if $A>0$, prove that there exists $xin{pm1}^n$, such that $(Ax)_ineq0$ for all $i$
$begingroup$
If $Ainmathbb{R}^{ntimes n}, A>0$ (symmetric, positive definite), prove that there exist $x=begin{bmatrix} sigma_1 \ vdots \ sigma_nend{bmatrix}$, $sigma_iin{pm1}$, such that $(Ax)_ineq0$ for all $i$.
Example: If $A=begin{bmatrix}2 & -1 & -1\-1 & 2 &1 \-1 & 1 &2end{bmatrix}>0$, then $Ax=begin{bmatrix}0\2\2end{bmatrix}$ for $x=begin{bmatrix}1\1\1end{bmatrix}$,
but $Ax=begin{bmatrix}4\-4\-4end{bmatrix}$ for $x=begin{bmatrix}1\-1\-1end{bmatrix}$.
From simulation of $3times3$ and $4times4$ matrices, it turns out to be true.
linear-algebra positive-definite
edited Jan 8 at 2:22
El borito
668216
asked Jan 8 at 2:03
LeeLee
330111
$endgroup$
add a comment |
$begingroup$
If $Ainmathbb{R}^{ntimes n}, A>0$ (symmetric, positive definite), prove that there exist $x=begin{bmatrix} sigma_1 \ vdots \ sigma_nend{bmatrix}$, $sigma_iin{pm1}$, such that $(Ax)_ineq0$ for all $i$.
Example: If $A=begin{bmatrix}2 & -1 & -1\-1 & 2 &1 \-1 & 1 &2end{bmatrix}>0$, then $Ax=begin{bmatrix}0\2\2end{bmatrix}$ for $x=begin{bmatrix}1\1\1end{bmatrix}$,
but $Ax=begin{bmatrix}4\-4\-4end{bmatrix}$ for $x=begin{bmatrix}1\-1\-1end{bmatrix}$.
From simulation of $3times3$ and $4times4$ matrices, it turns out to be true.
linear-algebra positive-definite
edited Jan 8 at 2:22
El borito
668216
asked Jan 8 at 2:03
LeeLee
330111
$endgroup$
$begingroup$
I am sure I have see a solution on MSE but have no idea how to search for it :-(.
$endgroup$
– copper.hat
Jan 9 at 19:15
add a comment |
$begingroup$
If $Ainmathbb{R}^{ntimes n}, A>0$ (symmetric, positive definite), prove that there exist $x=begin{bmatrix} sigma_1 \ vdots \ sigma_nend{bmatrix}$, $sigma_iin{pm1}$, such that $(Ax)_ineq0$ for all $i$.
Example: If $A=begin{bmatrix}2 & -1 & -1\-1 & 2 &1 \-1 & 1 &2end{bmatrix}>0$, then $Ax=begin{bmatrix}0\2\2end{bmatrix}$ for $x=begin{bmatrix}1\1\1end{bmatrix}$,
but $Ax=begin{bmatrix}4\-4\-4end{bmatrix}$ for $x=begin{bmatrix}1\-1\-1end{bmatrix}$.
From simulation of $3times3$ and $4times4$ matrices, it turns out to be true.
linear-algebra positive-definite
edited Jan 8 at 2:22
El borito
668216
asked Jan 8 at 2:03
LeeLee
330111
$endgroup$
If $Ainmathbb{R}^{ntimes n}, A>0$ (symmetric, positive definite), prove that there exist $x=begin{bmatrix} sigma_1 \ vdots \ sigma_nend{bmatrix}$, $sigma_iin{pm1}$, such that $(Ax)_ineq0$ for all $i$.
Example: If $A=begin{bmatrix}2 & -1 & -1\-1 & 2 &1 \-1 & 1 &2end{bmatrix}>0$, then $Ax=begin{bmatrix}0\2\2end{bmatrix}$ for $x=begin{bmatrix}1\1\1end{bmatrix}$,
but $Ax=begin{bmatrix}4\-4\-4end{bmatrix}$ for $x=begin{bmatrix}1\-1\-1end{bmatrix}$.
From simulation of $3times3$ and $4times4$ matrices, it turns out to be true.
linear-algebra positive-definite
linear-algebra positive-definite
edited Jan 8 at 2:22
El borito
668216
asked Jan 8 at 2:03
LeeLee
330111
edited Jan 8 at 2:22
El borito
668216
asked Jan 8 at 2:03
LeeLee
330111
edited Jan 8 at 2:22
El borito
668216
edited Jan 8 at 2:22
El borito
668216
edited Jan 8 at 2:22
El borito
668216
668216
asked Jan 8 at 2:03
LeeLee
330111
asked Jan 8 at 2:03
LeeLee
330111
asked Jan 8 at 2:03
LeeLee
330111
330111
$begingroup$
I am sure I have see a solution on MSE but have no idea how to search for it :-(.
$endgroup$
– copper.hat
Jan 9 at 19:15
add a comment |
$begingroup$
I am sure I have see a solution on MSE but have no idea how to search for it :-(.
$endgroup$
– copper.hat
Jan 9 at 19:15
$begingroup$
I am sure I have see a solution on MSE but have no idea how to search for it :-(.
$endgroup$
– copper.hat
Jan 9 at 19:15
$begingroup$
I am sure I have see a solution on MSE but have no idea how to search for it :-(.
$endgroup$
– copper.hat
Jan 9 at 19:15
add a comment |
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$begingroup$
I am sure I have see a solution on MSE but have no idea how to search for it :-(.
$endgroup$
– copper.hat
Jan 9 at 19:15