0

(I'm sorry for the vague title...please feel free to edit it, but I couldn't come up with a different title that wasn't super long and still captured the question).

Question:

I originally asked this...

Do commuting matrices with real and complex eigenvectors still share eigenvectors?

...question, which I think was well answered, but from it now I have a more general one.

Say we have some reason to believe that two matrices, $A$ and $B$, share eigenvectors. However, we end up finding out that $A$ has all real eigenvectors, while $B$ has eigenvectors with complex components.

Will vectors holding the real parts of the complex components eigenvectors of $B$ always be eigenvectors of $A$?

If they are, why? If not, what CAN we say about the relationship between the eigenvectors of $B$ and the eigenvectors of $A$?

Thanks!

Example:

As an example (like the one in the question linked above), say that two matrices commute. Commuting matrices share eigenvectors. However, the eigenvectors of one matrix are real, while the eigenvectors of the other are complex.

As a different example, consider a circulant $C$ matrix, formed via a linear combination of the single row permutation matrix $P$.

$P=\left[\begin{array}{llll}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0\end{array}\right]$

$C=\left[\begin{array}{llll}c_{0} & c_{1} & c_{2} & c_{3} \\ c_{3} & c_{0} & c_{1} & c_{2} \\ c_{2} & c_{3} & c_{0} & c_{1} \\ c_{1} & c_{2} & c_{3} & c_{0}\end{array}\right]$

$C=c_{0} P^{0}+c_{1} P^{1}+c_{2} P^{2}+c_{3} P^{3}$

From that last equation, we can see that they would share eigenvectors. However, as shown in the linked question, its possible that $C$ has real eigenvectors, while $P$ has complex ones.

enter image description here

(And then the eigenvectors of the permutation matrix hold the roots of unity, which is explained in this pdf linked here: http://perrylocal.org/meinkea/files/2010/10/Circulant-Matrices.pdf)

joshuaronis
  • 1,479
  • 10
  • 29
  • Are we given that $A$ and $B$ have real entries? – Ben Grossmann Aug 17 '20 at 18:19
  • @BenGrossmann thanks! What if they don't? Or in that case, could we not reach "a reason to believe that they have the same eigenvectors"? – joshuaronis Aug 17 '20 at 18:43
  • If $A$ does not have real entries, then it does not necessarily hold that the real and imaginary parts of the eigenvectors of $B$ are also eigenvectors of $A$. – Ben Grossmann Aug 17 '20 at 19:10
  • @BenGrossmann thank you! If u have the time, I'd appreciate an explanation! – joshuaronis Aug 17 '20 at 19:21
  • Randomly generated counterexample: $$ A = \left(\begin{array}{cc} 25 & 24-18{}\mathrm{i}\ -16-12{}\mathrm{i} & -24 \end{array}\right), \quad B = \left(\begin{array}{cc} -23{}\mathrm{i} & -18-24{}\mathrm{i}\ -12+16{}\mathrm{i} & 26{}\mathrm{i} \end{array}\right). $$ Both $A$ and $B$ have eigenvectors equal to the columns of $$ P = \left(\begin{array}{cc} 3+4{}\mathrm{i} & 6{}\mathrm{i}\ -4{}\mathrm{i} & 3-4{}\mathrm{i} \end{array}\right), $$ but $A$ has real eigenvalues $0,1$. – Ben Grossmann Aug 17 '20 at 19:57

1 Answers1

3

I assume that $A$ has real entries.

The answer is yes: the real and imaginary components of the complex eigenvector of $B$ will necessarily be eigenvectors of $A$.

Let $x + iy$ denote the complex eigenvector of $B$ (with $x$ and $y$ both vectors with real entries). We know that this must also be an eigenvector of $A$. However $A$ has only real eigenvalues. So, there must exist a real number $\lambda$ such that $$ A(x + iy) = \lambda(x + iy) = \lambda x + i\lambda y. $$ Because $A$ is a matrix with real entries, we find that $Ax$ and $Ay$ are necessarily vectors with real entries. So, we have $$ A(x + iy) = \lambda x + i \lambda y \implies (Ax) + i(Ay) = \lambda x + i \lambda y \implies \begin{cases} Ax = \lambda x,\\ Ay = \lambda y. \end{cases} $$ So, it is indeed the case that both $x$ and $y$ are eigenvalues of $A$ associated with the same real eigenvalue $\lambda$.

Ben Grossmann
  • 225,327