# About unsymmetric and symmetric matrices' eigenvalue problem

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? –  Raskolnikov Feb 20 '12 at 18:31
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.
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 $c' + D c = V^{-1} b$. –  Robert Israel Feb 21 '12 at 1:27