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The study of the properties of real and complex matrices that are more close to analysis and operator theory. For instance: the properties of positive definite matrices, matrix inequalities, perturbation analysis, matrix functions, inequalities between eigenvectors and singular values, majorization.
28
votes
Is the linear span of special orthogonal matrices equal to the whole space of $N\times N$ ma...
For $n>2$, the span of the matrices in $\mathrm{SO}(n)$ is the full space $M_n(\mathbb{R})$ of $n$-by-$n$ matrices with real entries.
One proof is using representation theory: If we let $S\subset M_ …
15
votes
Accepted
The space of positive definite orthogonal matrices
You may find the Cayley transform to be useful here:
As is well-known and easy to prove, every orthogonal $n$-by-$n$ matrix $R$ that does not have $-1$ as an eigenvalue can be written uniquely in the …
13
votes
Accepted
Is the set of real matrices with at least one real logarithm closed under multiplication?
This is already not true for $2$-by-$2$ matrices: Consider
$$
A = \begin{pmatrix}2 & 0 \\0 &\frac12\end{pmatrix}\quad
\text{and}\quad
B = \begin{pmatrix}-1 & 0 \\0 &-1\end{pmatrix}.
$$
$A = \exp\bigl …
8
votes
Accepted
Diagonalization of symmetric matrices of functions
In general, this cannot be done. For example, in dimension $2$ in coordinates $(x,y)$, let
$$
G(x,y) = \left[\begin{matrix}x&y\\y&-x\end{matrix}\right].
$$
If $G$ could be diagonalized by a different …
6
votes
Multiplicative Identity for all elements in SU(n)
New answer: I now have an answer for the subgroup case that the OP originally asked about. In fact, one has the following result: Let $G$ be a connected compact Lie group and let $p = (p_1,\ldots,p …
6
votes
Accepted
Is a pointwise " simple tensor" valued continuous map a tensor product of two continuous maps?
The answer is 'no'. For example, let
$$
X = \{\ B\otimes C\ | \det(B)\det(C) = 1\ \}\subset \mathrm{SL}(4,\mathbb{C}).
$$
Then $X$ is isomorphic to the group $\mathrm{SO}(4,\mathbb{C})$ and hence $\p …
6
votes
How to check whether a matrix is completely positive or not?
At least for $3$-by-$3$ matrices, the test for complete positivity of a matrix $A$ is not hard. Basically, you need that $A$ be positive-semi-definite and that the off-diagonal entries be non-negativ …
6
votes
Accepted
Analytical form for the nuclear norm of an $n \times n$ matrix
As Gro-Tsen pointed out, I had not computed the Galois group of the governing polynomial, and, in fact, my original answer was wrong. I believe that the following answer is correct, though.
If by 'li …
5
votes
How to solve a non-homogeneous quadratic matrix equation?
Yes, this is always possible, and $G$ is unique. Here is how you can see this:
Consider the pair of positive definite symmetric matrices $(-H^{-1}, M)$. By a well-known theorem (simultaneous diagon …
4
votes
Unitary transformations of Vandermonde matrices forms a smooth manifold?
The answer is 'not always', in particular, not when $(n,r)=(2,3)$.
The image is obviously is a smooth manifold when $n=0$, for then the image in $\mathbb{R}^{(n+1)r}=\mathbb{R}^r$ is the sphere $\Sigm …
2
votes
Rank of order-3 tensor with all slices being rank-1
The following comments may be useful to you, though you may not regard them as a complete answer.
To set the stage, first consider the case of an order 2 tensor $T\in V_1\otimes V_2$,
where $V_i$ a …