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

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Georgios Kirkinezos
Georgios Kirkinezos 2020-11-14
编辑： Bruno Luong 2020-11-15
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

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采纳的回答

Matt J
Matt J 2020-11-14
clear e
for i=numel(T):-1:1
e(:,:,i)=expm(Q*T(i));
end

更多回答（3 个）

James Tursa
James Tursa 2020-11-14
编辑：James Tursa 2020-11-15
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
Setsuna Yuuki.
Setsuna Yuuki. 2020-11-14
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 * --> .*
Bruno Luong
Bruno Luong 2020-11-14
编辑：Bruno Luong 2020-11-14
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))
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Bruno Luong
Bruno Luong 2020-11-14
编辑：Bruno Luong 2020-11-14
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
Bruno Luong 2020-11-15
编辑：Bruno Luong 2020-11-15
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

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