Hi, guys. How would one extract the lower triangle of a matrix without using the tril function and witout a for loop?

I tried

B = zeros(size(A)); n=size(A,2);

for(i=1:n) B(end+1:end+n-i+1)=A(i:n,i); end

but I seem to be getting some sort of error.

Thanks

[Jos (10584)]

What do you mean with "some sort of error"?

Answer 1

[Stephen23]

You could use bsxfun and rot90 to generate a matrix of indices:

>> N = 5;
>> K = 0; % select the diagonal
>> A = 1:N;
>> B = rot90(bsxfun(@plus,A,A(:)))-1;

Now lets try it on matrix of random values and extract only the lower diagonal elements:

>> X = rand(N);
>> X(B<=N+K)
ans =
    0.81472
    0.90579
    0.12699
    0.91338
    0.63236
     0.2785
    0.54688
    0.95751
    0.96489
    0.95717
    0.48538
    0.80028
    0.79221
    0.95949
    0.67874

And you can see that it extracts all of the lower-triangle values, matching those given by tril:

>> tril(X)
ans =
    0.81472            0            0            0            0
    0.90579       0.2785            0            0            0
    0.12699      0.54688      0.95717            0            0
    0.91338      0.95751      0.48538      0.79221            0
    0.63236      0.96489      0.80028      0.95949      0.67874

Note that this method also gives direct control over which diagonal you start with using the value K, just like the second argument of tril.

[ali92]

Thanks for the answer. How would you opt out the diagonal matrix from your list of outputs though?

[Stephen23]

What do you mean by "opt out the diagonal matrix"? If you want the same shape as tril, then try this:

>> N = 5;
>> K = 0; % select the diagonal
>> A = 1:N;
>> B = rot90(bsxfun(@plus,A,A(:)))-1;
>> X = rand(N)
X =
    0.7577    0.7060    0.8235    0.4387    0.4898
    0.7431    0.0318    0.6948    0.3816    0.4456
    0.3922    0.2769    0.3171    0.7655    0.6463
    0.6555    0.0462    0.9502    0.7952    0.7094
    0.1712    0.0971    0.0344    0.1869    0.7547
>> X(B>N+K) = 0
X =
    0.7577         0         0         0         0
    0.7431    0.0318         0         0         0
    0.3922    0.2769    0.3171         0         0
    0.6555    0.0462    0.9502    0.7952         0
    0.1712    0.0971    0.0344    0.1869    0.7547

And of course if you don't want to change the variable X then simply allocate it to another variable and change that one instead..

Answer 2

[Jos (10584)]

A = rand(5) ; %

[c,r] = meshgrid(1:size(A,1),1:size(A,2))

trilA = A ;
trilA(c>r) = 0 

diagA = A ;
diagA(c ~= r) = 0