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Mar 5 at 18:57 history edited kjetil b halvorsen CC BY-SA 4.0
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Mar 5 at 18:44 comment added Andreas Blass There seems to be a typo in the second row of matrix $D$. The diagonal entry should be $1$ and the other entry shjoould be $-1$. With this correction, I agree with @CarloBeenakker's first comment that the matrices differ only by an overall minus sign, which makes no difference.
Mar 5 at 12:49 review Close votes
Mar 22 at 3:10
Mar 5 at 12:33 comment added bing @Carlo Beenakker, Thanks for your time, the result i got with matlab has the same real part with your result. Maybe the imaginary parts i got is due to some numerical error, as one of the eigenvalue of (L'*L) is nearly zero. Thank you so much.
Mar 5 at 12:31 comment added Carlo Beenakker this is the square root in your case: $\left( \begin{array}{cccc} 1 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 1 \\ \end{array} \right)^{1/2}=\left( \begin{array}{cccc} 0.815493 & -0.544895 & -0.162212 & -0.108386 \\ -0.544895 & 1.19818 & -0.49107 & -0.162212 \\ -0.162212 & -0.49107 & 1.19818 & -0.544895 \\ -0.108386 & -0.162212 & -0.544895 & 0.815493 \\ \end{array} \right)$
Mar 5 at 12:18 comment added bing @Carlo Beenakker ,Thanks for your comment. I create a matrix in matlab first with L=[1 -1 0 0;0 1 -1 0;0 0 1 -1] . Then I use the command sqrtm(L'*L) and got a complex matrix but the imaginary parts of all the complex numbers are very small, maybe it's due to some numerical error?
Mar 5 at 12:07 comment added Carlo Beenakker your matrices $D$ and $L_1$ only differ by a minus sign; since $B=(D^\top D)^{1/2}$ the minus sign is irrelevant; the matrix $D^\top D$ is positive definite, so it has a real square root, you are probably making a mistake if you get an imaginary answer.
S Mar 5 at 11:53 review First questions
Mar 5 at 18:57
S Mar 5 at 11:53 history asked bing CC BY-SA 4.0