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.

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

2022年1月4日 · By definition, orthogonal matrix means its inverse is equal to its transpose, but I don't see where the row orthogonality would come from. And also, if A has orthogonal rows, is it correct that the matrix A is also orthogonal?

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.

2015年9月1日 · I've seen the statement "The matrix product of two orthogonal matrices is another orthogonal matrix. " on Wolfram's website but haven't seen any proof online as to why this is true. By orthogonal m...

2018年9月25日 · Orthogonal Block Matrix. Ask Question Asked 6 years, 4 months ago. Modified 6 years, 4 months ago. Viewed ...

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 …

2015年3月3日 · Hello fellow users of this forum: Show that for any orthogonal matrix Q, either det(Q)=1 or -1. Thanks

2017年4月3日 · Nearest semi-orthogonal matrix using the entry-wise $ {\ell}_{1} $ norm 3 For a positive definite symmetric matrix, the orthogonal diagonalization is a SVD of A.

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 $\pm 1$.