Skip to main content
MathOverflow

Return to Answer

replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Source Link
Community Bot
Community Bot
  • 1
  • 2
  • 3

This should be a comment. I think this question is not apt here. You should consider posting questions like these at math.stackexchangemath.stackexchange.

Given $U$ and if it is full rank, we can choose $$V = -2 \left(UU^T \right)^{-1}U$$ and the matrix $$W = I - 2 U^T \left(UU^T \right)^{-1}U$$ is an orthonormal matrix. A special case of this is provided by Jeff.

This should be a comment. I think this question is not apt here. You should consider posting questions like these at math.stackexchange.

Given $U$ and if it is full rank, we can choose $$V = -2 \left(UU^T \right)^{-1}U$$ and the matrix $$W = I - 2 U^T \left(UU^T \right)^{-1}U$$ is an orthonormal matrix. A special case of this is provided by Jeff.

This should be a comment. I think this question is not apt here. You should consider posting questions like these at math.stackexchange.

Given $U$ and if it is full rank, we can choose $$V = -2 \left(UU^T \right)^{-1}U$$ and the matrix $$W = I - 2 U^T \left(UU^T \right)^{-1}U$$ is an orthonormal matrix. A special case of this is provided by Jeff.

Source Link
user11000
user11000

This should be a comment. I think this question is not apt here. You should consider posting questions like these at math.stackexchange.

Given $U$ and if it is full rank, we can choose $$V = -2 \left(UU^T \right)^{-1}U$$ and the matrix $$W = I - 2 U^T \left(UU^T \right)^{-1}U$$ is an orthonormal matrix. A special case of this is provided by Jeff.