Eigenvector calculation, i have worked out everything for you! just need verification!

Question

anonymous · Answer

M=
  2 0  -2
  0  4  0  
-2  0  5


eigenvalues are 1, 4, 6,


need eigenvector for eigenvalue = 4 please!!

amistre64 · Answer

subtract 4 from the diags and rref

anonymous · Answer

i have done that but you get 

M- lambda = 
-2  0  -2
   0  0  0 
-2  0 -1

!!!

amistre64 · Answer

keep reducing ...

anonymous · Answer

M3,3 = 1  not -1... sorry

amistre64 · Answer

-2  0  -2
  0  0  0 
-2  0   1


  1  0   1
  0  0   0 
-1  0  1/2


  1  0   1
  0  0   0 
  0  0  3/2

  1  0   1
  0  0   0 
  0  0   1

  1  0   0
  0  0   0 
  0  0   1

right?

anonymous · Answer

which implies eigen vector of (0,0,0)!!!
can this be possible?   or am i wrong?

amistre64 · Answer

x1          0
x2 = x2  1
x3          0

anonymous · Answer

AHHHHHH   and (0,1,0)!!

amistre64 · Answer

the vectors we get from working the parametric form of the free variables become our Evectors

anonymous · Answer

Thanks amistre64!!!

amistre64 · Answer

youre welcome

anonymous · Answer

In a way, one has to solve
$$\left(
\begin{array}{ccc}
 2 & 0 & -2 \
 0 & 4 & 0 \
 -2 & 0 & 5 \
\end{array}
\right) \left(
\begin{array}{c}
 x \
 y \
 z \
\end{array}
\right)=4 \left(
\begin{array}{c}
 x \
 y \
 z \
\end{array}
\right)

$$

anonymous · Answer

yep,   got confused when i saw two zeros with my working out!

anonymous · Answer

You get x=0, z=0 and y can be anything different from 0

anonymous · Answer

what if x and z were not zeros?

anonymous · Answer

if x and y were not zero, 4 will not be an eigenvalue of the vector
(x,y,z)

anonymous · Answer

assuming it was, and x had a value of 1 and z had a value of 2, y= 1,       how would the matrix P and D be id P^-1  *  M*P  =D, where D is a diagonal matrix?

anonymous · Answer

The three eigenvectors are
{{-1, 0, 2}, {0, 1, 0}, {2, 0, 1}}

anonymous · Answer

$$
P=\left(
\begin{array}{ccc}
 -1 & 0 & 2 \
 0 & 1 & 0 \
 2 & 0 & 1 \
\end{array}
\right)\
P^{-1} M P =\left(
\begin{array}{ccc}
 6 & 0 & 0 \
 0 & 4 & 0 \
 0 & 0 & 1 \
\end{array}
\right)
$$

anonymous · Answer

yes but wouldnt that mean there would be four eigenvectors for the eigenvalues? 

what im trying to say is if you have a matrix M   that has  three eigenvalues but one of the eigenvalue has  two eigen vectors?     

lets say one of the eigenvalue was 4  and we get   M-(Lambda*I) = =
4 2 4
2 1 2
4 2 4

what would the eigen vector be?

anonymous · Answer

The maximum you can have is 3 independent eigenvectors.

anonymous · Answer

exactly!! so how would P  look like?  would there 5 columns?  and would this mean that D  would no more be a diagonal matrix?

anonymous · Answer

If Z is an eigenvector for 4 then
17 Z is an eigenvector for 4 then

anonymous · Answer

M (17 Z) = 17 M Z = 17(4 Z) = 4 ( 17 Z)

anonymous · Answer

P would look like
$$
P=\left(
\begin{array}{ccc}
 -1 & 0 & 2 \
 0 & 17 & 0 \
 2 & 0 & 1 \
\end{array}
\right)
$$
and it will still do the job.
$$P^{-1} M P=\left(
\begin{array}{ccc}
 6 & 0 & 0 \
 0 & 4 & 0 \
 0 & 0 & 1 \
\end{array}
\right)
$$

anonymous · Answer

Did you see what is going on?

anonymous · Answer

P Should be a 3x3 matrix whose rows are the independents eigenvectors.

anonymous · Answer

Usually we choose small numbers like 1 to make the computation easier.

anonymous · Answer

but what if an eigen value gives two eigenvectors?

anonymous · Answer

eigen space i think it was called

amistre64 · Answer

as long as the results and up in linearly INdependant vectors, your good

anonymous · Answer

If they are independent we take both. In this case, the eigenvalue is a double roots.

amistre64 · Answer

not all square matrixes are diagonalizable

anonymous · Answer

ahhhh okay!!

anonymous · Answer

thanks to all, well appreciated  !

anonymous · Answer

yw