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Edit: I correct the mistake pointed out by Matthew in his comment.

In fact, any matrix $X$ can be written as $X=\lambda(U+V)$ for unitary matrices $U,V$, and $\lambda$ can be taken as a half of the operator norm of $X$. For a proof, see thisthis question. This is slightly stronger than Matthew's comment, and the proof works in any finite von Neumann algebra (i.e. when the partial isometry in the polar decomposition can be taken as a unitary). In a $C^*$-algebra this is not possible (consider $z \mapsto z$ in the $C^*$-algebra of continuous functions on the unit disc of the complex plane) but, as noted by Matthew, you can get the same decomposition with $4$ unitaries.

Edit: I correct the mistake pointed out by Matthew in his comment.

In fact, any matrix $X$ can be written as $X=\lambda(U+V)$ for unitary matrices $U,V$, and $\lambda$ can be taken as a half of the operator norm of $X$. For a proof, see this question. This is slightly stronger than Matthew's comment, and the proof works in any finite von Neumann algebra (i.e. when the partial isometry in the polar decomposition can be taken as a unitary). In a $C^*$-algebra this is not possible (consider $z \mapsto z$ in the $C^*$-algebra of continuous functions on the unit disc of the complex plane) but, as noted by Matthew, you can get the same decomposition with $4$ unitaries.

Edit: I correct the mistake pointed out by Matthew in his comment.

In fact, any matrix $X$ can be written as $X=\lambda(U+V)$ for unitary matrices $U,V$, and $\lambda$ can be taken as a half of the operator norm of $X$. For a proof, see this question. This is slightly stronger than Matthew's comment, and the proof works in any finite von Neumann algebra (i.e. when the partial isometry in the polar decomposition can be taken as a unitary). In a $C^*$-algebra this is not possible (consider $z \mapsto z$ in the $C^*$-algebra of continuous functions on the unit disc of the complex plane) but, as noted by Matthew, you can get the same decomposition with $4$ unitaries.

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Edit: I correct the mistake pointed out by Matthew in his comment.

In fact, any matrix $X$ can be written as $X=\lambda(U+V)$ for unitary matrices $U,V$, and $\lambda$ can be taken as a half of the operator norm of $X$. For a proof, see this question. This is slightly stronger than Matthew's comment, and the proof works in any finite von Neumann algebra (i.e. when the partial isometry in the polar decomposition can be taken as a unitary). In a $C^*$-algebra this is not possible (consider $z \mapsto z$ in the $C^*$-algebra of continuous functions on the unit disc of the complex plane) but, as noted by Matthew, you can get the same decomposition with $4$ unitaries.

In fact, any matrix $X$ can be written as $X=\lambda(U+V)$ for unitary matrices $U,V$, and $\lambda$ can be taken as a half of the operator norm of $X$. For a proof, see this question. This is slightly stronger than Matthew's comment, and the proof works in any von Neumann algebra. In a $C^*$-algebra this is not possible (consider $z \mapsto z$ in the $C^*$-algebra of continuous functions on the unit disc of the complex plane) but, as noted by Matthew, you can get the same decomposition with $4$ unitaries.

Edit: I correct the mistake pointed out by Matthew in his comment.

In fact, any matrix $X$ can be written as $X=\lambda(U+V)$ for unitary matrices $U,V$, and $\lambda$ can be taken as a half of the operator norm of $X$. For a proof, see this question. This is slightly stronger than Matthew's comment, and the proof works in any finite von Neumann algebra (i.e. when the partial isometry in the polar decomposition can be taken as a unitary). In a $C^*$-algebra this is not possible (consider $z \mapsto z$ in the $C^*$-algebra of continuous functions on the unit disc of the complex plane) but, as noted by Matthew, you can get the same decomposition with $4$ unitaries.

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In fact, any matrix $X$ can be written as $X=\lambda(U+V)$ for unitary matrices $U,V$, and $\lambda$ can be taken as a half of the operator norm of $X$. For a proof, see this question. This is slightly stronger than Matthew's comment, and the proof works in any von Neumann algebra. In a $C^*$-algebra this is not possible (consider $z \mapsto z$ in the $C^*$-algebra of continuous functions on the unit disc of the complex plane) but, as noted by Matthew, you can get the same decomposition with $4$ unitaries.