how to create matrix c, where first column is vector a, and the last column is vector b without a loop

%if

a=[1;2;3;4;5];

%and

b=a+3;

%then how can i make a matrix c without a loop, but the result would be like the result of this loop?

for i = 1:5

c(i,:)=a(i,1):b(i,1);

end

Answer 1

[Voss]

a = [1;2;3;4;5];
N = 3;
c = a+(0:N)

c = 5×4

1 2 3 4 2 3 4 5 3 4 5 6 4 5 6 7 5 6 7 8

[Les Beckham]

Nice!

Expanding on this idea for an arbitrary dimension square result:

N = 10; % desired dimension
a = 1:N;
c = a' + (0:(N-1))

c = 10×10

1 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 9 10 11 3 4 5 6 7 8 9 10 11 12 4 5 6 7 8 9 10 11 12 13 5 6 7 8 9 10 11 12 13 14 6 7 8 9 10 11 12 13 14 15 7 8 9 10 11 12 13 14 15 16 8 9 10 11 12 13 14 15 16 17 9 10 11 12 13 14 15 16 17 18 10 11 12 13 14 15 16 17 18 19

[Renate]

great! thank you both very much!

Answer 2

[Les Beckham]

a = [1;2;3;4;5];
c = [a a+1 a+2 a+3] 

c = 5×4

1 2 3 4 2 3 4 5 3 4 5 6 4 5 6 7 5 6 7 8

If you are a beginner in using Matlab, I would suggest taking a couple of hours to go through the free tutorial that can be found here: Matlab Onramp

[Renate]

thank you very much!

however, what if i need to extend 3000 rows over 3000 columns? i.e. the rsulting size of the c matrix is not 5X4 but much larger

[Les Beckham]

I can't seem to come up with a way to do that without a loop. Assuming the result is supposed to be square, this should work and seems a bit more clear to me than your loop solution.

N = 10; % desired dimension
a = 1:N;
for i = 1:N
    c(i,:) = a + i - 1;
end
disp(c)

     1     2     3     4     5     6     7     8     9    10
     2     3     4     5     6     7     8     9    10    11
     3     4     5     6     7     8     9    10    11    12
     4     5     6     7     8     9    10    11    12    13
     5     6     7     8     9    10    11    12    13    14
     6     7     8     9    10    11    12    13    14    15
     7     8     9    10    11    12    13    14    15    16
     8     9    10    11    12    13    14    15    16    17
     9    10    11    12    13    14    15    16    17    18
    10    11    12    13    14    15    16    17    18    19

Note that this will create the upper triangular part of the matrix, but I couldn't figure out how to generate the lower half.

c = hankel(1:N)

c = 10×10

1 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 9 10 0 3 4 5 6 7 8 9 10 0 0 4 5 6 7 8 9 10 0 0 0 5 6 7 8 9 10 0 0 0 0 6 7 8 9 10 0 0 0 0 0 7 8 9 10 0 0 0 0 0 0 8 9 10 0 0 0 0 0 0 0 9 10 0 0 0 0 0 0 0 0 10 0 0 0 0 0 0 0 0 0