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Results tagged with matrix-theory
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user 36721
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.
4
votes
One observation of special type of square matrix exponentiation
(As is also now seen from your answer) I think in your question you actually wanted to impose the condition
$$\sum_{i=1}^j x_{ij} = 1\ \forall j \in \{2,\dots,n\} \tag{1}\label{1} $$
on the column sum …
- 128k
4
votes
Accepted
Matrix inequalities in series form
A counterexample is given by
$$A=\begin{bmatrix}
-1 & a \\
a & -1 \\
\end{bmatrix}$$
with (say) $a=2$.
Indeed, then the sum of your first series is
$$c(t):=\frac12\begin{bmatrix}
c_+(t) & c_-t) \\
…
- 128k
3
votes
If $x \ge 0$ and $\mathbf{1}^Tx \le \|x\|^2$ then $\mathbf{1}^T(I - xx^T / \|x\|^2) \mathbf{...
Another way to prove this: Rewrite your condition $\mathbf{1}^Tx\le\|x\|^2$ as $$a:=s_2/s_1\ge1$$
and your target inequality
$$\mathbf{1}^T(I-xx^T/\|x\|^2)\mathbf{1}\ge\|[\mathbf{1}-x]_+\|^2$$
as
$$s_ …
- 128k
3
votes
One observation of special type of square matrix exponentiation
This is to complete the nice answer by tsnao by showing that $A_1^k\to0$ as $k\to\infty$.
To get that conclusion it is enough to assume that the $x_{ij}$'s are any complex numbers such that $$t:=\max_ …
- 128k
3
votes
One observation of special type of square matrix exponentiation
Based on the previous answers by tsnao and myself, one gets another, more elementary proof of your desired conclusion (and actually of a more general statement).
Indeed, by those previous answers,
$$L …
- 128k
2
votes
Accepted
Matrix inequality with arbitrary large ratios
Consider the family of matrices $M:=M_t:=N^2$, where $N:=P+tI$, $P$ is the $n\times n$ matrix with all entries equal $1$ and $I$ is the $n\times n$ identity matrix, so that $M^{1/2}=N$. Let $t\downarr …
- 128k
1
vote
Accepted
An inequality regarding projection
The answer is no. E.g., let $a=-[1,0,0]^T$, $b=[0,1,0]^T$, $D=I_3$, and $C=[1,1,t]^T$ for $t>0$. Then $a^Tb=0<<1$ and the orthoprojector matrix onto the column space of $A=D^{-1/2}C$ is $P_A=CC^T/(2+t …
- 128k
1
vote
Inequality for matrix with rows summing to 1
$\renewcommand\bar\overline$We have this even stronger result:
\begin{equation}
g(A):=\sum_{m=1}^M \sum_{k=1}^K \frac{(a_{m,k})^2}{\sum_{i=1}^M a_{i,k}}\ge1. \tag{1}\label{1}
\end{equation}
Indeed …
- 128k
1
vote
Accepted
Faulty algorithm for simultaneous diagonalization?
I think you misunderstood the statement, which is probably that $U^{-1}AU$ and $U^{-1}BU$ are both Jordan, I guess with $U^{-1}AU=PD_A P^{-1}$ for some permutation matrix $P$ (since the Jordan form is …
- 128k