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suppose I have a symmetric matrix A in a differential equation,

$\displaystyle \frac{dx}{dt}+Ax=b$

Now, if $V=$ eigenspace of $A$ and $D=$ eigenvalue of $A$
we can write $x=V*c$, where $c=$ coefficients
then,

$\displaystyle \frac{d(V*c)}{dt}+A*(V*c)=b$

that is,

$\displaystyle \frac{d(c)}{dt}+D*c=V'*b$ where $V'*A*V=D$

I think this procedure is very well known. But in my work, my prof is saying that if $A$ is unsymmetric, this procedure is still true. I am confused because my simulation showed it is not possible. I need help to know whether it is possible or not? please.

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    Where do you use the symmetry in your argument?2012-02-20

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If $A$ is not symmetric, you want to use $V^{-1}$ rather than $V'$ (a real symmetric matrix can be diagonalized with an orthogonal matrix; a non-symmetric matrix may still be diagonalizable, but not with an orthogonal matrix). There may be complex eigenvalues, and not every square matrix is diagonalizable - in general you'll have to consider Jordan canonical form.

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    But could it be expanded at u=V*c..i mean solution (when unsymmetric matrix is used) can be expanded in eigenspace, it is possible or not?2012-02-20
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    Yes, if the matrix is diagonalizable, and $V$ is a matrix whose columns are the eigenvectors, then $A = V D V^{-1}$ where $D$ is diagonal, and you can write $x = V c$ where $c' + D c = V^{-1} b$.2012-02-21
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    thanks sir, now I think i have to recheck my code again. Thanks a lot.2012-02-21