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

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 $\mathbf{U}$?

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.
• What is the additional condition needed? Now I think more about it, $\mathbf{U^HA}$ should be a orthogonal matrix for this to be true. Is it?. But I got it from the definition of SVD. What does it actually imply? – dineshdileep Feb 8 '14 at 1:16
• If $\mathbf{U^HA}$ is orthogonal, then $\mathbf{U^HA}\mathbf{A^HU}$ is diagonal, which would indeed be from the definition of SVD. – adam W Feb 8 '14 at 1:23