How to index function-matrices?

Take for example: f =@(x) [x,1;1,x]

If you evaluate the function f, you get a matrix in return. Is there any way, to index this matrix before evaluating it?

Like f(1,1) and so forth.

Indexing the matrix while evaluating doesn't work either: f(1)(1,1)

You still need to refer to the result: f1 = f(1); f1(1,1)

=1

Answer 1

[Walter Roberson]

No, there is no way to index the matrix before evaluating it.

To index after evaluating it, define

INDEX2 = @(Matrix, R, C) Matrix(R,C);

Then

INDEX2(f(1), 1, 1)

[TheOpenfield]

Okay, so in this case, you can not keep the function characteristics of said entry of the matrix.

[Walter Roberson]

I do not understand what you mean about keeping the function characteristics ?

If you need to define an anonymous function do this is, you can do that

g = @(x) x*INDEX2(f(x),1,1)

[Walter Roberson]

MINDEX = @(x, R, C) INDEX2(M(x), R, C)

[TheOpenfield]

I see, works the same way as seen below. There is still another problem to solve:

In my case, my function is set up as a multiplication of matrices containing functions as entries. Like:

M = @(x) f(x)*f2(x)...

The multiplication of the matrices f, f2 is done while evaluating M at any point.

Is there any easy way to index this function too even though M doesn't know about its matrix properties before evaluation?

[Walter Roberson]

As I said,

MINDEX = @(x, R, C) INDEX2(M(x), R, C)

[TheOpenfield]

Ahhh, I see! That's it!

Answer 2

[Andrei Bobrov]

function out = f(x,ii,jj)
    a = [x,1;1,x];
    out = a(ii,jj);
end

use

>> f(1,1,1)
ans =
     1
>>

[TheOpenfield]

This might be it!

Now i can do further calulations, without loosing the function characteristics, like:

g = @(x) x*f(x,1,1)

There might be still another problem:

In my case, my function is set up as a multiplication of matrices containing functions as entries. Like:

M = @(x) f(x)*f2(x)...

The multiplication of the matrices f, f2 is done while evaluating M at any point.

Is there any easy way to index this function too even though M doesn't know about its matrix properties before evaluation?