SVD for a matrix with a given orthonormal $\mathbf{U}$

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Let:

  1. $\mathbf{A}$ be a $N\times N$ complex matrix.
  2. $\mathbf{u}\in \operatorname{span}(\mathbf{A})$ be a given unit norm vector, where $\operatorname{span}(\mathbf{A})$ denotes the column space of $A$.
  3. $\mathbf{U}$ be a $N\times N$ unitary matrix whose columns are an orthonormal basis for $\mathbf{A}$ and also its first column is $\mathbf{u}$ (if $\mathbf{A}$ is rank deficient, then the first $r$ columns are a basis for $\mathbf{A}$ and the rest for the null space of $\mathbf{A}$, where $r$ is its rank).

Is there a SVD of $\mathbf{A}$ with the left singular matrix as $\mathbf{U}$?

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The answer is no. You specify only that $\mathbf{u}$ is in the column space of the matrix. That is not enough to be a singular vector.