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Say we have a 2 by 2 matrix $A$ with real entries and $A$ has a complex eigenvector $V = a+bi$ with corresponding complex eigenvalue $\lambda$. How do I prove that the vectors $\mathrm{Re}(V) = a$ and $\mathrm{Im}(V) = b$ are linearly independent? This is a common fact that is used to produce real solutions for a system of differential equations with complex eigenvalues and vectors.

Assumptions and Facts:

  • We know that $\overline{V} = a-bi$ is also an eigenvector with eigenvalue $\overline{\lambda}$.
  • We obviously have that $V$ and $\overline{V}$ are linearly independent. (For one thing, eigenvectors of distinct eingenvalues must be LI.)

I have started a few ways by trying to find a contradiction, assuming that there is a nonzero $k \in \mathbb{R}$ such $\mathrm{Re}(V)$ = $k\mathrm{Im}(V)$. I tried using the definitions of $\mathrm{Re}(V)$ and $\mathrm{Im}(V)$ in terms of $V$ and $\overline{V}$ but that didn't get me anywhere.

I tried starting with the fact that $V$ and $\overline{V}$ are LI to show this directly implies that $V$ and $\overline{V}$ line must be LI. I also couldn't work this through. Please help me out!

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  • $\begingroup$ I'm assuming $A$ is real, otherwise it's not true. $\endgroup$
    – Git Gud
    Dec 6, 2014 at 0:01
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    $\begingroup$ If $A=I$, then every nonzero vector is an eigenvector, including ones where the real and imaginary parts are dependent. $\endgroup$
    – Nishant
    Dec 6, 2014 at 0:17
  • $\begingroup$ @GitGud: Yes, thanks, that's what I intended. $\endgroup$
    – mb7744
    Dec 6, 2014 at 1:20
  • $\begingroup$ @Nishant: Shoot, another implicit assumption I missed. The corresponding $\lambda$ and $\overline{\lambda}$ should be assumed to be complex. Now, these points aside, does anyone have a hint they can give me? $\endgroup$
    – mb7744
    Dec 6, 2014 at 1:22
  • $\begingroup$ If V and its conjugate are independent, the isn't it just enough to consider V + V* and i(V - V*)? Apologies, I can't type latex off my tablet. $\endgroup$
    – user27182
    Dec 6, 2014 at 1:27

3 Answers 3

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I think I may have found the proof, let me know if this is solid:

Using the knowledge that $V = a+ib$ and $\overline{V} = a-ib$ are LI, we know that for any $k_1, k_2 \in \mathbb{C}$: $$ k_1 (a+ib) + k_2 (a-ib) = 0 \implies k_1 = k_2 = 0$$ Thus $$ (k_1+k_2)a + (k_1-k_2)ib = 0 \implies k_1 = k_2 = 0$$ So for any two numbers $c_1$ and $c_2$, assume: $$ c_1 a + c_2 b = 0 $$ Then set $$k_1 = \frac{c_1+c_2 i}{2} \; ; \; k_2 = \frac{c_1-c_2 i}{2} $$ so the statement becomes $$ (k_1+k_2)a + (k_1-k_2)ib = 0 $$ which implies $ k_1 = k_2 = 0$ which in turn implies $ c_1 = c_2 = 0$.

So we have linear independence of $a$ and $b$.

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  • $\begingroup$ It's fine. ${}$ $\endgroup$
    – Git Gud
    Dec 6, 2014 at 4:13
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Let one of the complex eigenvectors be $\vec{v} = \vec{u_1} + i\vec{u_2}$.

Let us suppose, by way of contradiction, that $\exists k_1,k_2$ both non-zero such that $k_1\vec{u_1} + k_2\vec{u_2} = 0$ (i.e. $\vec{u_1}$ $\vec{u_2}$ are lin. dependent).

Then, if $\vec{w}$ is the complex conjugate of $\vec{v}$, consider the following:

$$k_1\vec{v} + ik_2\vec{w} = k_1\vec{u_1} + ik_1\vec{u_2} + ik_2\vec{u_1} + k_2\vec{u_2} = 0$$

Which is a contradiction since eigenvectors corresponding to different eigenvalues are linearly independent.

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More generally, if $\lambda_1,\overline{\lambda}_1,\dotsc,\lambda_r,\overline{\lambda}_r$ are distinct complex eigenvalues of a real matrix, with corresponding eigenvectors $a_k \pm i b_k$, then $a_1,b_1,a_2,b_2,\dotsc,a_r,b_r$ are linearly independent because they have same $(2r)$-dimensional span as the original list of eigenvectors.

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    $\begingroup$ I was just about to state a more specialized version of this before I read your answer. Love the generality! One question: if it's already stated that the eigenvalues occur in conjugate pairs, is it necessary to restrict the entries of the matrix to real values? $\endgroup$
    – Ugo
    Nov 4, 2021 at 9:32
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    $\begingroup$ @Ugo If I understand your question, it is: "If a square complex matrix has eigenvalues in conjugate pairs, then are the real and imaginary parts of its eigenvectors linearly independent?" No, without assuming the matrix has real entries this can fail. When the matrix has real entries it implies that the corresponding eigenvectors are also conjugate. Without assuming the matrix has real entries, the eigenvectors might not be conjugate. For example form the matrix with eigenvalues $1 \pm i$ and eigenvectors $(1,1+i)$ and $(1+i,-1+2i)$: then the parts are $(1,1), (0,1), (1,-1), (1,2)$. $\endgroup$ Nov 29, 2021 at 4:35
  • $\begingroup$ I'm not sure I understand why this is true. Assuming $x_1a_1+y_1b_1+\ldots+x_ra_r+y_rb_r = 0$, why does $x_1=y_1=\ldots=x_r=y_r=0$ follow? I understand that the set $\{a_k\pm ib_k: k=1,\ldots,r\}$ is linearly independent since the eigenvectors of this set correspond to different eigenvalues, but I don't see how this implies what I want. $\endgroup$
    – Simon SMN
    Apr 23, 2023 at 10:35

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