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what type of decomposition are you doing here ?

you should get D=dia(-3,-1,-3)

ok I will try!

well, as long as the eigen vector columns are arranged in proper order, you should be good.

still no luck...

no

eigen vectors=?
for t=-1:
\[-2x+2y+6z=0\\6z=0\\-3z=0\implies v_1=(1,1,0)^T\]

yup

and v2 and v3 I put at (0,-3,1)

for t= -3:
\[
2y+6z=0\\
2y+6z=0\\
v_{2,3}=(\alpha,-3\beta,\beta)^T
\]

oh,ok..

you can use
v2 when a=0,b=1
v3 when a=1,b=1

or v3 when a=1,b=0
ANY combination of alpha and beta
that do not give the same vectors..

can you give me an example?

got it!

I forgot to set alpha!

thanks @electrokid, great help!

I did nothing guy!!!