CODEWITHSUNDEEP
matlabmatrix
Marta Sampietro
Marta Sampietro

Reputation: 339

Identity matrix in Matlab

I have to create a function in Matlab that given a parameter N, it returns the N-by-N identity matrix. I cannot use loops, nor built-in functions like eye or diag. I have tried the following:

function I = identity( n )
    I = zeros( n,n );
    p = [1:n;1:n]';
    I( p ) = 1;
end

But, when I call it with I = identity(3); I obtain the following result:

I =

 1     0     0
 1     0     0
 1     0     0

And I don't understand why, because I thought Matlab could use a vector as a matrix index, and the way I did it, I have that:

p =

 1     1
 2     2
 3     3

So when I do I( p ) = 1, the first step should be I( 1,1 ) = 1 then I( 2,2 ) = 1 and so on. What am I not seeing?

Upvotes: 3

Views: 2499

Answers (4)

Benoit_11
Benoit_11

Reputation: 13945

The way in which MATLAB indexes is column-major, therefore it fills matrix I with linear indices contained in p, starting at (1,1) then going to (2,1) and so on. Therefore it "sees" the indices as [1 2 3] and then again [1 2 3].

What you could do is change p into a 1xn vector containing the appropriate linear indices.

For instance:

p = 1:n+1:n^2

gives rise to those indices:

p =

     1     5     9

and the following matrix I:

I =

     1     0     0
     0     1     0
     0     0     1

yay!

Upvotes: 3

rayryeng
rayryeng

Reputation: 104565

Going with thewaywewalk's answer, we can achieve the same thing with the bsxfun approach but without using any built-in functions... except for ones. Specifically, we can use indexing to replicate rows and columns of vectors, then use the equality operator when finished. Specifically, we would first generate a row vector and column vector from 1:n, replicate them so that they're respectively n x n matrices, then use equality. The values in this matrix should only be equal along the diagonal elements and hence the identity is produced.

As such:

row = 1:n;
col = row.';
row = row(ones(n,1),:);
col = col(:, ones(n,1));
I = (row == col) + 0;

We need to add 0 to the output matrix to convert the matrix to double precision as row == col would produce a logical matrix. I didn't cast using the double function because you said you can't use any built-in functions... but I took the liberty of using ones, because in your solution you're using zeros, which is technically a built-in function.

Upvotes: 0

Robert Seifert
Robert Seifert

Reputation: 25242

Is bsxfun allowed?

function I = identity(n)

I = bsxfun(@eq,1:n,(1:n).');

end

Upvotes: 3

Divakar
Divakar

Reputation: 221754

Using no functions, just matrix indexing -

A(N,N) = 0;
A((N+1)*(0:N-1)+1) = 1

Thus, the function becomes -

function A = identity( N )
A(N,N) = 0;
A((N+1)*(0:N-1)+1) = 1;
end

Upvotes: 4

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