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?
Best Answer
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.