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15m
comment PSD matrix and non-negative polynomial
Are you using 0-based indicing for the rows and columns of $Q$? If 1-based indicing is used, you can't have $i+j=1$. So, how is $a_0$ defined?
21m
answered Boolean Least Squares semidefinite relaxation
57m
answered Derivative of a Matrix to a Power
8h
answered equivalent condition for interpolation polynomial
10h
comment Why does a positive definite matrix with a repeated eigenvalue have infinitely many square roots?
As a remark, please note that while $A$ has infinitely many Hermitian square roots (and hence infinitely many square roots), only one of them is positive definite.
10h
answered proof of positive semi-definiteness of the precision matrix (inverse of the covariance matrix)
18h
comment Notation - Transpose of Block Matrices [Lay P121 Q2.4.12]
As a side remark, $\pmatrix{M&N}^T$ can be $\pmatrix{M^T\\ N^T}$ or $\pmatrix{M\\ N}$, depending on the meaning of $\pmatrix{M&N}$: if it is the augmented matrix of $M$ and $N$, its transpose is $\pmatrix{M^T\\ N^T}$; if it is a matrix over the ring of matrices (i.e. if each of $M$ and $N$ is by itself a matrix element) then its transpose is $\pmatrix{M\\ N}$. However, I presume that you take $\pmatrix{M&N}$ to mean an augmented matrix here.
18h
comment Is the smallest singular value able to measure the similarity between two matrices?
Please clarify in your question the followings: (a) whether $A,B$ are square matrices, (b) what does $\operatorname{span}\,B$ means (a unit multiple of $B$, the column space of $A$ is a subspace of the column space of $B$, etc.).
1d
answered The solution for the matrix system $(A-X)X=0$
1d
comment Centralizer of $SO(n)$
@Robrt Yes. Now, by choosing $P$ appropriately, you may set the 2x2 principal submatrix of $R$ corresponding to the $i$-th and $j$-th rows and columns to $\pmatrix{0&-1\\ 1&0}$. By the same argument, we can prove that $a_{ik}=0$ for all $k\notin\{i,j\}$. Since $i$ and $j$ are arbitrary, it follows that all off-diagonal entries of $A$ are zero.
1d
comment Centralizer of $SO(n)$
@Robrt For the moment, let $P=I$ and partition $A$ as $\pmatrix{X&Y\\ Z&W}$, where $X$ is 2x2. Can you show that $AR=RA$ implies $Y,Z$ are zero matrices?
1d
answered Centralizer of $SO(n)$
1d
comment Where am I going wrong with this limit?
If $n=1$, the fraction and its limit are zero. If $n\ne1$, L'Hospital rule does work.
2d
comment If all eigenvalues are 1 or -1, is then $A^{12}=I$?
Why can you substitute $\lambda=1$? Are you suggesting that $\begin{pmatrix}1&0\\0&-1\end{pmatrix}=\begin{pmatrix}0&0\\0&2\end{pmatrix}$?!
2d
answered Prove trace inequality $\mathrm{tr}\{ABCBAD-ABCD-ADCB+CD\} \geq 0$
2d
answered Max determinant
2d
comment How does one show a matrix is irreducible and reducible?
@npisinp Let the matrix in question be $A$ and let $B=(b_{ij})$ be the matrix such that $b_{ij}=1$ if $a_{ij}\ne0$, and $b_{ij}=0$ if $a_{ij}=0$. Then $B$ is the adjacency matrix of a directed graph, and $A$ is reducible iff this directed graph has proper strongly connected components.
2d
comment pseudo-inverse by SVD decomposition has not accurate results?
@Pey Again, "optimal" in what sense? For a usual linear equation $Mx=y$ where $M$ has deficient rank, the "optimal" (least-norm) solution $x$ is given by $x=M^+y$. In this sense, your optimal $\operatorname{vec}(\mathbf A)$ should be $-(\mathbf p^T\otimes{\boldsymbol{\alpha}}^T)^+\eta$.
Apr
11
comment pseudo-inverse by SVD decomposition has not accurate results?
@Pey "Accurate" in what sense? As far as the value of $f$ is concerned, every solution of $(\mathbf p^T\otimes{\boldsymbol{\alpha}}^T)\operatorname{vec}(\mathbf A)=-\eta$ is as "accurate" as the others.
Apr
11
answered pseudo-inverse by SVD decomposition has not accurate results?