MedVision ad

Eigenvalues question (1 Viewer)

undalay

undalay

Active Member
Joined
Dec 14, 2006
Messages
1,002
Location
Ashfield
Gender
Male
HSC
2008
How do you do part (ii) guys ) :

edit: free cookie if u get it
 
Last edited:
I

Iruka

Member
Joined
Jan 25, 2006
Messages
544
Gender
Undisclosed
HSC
N/A
Bx is also an eigenvector, hence it is also in the eigenspace of lambda. What does this tell you about the form of Bx?
 
undalay

undalay

Active Member
Joined
Dec 14, 2006
Messages
1,002
Location
Ashfield
Gender
Male
HSC
2008
Iruka said:
Bx is also an eigenvector, hence it is also in the eigenspace of lambda. What does this tell you about the form of Bx?
Click to expand...
Bx = mu x ?

Hence x is an eigen vector?

edit: thank you : )
 
Last edited:
You must log in or register to reply here.

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top