Use kron to build a matrix
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Hi,
I would like to build a matrix that resembles something like
A = [-1 1 0 0 0 0 0 0;
0 -1 1 0 0 0 0 0;
0 0 -1 1 0 0 0 0;
0 0 0 -1 1 0 0 0;
0 0 0 0 -1 1 0 0;
0 0 0 0 0 -1 1 0;
0 0 0 0 0 0 -1 1];
Can I use kron function to build something like this? One of the things that needs to be a variable will be the number of rows, so I need my kron function to be flexible such that it can build more such matrices, for eg. a matrix with four such rows would look like
A = [-1 1 0 0 0;
0 -1 1 0 0;
0 0 -1 1 0;
0 0 0 -1 1;]
Also happy to do using any other method than kron. Thank you for your help.
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採用された回答
Matt J
2021 年 5 月 24 日
No, the matrix you've shown is not a Kronecker product, but the matrix is easily generated using diff, e.g.,
A=full(diff(speye(7)))
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その他の回答 (1 件)
Sulaymon Eshkabilov
2021 年 5 月 24 日
Hi,
In you exercise, use eye() and kron(). and then change the necessary element of the generated matrix if necessary.
Good luck
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