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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.
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 …
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 …
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
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 …
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 …
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 …
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_ …
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 …
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 …