Generally, those matrices that are both orthogonal and have determinant $1$ are referred to as special orthogonal matrices or rotation matrices. If I read "orthonormal matrix" somewhere, I would assume it meant the same thing as orthogonal matrix. Some examples: $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ is not orthogonal.

This is because SVD works by finding a right and a left orthogonal (rotation) matrix, which rotates the matrix in question to column resp row orthogonality. But the nxn orthogonal matrix is already row and column-orthogonal. Thus the SVD routine has no rotation-criterion.

I recently took linear algebra course, all that I learned about orthogonal matrix is that Q transposed is Q inverse, and therefore it has a nice computational property. Recently, to my surprise, I learned that transformations by orthogonal …

The eigenvalues of an orthogonal matrix needs to have modulus one. If the eigenvalues happen to be real, then they are forced to be $\pm 1$. If the eigenvalues happen to be real, then they are forced to be $\pm 1$.

The original question was asking about a matrix H and a matrix A, so presumably we are talking about the operator norm.

Well, if the columns are orthonormal (i.e. norm 1), then the matrix is orthogonal, and has many beautiful properties. If not, see Name for matrices with orthogonal (not necessarily orthonormal) rows. I suppose the right way to think about it is that this matrix maps the standard basis vectors to an orthogonal basis.

Jan 15, 2017 · If you sample elements from a uniform distribtution over $[-1,1]$ and apply the Gram Schmidt procedure, you can generate every possible orthogonal matrix (note that orthogonal matrices necessarily have elements within $[-1,1]$). However, I don't believe that it will generate all matrices with equal probability.

A matrix can very well be invertible and still not be orthogonal, but every orthogonal matrix is ...

May 2, 2015 · The term "orthogonal matrix" probably comes from the fact that such a transformation preserves orthogonality of vectors (but note that this property does not completely define the orthogonal transformations; you additionally need that the length is not changed either; that is, an orthonormal basis is mapped to another orthonormal basis).

Thr characteristic polynomial is of degree 2 which tells you the eigenvalues, and since your matrix is symmetric, this tells you up to an orthogonal transformation it is a diagonal matrix with plus or minus ones on the diagonal. $\endgroup$