpha_1^{(s)} V_1 +\dots +\alpha_n^{(s)} V_n =0, \quad s=1, \ldots, r.$

NOTE: if $\alpha_1^{(s)},\dots, \alpha_n^{(s)} \in \mathbb{R}, \quad s=1, \ldots, r$, than conditions are equivalent.

Indeed, assume that 1 holds. Let $v_i=U_i e_1$ be the first column of $U_i$, then there is a unitary operator $W$ such that $v'_i=Wv_i\in \mathbb{C}^k$ for some $k$. Obviously we have

$\alpha_1^{(s)} v'_1 +\dots +\alpha_n^{(s)} v'_n =0, \quad s=1, \ldots, r.$

Let $Re(v)$ and $Im(v)$ be vectors that are obtained taking real and imaginary part of the entries of $v\in\mathbb{C}^k$.

Then $w_i=(Re(v'_i), Im(v'_i))\in \mathbb{R}^k$ satisfying again equations. Since the matrix $A=(\langle w_i,w_j\rangle)$ is positive with $1$ on the diagonal and real coefficients, we have $A=(\langle w_i,w_j\rangle)=(tr(V_iV_j^*))$, where $V_i\in M_{2^k}(\mathbb{C})$ can be chosen as generators of Clifford algebra.

Let $\alpha_s=(\alpha_1^{(s)},...,\alpha_n^{(s)})$, then $A\alpha_s^t=0$, thus $\alpha_s(tr(V_iV_j^*))\alpha_s^t=0$ for every $s$. Therefore $tr((\alpha_1^{(s)} V_1 +\dots +\alpha_n^{(s)} V_n)^*(\alpha_1^{(s)} V_1 +\dots +\alpha_n^{(s)} V_n))=0$ and thus $\alpha_1^{(s)} V_1 +\dots +\alpha_n^{(s)} V_n=0$ for all $s$.

UPDATE: the question is related to http://mathoverflow.net/questions/53359/commuting-unitaries

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Let $\alpha_1^{(s)},\dots, \alpha_n^{(s)} \in \mathbb{C}, \quad s=1, \ldots, r.$

Question:

Are the following conditions equivalent:

1. There are unitary operators $U_1, \ldots, U_n\in B(H)$ such that $\alpha_1^{(s)} U_1 +\dots +\alpha_n^{(s)} U_n =0, \quad s=1, \ldots, r.$

2. There are $k$ and unitary operators $V_1, \ldots, V_n \in M_{k}(\mathbb{C})$ such that $\alpha_1^{(s)} V_1 +\dots +\alpha_n^{(s)} V_n =0, \quad s=1, \ldots, r.$

NOTE: if $\alpha_1^{(s)},\dots, \alpha_n^{(s)} \in \mathbb{R}, \quad s=1, \ldots, r$, than conditions are equivalent.

Indeed, assume that 1 holds. Let $v_i=U_i e_1$ be the first column of $U_i$, then there is a unitary operator $W$ such that $v'_i=Wv_i\in \mathbb{C}^k$ for some $k$. Obviously we have

$\alpha_1^{(s)} v'_1 +\dots +\alpha_n^{(s)} v'_n =0, \quad s=1, \ldots, r.$

Let $Re(v)$ and $Im(v)$ be vectors that are obtained taking real and imaginary part of the entries of $v\in\mathbb{C}^k$.

Then $w_i=(Re(v'_i), Im(v'_i))\in \mathbb{R}^k$ satisfying again equations. Since the matrix $A=(\langle w_i,w_j\rangle)$ is positive with $1$ on the diagonal and real coefficients, we have $A=(\langle w_i,w_j\rangle)=(tr(V_iV_j^*))$, where $V_i\in M_{2^k}(\mathbb{C})$ can be chosen as generators of Clifford algebra.

Let $\alpha_s=(\alpha_1^{(s)},...,\alpha_n^{(s)})$, then $A\alpha_s^t=0$, thus $\alpha_s(tr(V_iV_j^*))\alpha_s^t=0$ for every $s$. Therefore $tr((\alpha_1^{(s)} V_1 +\dots +\alpha_n^{(s)} V_n)^*(\alpha_1^{(s)} V_1 +\dots +\alpha_n^{(s)} V_n))=0$ and thus $\alpha_1^{(s)} V_1 +\dots +\alpha_n^{(s)} V_n=0$ for all $s$.

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# Linear equations inunitaries on infinitedimensionalHilbertspacevsfiniteoneunitaryoperators

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