0
$\begingroup$

Given a basis of complex eigenvectors for, say, a $2 \times 2$ symmetric matrix $A$ (which hence has real eigenvalues).

Can one generate a basis of real(-valued) eigenvectors from the real and imaginary parts of the given basis of complex eigenvectors?

Say $v = ((a_1 + b_1i),(a_2 + b_2i))$ is a complex eigenvector with real eigenvalue $\lambda_1$ and $w = ((c_1 + d_1i), (c_2 + d_2i))$ is a complex eigenvector with real eigenvalue $\lambda_2$

and these two complex eigenvectors $v$ and $w$ are orthogonal.

Then $(a1,a2)$ and $(b1,b2)$ are real eigenvectors with real eigenvalue $\lambda_1$ (for matrix $A$) And $(c1,c2)$ and $(d1,d2)$ are real eigenvectors with real eigenvalue $\lambda_2$ (for matrix $A$)

Can these real-valued eigenvectors be used to construct an orthogonal basis of real-valued eigenvectors for A? How does the construction proceed?

$\endgroup$
6
  • $\begingroup$ A complex eigenvector in this case would be a scalar complex multiple of a real eigenvector. Once you factor out that scalar complex multiple you will have your real eigenvector. $\endgroup$ – Paul May 7 '20 at 18:48
  • $\begingroup$ How does orthogonality of the real eigenvectors follow? Or independence at least? $\endgroup$ – Michel May 7 '20 at 18:50
  • $\begingroup$ @Paul Not quite. $A$ could be the identity matrix. $\endgroup$ – Robert Israel May 7 '20 at 18:50
  • $\begingroup$ There can only be 2 linearly independent eigenvectors, one for each eigenvalue. Each eigenvalue will not have 2 different eigenvectors, like you state above. $\endgroup$ – Paul May 7 '20 at 18:52
  • $\begingroup$ I am not insisting that the eigenvalues are different. And I did not state that each eigenvalue has 2 different eigenvectors? Merely that some of the real (or imaginary parts) should be used to form (or construct) a basis of real eigenvectors. $\endgroup$ – Michel May 7 '20 at 19:04
1
$\begingroup$

If $x$ is a complex eigenvector of a real matrix $A$ for a real eigenvector $\lambda$, i.e. $A x = \lambda x$, then taking real and imaginary parts in this equation we find that $\text{Re}(x)$ and $\text{Im}(x)$, if nonzero, are also eigenvectors of $A$ for eigenvalue $\lambda$. Since $x$ is nonzero, at least one of $\text{Re}(x)$ and $\text{Im}(x)$ is nonzero, and of course $x$ is in the linear span of $\text{Re}(x)$ and $\text{Im}(x)$.

If you have a basis consisting of complex eigenvectors, take their real and imaginary parts and you still have a set that spans the whole space. Take a maximal linearly independent subset and you have a basis. It might not be orthogonal: eigenvectors for different eigenvalues are automatically orthogonal, but eigenvectors for the same eigenvalue might not be. So you might use the Gram-Schmidt process to find an orthonormal basis.

$\endgroup$
1
  • $\begingroup$ To conclude the argument, I assume that one still needs to argue that if you take a maximal linearly subset of real and imaginary parts, forming a basis in C, it will form a basis in R since the dimension of the nulspace (A - lambda I ) is invariant when working in R or working in C (?). I.e. there are always sufficiently many linearly independent real parts of complex eigenvalues to form a basis in the R-case. This seems to conclude matters. $\endgroup$ – Michel May 8 '20 at 6:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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