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?

  • $\begingroup$ 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, 2020 at 18:48
  • $\begingroup$ How does orthogonality of the real eigenvectors follow? Or independence at least? $\endgroup$ May 7, 2020 at 18:50
  • $\begingroup$ @Paul Not quite. $A$ could be the identity matrix. $\endgroup$ May 7, 2020 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, 2020 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. $\endgroup$ May 7, 2020 at 19:04

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ May 8, 2020 at 6:00

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