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I am reading a paper (Parameter Choice Strategies for Multipenalty Regularization Massimo Fornasier, Valeriya Naumova, and Sergei V. Pereverzyev SIAM Journal on Numerical Analysis 2014 52:4, 1770-1794)

In this paper first derivative operator is a matrix like the pics below

image of matrix

is this right?

because most of the first-order derivative discrete operator matrices I've seen are like this

enter image description here

from(Discrete inverse problems: insight and algorithms,Hansen)

Could someone please explain this to me? And how to calculate the matrix B in the first pics with matlab? The B matrix I obtained using the "sqrtm" command is a complex matrix, is this right?\

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    $\begingroup$ 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. $\endgroup$
    – Carlo Beenakker
    Commented Mar 5 at 12:07
  • $\begingroup$ @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? $\endgroup$
    – bing
    Commented Mar 5 at 12:18
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    $\begingroup$ this is the square root in your case: $\left( \begin{array}{cccc} 1 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 1 \\ \end{array} \right)^{1/2}=\left( \begin{array}{cccc} 0.815493 & -0.544895 & -0.162212 & -0.108386 \\ -0.544895 & 1.19818 & -0.49107 & -0.162212 \\ -0.162212 & -0.49107 & 1.19818 & -0.544895 \\ -0.108386 & -0.162212 & -0.544895 & 0.815493 \\ \end{array} \right)$ $\endgroup$
    – Carlo Beenakker
    Commented Mar 5 at 12:31
  • $\begingroup$ @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. $\endgroup$
    – bing
    Commented Mar 5 at 12:33
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    $\begingroup$ 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. $\endgroup$
    – Andreas Blass
    Commented Mar 5 at 18:44

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