Replace diagonals in a matrix

Hello every one,

How to replace the upper and lower part of a n*n matrix with zeros (The upper part starts from "diagonal+2" until n while the lower part starts from "diagonal-2" until n)?

[Image Analyst]

Original question is below in case he deletes it like he's done before:

Replace diagonals in a matrix

Hello every one,

How to replace the upper and lower part of a n*n matrix with zeros (The upper part starts from "diagonal+2" until n while the lower part starts from "diagonal-2" until n)?

Answer 1

[Matt J]

For example,

A=rand(10),

A = 10×10

0.3632 0.5683 0.2698 0.2133 0.4828 0.8948 0.0478 0.7711 0.5088 0.7301 0.8024 0.1100 0.1475 0.6313 0.8957 0.8629 0.1435 0.8356 0.1381 0.8203 0.8949 0.0767 0.3939 0.2073 0.3055 0.5721 0.7656 0.1443 0.3270 0.6013 0.5722 0.1502 0.8491 0.7124 0.9746 0.7291 0.9774 0.9558 0.4083 0.0934 0.1948 0.7491 0.7331 0.3686 0.4528 0.2436 0.2373 0.4254 0.9163 0.6997 0.4224 0.4079 0.6316 0.8894 0.8130 0.2013 0.0054 0.5217 0.1405 0.7939 0.0219 0.5942 0.3860 0.2447 0.6384 0.9649 0.3000 0.2048 0.2233 0.1450 0.4361 0.2666 0.2646 0.3323 0.6236 0.6866 0.9302 0.3413 0.4398 0.0531 0.3450 0.6491 0.0609 0.1894 0.8414 0.7489 0.1080 0.4159 0.2769 0.3304 0.0411 0.9652 0.9593 0.3298 0.9979 0.3895 0.2349 0.0686 0.9871 0.3078

mask=tril(  triu(  true(size(A)), -2  ), +2);
B=A.*mask

B = 10×10

0.3632 0.5683 0.2698 0 0 0 0 0 0 0 0.8024 0.1100 0.1475 0.6313 0 0 0 0 0 0 0.8949 0.0767 0.3939 0.2073 0.3055 0 0 0 0 0 0 0.1502 0.8491 0.7124 0.9746 0.7291 0 0 0 0 0 0 0.7331 0.3686 0.4528 0.2436 0.2373 0 0 0 0 0 0 0.8894 0.8130 0.2013 0.0054 0.5217 0 0 0 0 0 0 0.6384 0.9649 0.3000 0.2048 0.2233 0 0 0 0 0 0 0.6866 0.9302 0.3413 0.4398 0.0531 0 0 0 0 0 0 0.1080 0.4159 0.2769 0.3304 0 0 0 0 0 0 0 0.0686 0.9871 0.3078

[Hasan Hassoun]

Thank you!

Answer 2

[David Goodmanson]

Hi Hasan,

here is one way

r = rand(7,7)

n = size(r,1);
m = (n-1)/2;
a = (-m:m)-(-m:m)';
r(abs(a)>1)=0

assuming the main diagonal is number 0, and the zeros start on the +-2nd diagonal, otherwise adjust the '1' on the last line of code accordingly

[Hasan Hassoun]

Thank you!