Mirror pattern of the leading diagonal about the middle element of the matrix

Hi All,

I am interested to create the following matrix but unable to think about the logic to create it.

for example : for n = 5 ; the elememts of leading diagonal are 2 1 0 1 2 and rest elements 0

for n = 7, the elements of the leading diagonal are 3 2 1 0 1 2 3 and rest elements are 0.

Any help will be highly appreciated.

Answer 1

[Star Strider]

This works for odd ‘n’:

dv = @(n) abs(diag(unique(fix(-n/2:n/2))));                     % Create Diagonal Matrix

n = 5;
m5 = dv(n)

n = 7;
m7 = dv(n)

producing:

m5 =
     2     0     0     0     0
     0     1     0     0     0
     0     0     0     0     0
     0     0     0     1     0
     0     0     0     0     2

m7 =
     3     0     0     0     0     0     0
     0     2     0     0     0     0     0
     0     0     1     0     0     0     0
     0     0     0     0     0     0     0
     0     0     0     0     1     0     0
     0     0     0     0     0     2     0
     0     0     0     0     0     0     3

.

[Swati Sarangi]

Thank you so much! This is so perfect! :)

[Star Strider]

As always, my pleasure!

Thank you!

[Swati Sarangi]

@Star Strider

One more pattern that comes to my mind, in this format

What should be the code for the leading diagonal elements to be -2 -1 0 1 2 ?

[Star Strider]

Leave out the abs() call:

dv = @(n) diag(unique(fix(-n/2:n/2)));                          % Create Diagonal Matrix

n = 5;
m5 = dv(n)

n = 7;
m7 = dv(n)

producing:

m5 =
    -2     0     0     0     0
     0    -1     0     0     0
     0     0     0     0     0
     0     0     0     1     0
     0     0     0     0     2

m7 =
    -3     0     0     0     0     0     0
     0    -2     0     0     0     0     0
     0     0    -1     0     0     0     0
     0     0     0     0     0     0     0
     0     0     0     0     1     0     0
     0     0     0     0     0     2     0
     0     0     0     0     0     0     3

.

[Swati Sarangi]

wow! You're a Genius! 

[Star Strider]

Thank you!