Question

eigenvalues and vectors

Answer 1

?

Answer 2

good you're here

Answer 3

in order to understand what eigenvalues and eigenvectors are, you need to have some sort of understanding of linear algebra
or what a vector is

Answer 4

KK

Answer 5

ok
assuming you know this, lets jump into eigenvalues

Answer 6

ok

Answer 7

let A equal some square matrix m (mxm)
and v equal some vector (mx1)

Answer 8

and we denote an eigenvalue as lambda
\[\lambda\]

Answer 9

ok we have to assume this equation?

Answer 10

\[Av=\lambda v\]

Answer 11

there's nothing to assume, this is just a condition that true for eigenvalue
its basically the very definition of one
lambda is a real number
Av is always equal to a multiple of v

Answer 12

i cant understand pls once more

Answer 13

A is a matrix
v is a vector
if you multiply those 2 together, you will get another vector

if the answer if a multiple of v
then that multiple is considered the eigenvector

Answer 14

but normally, you dont need to know what it is or where it came from, you just need to know how to solve them

Answer 15

you need to know why you use eigenvalues

Answer 16

there's a specific reason

Answer 17

kk tell

Answer 18

you're welcome to explain it in my place

Answer 19

ok

Answer 20

Eigenvalues are used to find nontrivial solutions of equations. In nontrivial I mean c1,c2,c3=0

for example you have the equation

\[0=c_1x_1+c_2x_2+c_3x_3\]

now you know the trivial answer is if

\[c_1=c_2=c_3=0\]
hence eigenvalues are used to find the nontrivial or nonzero numbers to such equations

Answer 21

ok

Answer 22

what class ar eyou in balua

Answer 23

engineer 1st yr

Answer 24

ahh so theris two different ways of doing this one way is as completeidiot described. You can also find eigenvalues through this equation

\[Det(A-\lambda I)K=0\]

Answer 25

havent gotten there yet

Answer 26

alright go on

Answer 27

now if you want to find the eigenvalue
\[IAv=I\lambda v\]
I is the identity matrix
IA = A
\[Av-I\lambda v=0\]
\[(A-I\lambda)v =0\]

in order for the previous statement (definition of linear independence) outkast stated to be true
c1=c2=c3=0


determinant of A-lambda*I must be equal to 0
\[\det(A-I\lambda)=0\]

Answer 28

if A was a 2x2 matrix
the the determinant might look like \[\lambda^2+b\lambda +c =0\]
where you solve for lambda using quadratic equation or by factoring if you can

Answer 29

you might have more than one eigenvalue, if you do, then that means there will be more than 1 eigenvector

once you find the eigenvalue, you substitute that back into the equation
\[Av-I\lambda v=0\]
or
\[(A-I\lambda)v=0\]
and solve for v by using gaussian elimination or whatever method you're taught

for 2x2 matrices, you will get a matrix like \[\left[\begin{matrix}a & b \\ 0 & 0\end{matrix}\right]v=0\]
and \[v =\left(\begin{matrix}x \\ y\end{matrix}\right)\]
so ax+by= 0
the eigenvector will be any x and y as long as the above equation is true

do the same thing for the other eigenvalue to find the eigenvector for that specific eigenvalue

Answer 30

and because you are offline, good night

if i made any mistakes here, feel free to correct them
@Outkast3r09