create random diagonalisable matrix

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hi.. I would like to create a random diagonalisable integer matrix. Is there any code for that? thereafter I would want to create matrix X such that each the columns represent the eigenvectors.

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David Goodmanson
David Goodmanson 2018-9-19
Hi Gary,
another way:
n = 7 % A is nxn
m = 9 % random integers from 1 to m
X = randi(m,n,n)
D = round(det(X))
lam = 1:n % some vector of unique integer eigenvalues, all nonzero
lamD = lam*D % final eigenvalues
A = round(X*diag(lamD)/X)
A*X - X*diag(lamD) % check
If n is too large and m is too small, this doesn't work sometimes because X comes up as a singular matrix.
  3 个评论
Bruno Luong
Bruno Luong 2018-9-19
编辑:Bruno Luong 2018-9-19
Actually there is no problem of lam to have null element(s). One can also select it randomly in the above code if the spectral probability is matter.
p = 5; % eg
lam = randi(p,1,n)
David Goodmanson
David Goodmanson 2018-9-19
spectral variation does seem like a good idea.

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更多回答(2 个)

Bruno Luong
Bruno Luong 2018-9-18
编辑:Bruno Luong 2018-9-19
Code for both A and X are integer.
I edit the 1st version of the code (if you happens to see t) essentially a bug correction and better generation and simplification. Second edit: fix issue with non-simple eigen-value.
% Building A random (n x n) integer matrix
% and X (n x n) integer eigen-matrix of A
% meaning A*X = diag(lambda)*X
n = 4;
m = 5;
p = 5;
d = randi(2*m+1,[1,n])-m-1;
C = diag(d);
while true
P = randi(2*p+1,[n,n])-p-1;
detP = round(det(P));
if detP ~= 0
break
end
end
Q = round(detP * inv(P));
A = P*C*Q;
g = 0;
for i=1:n*n
g = gcd(g,abs(A(i)));
end
A = A/g;
lambda = sort(d)*(detP/g);
I = eye(n);
X = zeros(n);
s = 0;
for k=1:n
Ak = A-lambda(k)*I;
r = rank(Ak);
[~,~,E] = qr(Ak);
[p,~] = find(E);
j1 = p(r+1:end);
j2 = p(1:r);
[~,~,E] = qr(Ak(:,j2)');
[p,~] = find(E);
i1 = p(r+1:end);
i2 = p(1:r);
Asub = Ak(i2,j2);
s = mod(s,length(j1))+1;
x = Ak(:,j2) \ Ak(:,j1(s));
y = zeros(n-r,1);
y(s) = -1;
x = round([x; y]*det(Asub));
g = 0;
for i=1:n
g = gcd(g,abs(x(i)));
end
X([j2;j1],k) = x/g;
end
D = diag(lambda);
A
X
% % Verification A*X = X*D
A*X
X*D

Matt J
Matt J 2018-9-18
编辑:Matt J 2018-9-18
How about this,
A=randi(m,n);
A=A+A.';
[X,~]=eig(A,'vector');
  4 个评论
Matt J
Matt J 2018-9-18
编辑:Matt J 2018-9-18
I don't think the problem is specified well enough. Eigenvectors are always unique only up to a scale factor and, in finite precision computer math, can always be made integer if you multiply them by a large enough scaling constant.

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