Complex Conjugate Eigenvectors Proof

a.) Prove that if $x ^+_- iy$ is a complex conjugate pair of eigenvectors of a real matrix A corresponding to complex conjugate eigenvalues $\mu ^+_- iv$ with $v\ne 0$, then x and y are linearly independent real vectors.

b.) More generally, if $v_j=x_j {^+_-} iy_j$, $j=1,...,k$ are complex conjugate pairs of eigenvectors corresponding to distinct pairs of complex conjugate eigenvalues $\mu_j {^+_-} iv_j,v_j \ne 0$, then the real vectors $x_1,...,x_k,y_1,...,y_k$ are linearly independent.

How will I be able to prove these?

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This is straightforward by definition. –  Bombyx mori Nov 12 '12 at 1:13
@user32240 what do you mean by that? –  diimension Nov 12 '12 at 1:28
I mean this is straightforward; different eigenvalues correspond to linearly independent eigenvectors. Now if $a,b$ is linearly independent, so is $\frac{a+b}{2}$ and $-i\frac{a-b}{2}$. –  Bombyx mori Nov 12 '12 at 1:54

a) Assume that $cx = y$ and $c\in {\bf R}$

$A(x + icx) = (\mu + i \nu)(x+icx)$ so that $Ax = (\mu + i \nu)x$

Hence $\nu=0$. Contradiction !

b) $Av=cv$ and $Aw = dw$ where $A$ is real, $c$, $d\in {\bf C}$, and $v\neq 0$, $w\neq 0\in {\bf C}^n$

If $c\neq d$, and if $av + bw=0$, then $0= aAv + b Aw = acv + bdw$ so that $-av = bw=bd/cw$ Hence $b=0$ implies $a=0$. $b\neq 0$ implies $d/c=1$ It is a contradiction. So we conculde that eigenvectors corresponded to differen eigenvalues are independent.

Similarly we can show that $n$ eigenvectors corresponded to different $n$ eigenvalues are independent.

Assume that $a_i x_i + b_i y_i =0$. So $(a_k -i b_k) (x_k + iy_k) + (a_k + ib_k)(x_k-iy_k) = 0$ This implies that $a_k=b_k=0$ So we finished the proof.

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Thank you very much! –  diimension Nov 12 '12 at 4:29