Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

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.

share|cite|improve this question
Where do you use the symmetry in your argument? – Raskolnikov Feb 20 '12 at 18:31
up vote 1 down vote accepted

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.

share|cite|improve this answer
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? – gman Feb 20 '12 at 21:44
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
thanks sir, now I think i have to recheck my code again. Thanks a lot. – gman Feb 21 '12 at 15:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.