multiplying matrix by another matrix element and using it in a command

Q=[-0.0905 0.0604 0.0301;0.142 -0.239 0.097;0 0 0];
T =[ 1, 3, 5, 7, 10];
e=expm(Q*T);   %%% the problem is here

I have a problem with calculating e, i want to multiply Q with each value of T and still use the expm command

Answer 1

[Matt J]

clear e
for i=numel(T):-1:1
   e(:,:,i)=expm(Q*T(i)); 
end

Answer 2

[James Tursa]

You can use a loop

QT = Q .* reshape(T,1,1,[]);
e = zeros(size(QT));
for k=1:size(QT,3)
    e(:,:,k) = expm(QT(:,:,k)); % fixed
end

[Georgios Kirkinezos]

ALSO WORKS!!!

Giving same result like the other accepted answer!

(note:it is missing a clossing parenthesis in the end )

Answer 3

[Setsuna Yuuki.]

You need matrix T to have the same number of rows or columns as matrix Q

Q=[-0.0905 0.0604 0.0301;0.142 -0.239 0.097;0 0 0];
%T =[ 1, 3, 5, 7, 10]; 
T =[1,3,5]
e=expm(Q*T); %change * --> .*

Answer 4

[Bruno Luong]

This must be the fastest if your T is a large vector

[P,D] = eig(Q);
e = exp(diag(D)*T(:).');
e = reshape(e,1,size(Q,1),length(T));
e = pagemtimes(P.*e,inv(P))

[Matt J]

Does this still work if Q is not diagonalizable?

[Bruno Luong]

You mean the Jordan form? It does not exist in numerical world. ;-)

I'm kidding. Indeed Q supposes to be diagonalizable.

Q = P*D/P

D diagonal.

[Bruno Luong]

Speed comparison between expm for loop and eigen-value methods. It speed up about 100 fold.

Q = [-0.0905 0.0604 0.0301;
     0.142 -0.239 0.097;
     0 0 0];
T = linspace(0,10,10000);

tic
clear E1
for i=numel(T):-1:1
E1(:,:,i)=expm(Q*T(i));
end
toc % Elapsed time is 0.227108 seconds.

tic
[P,D] = eig(Q);
E2 = exp(diag(D)*T(:).');
E2 = reshape(E2,[1,size(E2)]);
E2 = pagemtimes(P.*E2,inv(P));
toc % Elapsed time is 0.001398 seconds.

norm(E1(:)-E2(:),Inf) % 4.7531e-16