MATLAB Problem. How to design matrix from eigenvalues

This is a previously posted problem that I am working on but I can't find any solutions online. Here is the description:

Write a function that takes one input argument n and outputs a (n x n) square matrix A with the following properties:

• A has an eigenvalue of 3
• all elements of A differ more than .5 from each other

Here is what I have tried so far. I am having difficulty with setting up the specific eigenvalue.

function A = matrix_design(n)
A=randn(n); % set up the matrix
[V,D]=eig(A) % use built-in function for eigendecomposition
abs(A-A.')>0.5 % specify elements of A to differ by 0.5
end
disp(A)

[Torsten]

I don't understand how the code could lead to a solution of the problem.

[Walter Roberson]

Is it possible at all?

d = sym(2)/3

d = 

syms c
M = [c+0*d, c+1*d, c+2*d
     c+3*d, c+4*d, c+5*d
     c+7*d, c+9*d, c+8*d]

M = 

[V,D] = eig(M);
c0 = solve(D(1)==3, c)

c0 = 

Mnum = double(subs(M, c, c0))

Mnum = 3×3

-2.6566 -1.9899 -1.3232 -0.6566 0.0101 0.6768 2.0101 3.3434 2.6768

eig(Mnum)

ans = 3×1

-2.6320 -0.3377 3.0000

... I guess so, at least for the 3 x 3 case -- and assuming that "A has an eigenvalue of 3" means that at least one of the eigenvalues of the matrix is 3

Answer 1

[Matt J]

mindist=@(A) min(sqrt(pdist(real(A(:))).^2+pdist(imag(A(:))).^2));

n=5;

Q=rand(n); Q(1)=0;

Q=(1-eye(n)).*Q*0.5/mindist(Q);

r=max(min(sum(Q,2)),0.5);  

Q=diag((1:n)*2*r)-Q;

emin=min(eig(Q));

A=Q+speye(n)*(3-emin);



%%CHECK%%%
min(eig(A))

ans = 3.0000

mindist(A)

ans = 0.5000

Answer 2

[Matt J]

mindist=@(A) min(sqrt(pdist(real(A(:))).^2+pdist(imag(A(:))).^2));

n=5;

P=rand(n);
D=diag(rand(1,n));
A=P*D/P;

A=A/mindist(A)*0.5;

e=eig(A);

A=A+(3-e(1))*speye(n);


%%Check

mindist(A)

ans = 0.5000

eig(A)

ans = 5×1

3.0000 16.7245 25.2124 29.3054 28.3224