Why is the intersection of spans zero? 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.
 A: 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:


*

*$u_1=\bar{u}_1=0$. 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 $x=0$.

*$u_1,\bar{u}_1\ne0$. 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.
