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As a part of a larger proof, my text claims that if

$$A\begin{bmatrix}u_1&u_2\\ \end{bmatrix}=\begin{bmatrix}u_1&u_2\\ \end{bmatrix}\begin{bmatrix}\lambda&1\\0&\lambda\\ \end{bmatrix}$$

where $A\in \mathbb R^{n\times n}, u_1, u_2$ are complex vectors, and $\lambda \in \mathbb C \setminus \mathbb R$, then

$$\text{span}\{u_1, u_2 \}\cap \text{span}\{ \overline{u_1}, \overline{u_2} \}=\{ 0 \}$$

I don't quite understand why. So we conjugate both sides:

$$A\begin{bmatrix}\overline{u_1}&\overline{u_2}\\ \end{bmatrix}=\begin{bmatrix}\overline{u_1}&\overline{u_2}\\ \end{bmatrix}\begin{bmatrix}\overline{\lambda}&1\\0&\overline{\lambda}\\ \end{bmatrix}$$

And we know that $\lambda \neq \overline{\lambda}$.

Naively taking an element in the intersection of spans and writing it out as a combination doesn't seem to help.

There must be some simple argument I don't see.

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  • $\begingroup$ What are $u_1$ and $u_2$? numbers? vectors? real? complex? $\endgroup$ – Gerry Myerson May 4 '14 at 7:20
  • $\begingroup$ @GerryMyerson: complex vectors. $\endgroup$ – Leo May 4 '14 at 7:24
  • $\begingroup$ What does the notation $A [u_1 u_2]$ mean? I am assuming $u_i \in \mathbb R^{1 \times n}$. $\endgroup$ – user89987 May 4 '14 at 15:43
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    $\begingroup$ @user89987: $u_i \in \mathbb C^{n \times 1}$ and $A[u_1 \ u_2]=[Au_1 \ Au_2]\in \mathbb C^{n \times 2}$, as far as I understood. $\endgroup$ – Leo May 4 '14 at 16:03
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Edit. OK, I have misread the question. Here is a new answer. I've just rushed for a quick fix and haven't spent much time on it. So, I guess the proof below is very clumsy, and more elegant proofs can be found in those textbooks that discuss real Jordan forms.

By assumption, we have \begin{cases} (A-\lambda I)u_1 = 0,\\ (A-\lambda I)u_2 = u_1,\\ (A-\bar{\lambda}I)\bar{u}_1 = 0,\\ (A-\bar{\lambda}I)\bar{u}_2 = \bar{u}_1. \end{cases} Suppose inside $\operatorname{span}\{u_1, u_2\}\cap\operatorname{span}\{\bar{u}_1, \bar{u}_2\}$ there lies a vector $$x = au_1+bu_2=c\bar{u}_1+d\bar{u}_2\tag{$\ast$}$$ where $a,b,c,d$ are some complex numbers. Apply $(A-\lambda I)(A-\bar{\lambda}I)$ on both sides (note that the two multiplicands of this matrix product commute), we get \begin{align} (A-\bar{\lambda}I)bu_1 &= (A-\lambda I)d\bar{u}_1,\\ b(\lambda -\bar{\lambda})u_1 &= d(\bar{\lambda}-\lambda )\bar{u}_1,\\ bu_1 &= -d\bar{u}_1.\tag{$\dagger$} \end{align} Now there are two possibilities:

  1. $u_1=\bar{u}_1=0$. So we have $x=bu_2=d\bar{u}_2$ in $(\ast)$. If $x$ is nonzero, then $u_2=\frac{d}{b}\bar{u}_2$, i.e. $u_2$ is a nonzero multiple of a nonzero real vector, but this is impossible because $A$ is real, $\lambda$ is nonreal and $(A-\lambda I)u_2 = 0$. Hence $x=0$.
  2. $u_1,\bar{u}_1\ne0$. From $(\dagger)$, $b$ and $d$ must be both zero or both nonzero. If they are both nonzero, then $u_1=\frac{-d}{b}\bar{u}_1$. By a similar reasoning to the above, we see that this is impossible. Therefore $b=d=0$ and we get $x = au_1=c\bar{u}_1$ in $(\ast)$. Again, $a$ and $c$ must be zero. Hence $x=0$.

In other words, $x$ must be zero, i.e. the two spans in question have a zero intersection.

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  • $\begingroup$ $u_1$ and $u_2$ are in ${\bf C}^n$, not ${\bf C}^2$. $\endgroup$ – Gerry Myerson May 5 '14 at 13:20
  • $\begingroup$ @GerryMyerson Thanks for catching my error. $\endgroup$ – user1551 May 5 '14 at 17:31

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