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 $\mathbf{U}$?


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Commented Feb 8, 2014 at 1:16
  • $\begingroup$ If $\mathbf{U^HA}$ is orthogonal, then $\mathbf{U^HA}\mathbf{A^HU}$ is diagonal, which would indeed be from the definition of SVD. $\endgroup$
    – adam W
    Commented Feb 8, 2014 at 1:23
  • $\begingroup$ Though the word orthogonal as it is commonly used implies that the diagonal would be ones. That is not true unless all the singular values of A have magnitude one. $\endgroup$
    – adam W
    Commented Feb 8, 2014 at 1:48
  • $\begingroup$ Isn't orthonormal the correct word? I think orthogonal doesn't necessarily mean that gramian matrix is identity? Isn't it so? $\endgroup$ Commented Feb 8, 2014 at 4:05
  • $\begingroup$ hello adam, I request you to go through this question I am struggling with math.stackexchange.com/questions/654817/… $\endgroup$ Commented Feb 8, 2014 at 4:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .