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Indian Institute of Science


Oct
17
comment On Neumann-series of matrices
Does this proof cover the cases when spectral radius is different from the absolute value of the maximum eigenvalue? For instance, how about a negative definite $A$.
Oct
16
comment On Neumann-series of matrices
maybe you should also include the case when $A$ is not diagonalizable. Did you mean "$Q$ is a invertible matrix" or unitary?
Oct
16
comment Is this argument on positive definite matrices correct?
:) :) @Karthik Happy to see you here!!.
Oct
16
accepted Is this argument on positive definite matrices correct?
Oct
15
asked Is this argument on positive definite matrices correct?
Oct
14
comment Show that the inverse of a strictly diagonally dominant matrix is monotone
if you want to prove $A^{-1}>0$ (not the individual entries of $A$), observe that strict diagonal dominance implies positive definitiness.
Sep
30
awarded  Explainer
Sep
30
revised Is it true that $\biggl\|I-\frac{vv^T}{v^Tv}\biggr\|=1$?
added 4 characters in body
Sep
30
comment Is it true that $\biggl\|I-\frac{vv^T}{v^Tv}\biggr\|=1$?
@AlexR Please note that $\sigma_1$ is the highest singular value (as given in the explanation on the right end of the equation in the answer) and not the highest eigenvalue. For a symmetric matrix, the modulus of the eigenvalues gives the singular values. Thanks for the typo about the unit norm on the maximization.
Sep
27
answered Is it true that $\biggl\|I-\frac{vv^T}{v^Tv}\biggr\|=1$?
Sep
24
awarded  Autobiographer
Sep
19
asked A Difficult combinatorial optimization problem
Sep
18
asked Verification of the Approach to a given non-convex integer programming problem
Sep
17
awarded  Yearling
Sep
11
comment Proof for a rank-one Decomposition theorem
oh I see!!. does this happen because $trace(AX)=0$? I mean, if you consider the space of hermitian matrices, they are sort of orthogonal to each other.
Sep
5
revised Proof for a rank-one Decomposition theorem
changed title, stated the result first
Sep
5
revised Proof for a rank-one Decomposition theorem
changed title, stated the result first
Sep
5
comment Proof for a rank-one Decomposition theorem
Hi, the algo takes eigen decomposition of $R_1^TAR_1$ and not that of $R_1AR_1^T$ which is what you have considered. If you take $R_1^TAR_1$, the conclusions follow easily.
Sep
2
comment Minimum eigen value of this operator valued self-adjoint matrix
if $e_1$ is not an eigenvector corresponding to the minimum eigenvalue, strict-inequality is guaranteed.
Sep
2
answered Minimum eigen value of this operator valued self-adjoint matrix