Hi. The question I am battling with is very simple in a way. How can one show that the given vectors w1 and w2 which are all 4x1 matrices are aactually eigenvectors of a matrix A given?

Question

anonymous · Answer

attachment

unklerhaukus · Answer

Have you computed $Aw_1?$

michele_laino · Answer

hint:
we have to compute Aw1, and Aw_2

anonymous · Answer

yes

anonymous · Answer

i have got Aw1 and Aw2

michele_laino · Answer

if w_1 and w_2 are actually eigenvectors, then Aw_1 has to be a multiple of w_1 and similarly for w_2

anonymous · Answer

so, do you divide ? how do you show that in the case of matrices and columnvectors like these?

michele_laino · Answer

is Aw_1 a column-vector in this form?
$$A{w_1} = k{w_1}$$

anonymous · Answer

yes, a 4x1 column. i want to assume that k is the eigenvalue?

michele_laino · Answer

yes!

anonymous · Answer

i got to that stage, but still,I can't really really separate k and w1 to show which is which after I have multiplied the l.h.s . Becasue they are equal, how do you take out your w1 ( and possibly remain with your k as well?

michele_laino · Answer

for example, I got this:
$$\begin{gathered}
  A{w_1} = \left( {\begin{array}{*{20}{c}}
  {1 + i}&i&0&0 \ 
  i&{1 + i}&0&0 \ 
  0&0&{1 + i}&{ - i} \ 
  0&0&{ - i}&{1 + i} 
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  i \ 
  i \ 
  {1 - i} \ 
  { - 1 + i} 
\end{array}} \right) =  \hfill \
   = \left( {\begin{array}{*{20}{c}}
  {i + {i^2} + {i^2}} \ 
  {{i^2} + i + {i^2}} \ 
  {1 - {i^2} + i - {i^2}} \ 
  { - i + {i^2} + {i^2} - 1} 
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {i - 2} \ 
  {i - 2} \ 
  {3 + i} \ 
  { - 3 - i} 
\end{array}} \right) \hfill \ 
\end{gathered} $$

michele_laino · Answer

please wait a moment I will come back as soon as possible

anonymous · Answer

me  too i got [-2+i,-2+i,3+i,-3-i] in column for like you did. I will wait

rational · Answer

rest should be easy
just conclude by showing that the resultant vector you got is parallel to the eigen vector

rational · Answer

multiply the scalar  $\large \dfrac{i}{i-2}$ to your resultant vector and see what you get

anonymous · Answer

thank you too rational, let me see what I get, I will tell you just now

anonymous · Answer

so @rational , how did we get this magic i/(i-2)  because its giving me back my w1

rational · Answer

haha it is magic!

rational · Answer

two vectors are parallel if their components are proportional, yes ?

rational · Answer

if the vectors

v1 = (a, b, c)  
v2 = (d, e, f)

are parallel then we have
a/d = b/e = c/f

anonymous · Answer

oh, i see, said the blindest man. thank you guys!

anonymous · Answer

thank you @Michele_Laino ! toooo

rational · Answer

yw :) please medal michele, not me :)

anonymous · Answer

he has 99 the maximum number, that how I passed it to you lol

michele_laino · Answer

thank you! @MERTICH

anonymous · Answer

oh pardon me, it's 91. but your help exceeds  medals so i am equally grateful