Real and imaginary parts of basis of complex eigenvectors to form base of real eigenvectors?

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?

• 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. – Paul May 7 '20 at 18:48
• How does orthogonality of the real eigenvectors follow? Or independence at least? – Michel May 7 '20 at 18:50
• @Paul Not quite. $A$ could be the identity matrix. – Robert Israel May 7 '20 at 18:50
• There can only be 2 linearly independent eigenvectors, one for each eigenvalue. Each eigenvalue will not have 2 different eigenvectors, like you state above. – Paul May 7 '20 at 18:52
• 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. – Michel May 7 '20 at 19:04

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 $$\text{Im}(x)$$.