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Matrix theory is the study of matrices as concrete objects, rather than as abstract linear operators between vector spaces (whose study belongs to linear algebra). For instance, this involves matrix factorizations and decompositions, nonnegative matrices and Perron-Frobenius theory, Schur complements, structured and special matrices, matrix functions and equations.
2
votes
Accepted
Is the trace of a Lyapunov transform of a semistable matrix always nonpositive?
Here is a counterexample. Notice that $\mathrm{Tr}(A^T P)=\mathrm{Tr}(PA)$, so it suffices to calculate $\mathrm{Tr}(PA)$. Suppose $n=2$ and $A$ triangular:
$$P=\begin{pmatrix} x & y \\ y & z \end{pma …
6
votes
Accepted
When are cones of matrices "generated" by vectors?
Let $K$ be a closed convex cone in ${\bf Sym}_n({\mathbb R})$. I assume a generic cone: non void interior, strictly convex. Let
$$K^0=\{ S\in{\bf Sym}_n({\mathbb R})\quad|\quad{\rm Tr}(SH)\ge0,\quad\ …
3
votes
inverse-closed matrix spaces
Take any matrix $n\times n$ $A$, then $E$ the sub-algebra spanned by $A$ ,that is the set of $P(A)$ with $P$ polynomials. This space is inverse-closed (Caylet-Hamilton). Its dimension may be $n$.
3
votes
A question about matrices with more details
To my taste, it seems more natural to let $A$ and $B$ play symmetric role, by asking whether there exists non-trivial factors $s_jA+t_jB$ such that
$$A\left(\prod_{j=1}^p(s_jA+t_jB)\right)B=B\left(\p …
19
votes
Is the determinant the only multiplicative matrix function?
It depends on what is the target space. Linear representations of ${\bf Gl}_n(k)$ do satisfy $\rho(AB)=\rho(A)\rho(B)$ by definition, and they often extend in a natural way to ${\bf M}_n(k)$.
On anot …
7
votes
Accepted
When is the matrix norm multiplicative
At least, there is this important case: in $C^*$-algebras, there is an involution and the norm has the property that
$$\|x\|=\|x^*\|=\|xx^*\|^{1/2}.$$
In the special case of ${\bf M}_n({\mathbb C})$, …
11
votes
Accepted
Generalized Hölder's inequality for operator (subordinate) norms
Actually, there is a much stronger result, known as the Riesz-Thorin Theorem:
The subordinate norm $\|A\|_p$ is a log-convex function of $\frac1p$.
In other words,
$$\left(\frac1r=\frac\theta{p}+\fr …
5
votes
Accepted
Eigenvalues of product of symmetric positive definite matrices
It is true that the product $M=T_1T_2$ of two positive definite symmetric matrices has real and positive eigenvalues. And conversely, every matrix $M$ with real positive eigenvalues can be factored $M …
3
votes
system of homogeneous matrix equations
A partial answer: Let $\Sigma$ denote the manifold $x^n+y^n=0$. Away from $\Sigma$, the equation and the fact that the roots of the polynomial $X^n-x^n-y^n$ are simple, tell you that $xA+yB$ is diagon …
5
votes
A question of invertibility of matrices
What about
$$A=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\qquad B=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\quad ?$$
3
votes
On the existence of integer square root of a $3 \times 3$ positive definite matrix
Besides the fact that $\det M$ needs to be a perfect square, there is an other set of necessary conditions: among the diagonal entries $m_{ii}$ and the principal $2\times2$ minors $m_{ii}m_{jj}-m_{ij} …
0
votes
Determine unknown matrix function of particular form from known points
Not an answer, but the formula is too wide for a comment.
Why didn't you say that $f(X)$ is an entry of an inverse matrix
$$\begin{pmatrix}
X-A & -B \\ - C & -D \end{pmatrix}^{-1}=\begin{pmatrix}
\cdo …
13
votes
One observation of special type of square matrix exponentiation
The answer is quite simple. First observe that $A$ is triangular, hence the spectrum is on the diagonal. From your assumptions, $1$ is a simple eigenvalue and the other eigenvalues belong to $[0,1)$. …
15
votes
Accepted
Is the linear span of special orthogonal matrices equal to the whole space of $N\times N$ ma...
Elementary proof. The linear space $E$ spanned by $SO_n$ is the orthogonal of those matrices $M$ such that $\langle M,Q\rangle:={\rm Tr}(MQ)=0$ for every $Q\in SO_n$. Let $M=SR$ be a polar decompositi …
8
votes
Relation between eigenvalues of $A$ and $A^TA$?
Let $\lambda_1,\ldots,\lambda_n$ be the eigenvalues of $A$ and $s_1,\ldots,s_n$ be those of $\sqrt{A^*A}$, ordered by
$$|\lambda_1|\ge\cdots\ge|\lambda_n|,\qquad s_1\ge\cdots\ge s_n.$$
Then it holds
$ …