Question

In a 3x3 matrix, how do I calculate the eigenvalues and eigenvectors, given that Ax=λx, λ=4? How do I find a, b or c? Thank?

Answer 1

and what is a, b and c?

Answer 2

To find eigen values: \[
\det(A-\lambda I)=0
\]Solve for \(\lambda\).

Answer 3

Basically:
1) subtract \(\lambda\) from the diagonals
2) find the determinate, giving you a polynomial
3) find the roots of that polynomial

Answer 4

Thx, wio. Let me try that.

Answer 5

If you have repeated or imaginary eigen values, it gets really hairy. Hopefully they're all unique and real.

Answer 6

Okay, I have -2, -1, -2 on the diagonal. Do i subtract λ diagonally and proceed or?

Answer 7

In that case the eigen vector \(\mathbf v_i\) corresponding to the eigen value \(\lambda_i\) can be found by solving: \[
(A-\lambda_i I)\mathbf v_i = \mathbf 0
\]

Answer 8

I think? Actually I'm pretty sure it's just finding the null space of: \[
(A-\lambda_i)
\]

Answer 9

Which may be the same thing as that equation above...

Answer 10

\[
A-\lambda_i I
\]

Answer 11

I'm gettting confused here. Did you see the pic I attached?

Answer 12

Yes, why?

Answer 13

I understand I need to subtract λ diagonally, but why is it given?

Answer 14

I don't know. It isn't even an Eigen value for the picture you uploaded.

Answer 15

So, are you suggesting the question was wrong?

Answer 16

That is a possibility.

Answer 17

I think we might be interpreting it wrong, or you could have written it down wrong.

Answer 18

Okay. Here is the full thing.

Answer 19

Hello wio. Still there?

Answer 20

question looks fine to me,
to make ur life simple, they gave u one eigen value. you're supposed to solve other two roots

Answer 21

@d3Xter you able to find the eigenvectors ?

Answer 22

Well, using \[ \lambda I _{3}-A.\] I got \[\lambda ^{3}-7\lambda ^{2}+14\lambda-8\]

Answer 23

ok, \(\lambda = 2 \) satisfies that equation,
so it is clar the thw question is asking us to find the eigenvector corresponding to \(\lambda = 2 \)

Answer 24

you knw how to find the corresponding eigenvector right ?

Answer 25

Okay.

Answer 26

How do I go about that?

Answer 27

\[ \left( \begin{array}{ccc}
2-\lambda & 2 & -2 \\
1 & 3-\lambda & 1 \\
1 & 2 & 2-\lambda
\end{array} \right)

\left( \begin{array}{c}
x \\
y \\
z \end{array} \right)

= 0

\]

Answer 28

put \(-\lambda = 2\) and solve above

Answer 29

\[

\left( \begin{array}{ccc}
2-2 & 2 & -2 \\
1 & 3-2 & 1 \\
1 & 2 & 2-2
\end{array} \right)

\left( \begin{array}{c}
x \\
y \\
z \end{array} \right)

= 0
\]

\[

\left( \begin{array}{ccc}
0 & 2 & -2 \\
1 & 1 & 1 \\
1 & 2 & 0
\end{array} \right)

\left( \begin{array}{c}
x \\
y \\
z \end{array} \right)

= 0
\]

Answer 30

reduce the matrix

Answer 31

\[

\left( \begin{array}{ccc}
0 & 2 & -2 & 0 \\
1 & 1 & 1 & 0\\
1 & 2 & 0 & 0
\end{array} \right)

\]


\[

\left( \begin{array}{ccc}
1 & 1 & 1 & 0\\
0 & 2 & -2 & 0 \\
1 & 2 & 0 & 0
\end{array} \right)

\]

\[

\left( \begin{array}{ccc}
1 & 1 & 1 & 0\\
0 & 2 & -2 & 0 \\
0 & 1 & -1 & 0
\end{array} \right)

\]

\[

\left( \begin{array}{ccc}
1 & 1 & 1 & 0\\
0 & 1 & -1 & 0 \\
0 & 1 & -1 & 0
\end{array} \right)

\]

\[

\left( \begin{array}{ccc}
1 & 1 & 1 & 0\\
0 & 1 & -1 & 0 \\
0 & 0 & 0 & 0
\end{array} \right)

\]

\[

\left( \begin{array}{ccc}
1 & 0 & 2 & 0\\
0 & 1 & -1 & 0 \\
0 & 0 & 0 & 0
\end{array} \right)

\]

Answer 32

that gives you
x+2z = 0
y-z = 0
z is free

Answer 33

corresponding eigen vector for \(\lambda = 2\)


\[

\left( \begin{array}{ccc}
-2z \\
z \\
z
\end{array} \right)

\]

Answer 34

\[

\left( \begin{array}{ccc}
-2 \\
1 \\
1
\end{array} \right)

\]

Answer 35

how did you get this?
\left( \begin{array}{ccc}
1 & 0 & 2 & 0\\
0 & 1 & -1 & 0 \\
0 & 0 & 0 & 0
\end{array} \right)

Answer 36

The first row.

Answer 37

R1-R2

Answer 38

Okay, understood, but why did you do that?

Answer 39

i wanted to express x and y in terms of free variable z

Answer 40

so, in the x row, ive simply eliminated y

Answer 41

hmm, that makes sense.

Answer 42

cool/

Answer 43

How about this?
\left( \begin{array}{ccc}
-2z \\
z \\
z
\end{array} \right)

Answer 44

Why the z-s?

Answer 45

cuz our last row is all 0s. z is the free variable here

Answer 46

you will get infinite multiples for that vector by substituting a value for z

Answer 47

So, that makes it one since x=2z, right?

Answer 48

x = -2z

dint get ur q ?

Answer 49

oops, that was an oversight

Answer 50

hey im not getting u hmm

Answer 51

Since x = -2z, z=1, right?

Answer 52

we have the vector in terms of z like below :-

-2z
z
z

Answer 53

put z = 1, 2, 3, 4.... u wil get different vectors all corresponding to eigen value of 2

Answer 54

put z =1

-2
1
1


is the eigen vector.

Answer 55

Also every non-zero multiple of this vector is an eigen vector corresponding to \(\lambda = 2 \)

Answer 56

Thanks a lot. This really got me crazy for the past week.

Answer 57

np :) you may use wolframalpha when you're stuck

Answer 58

here is eigen values and eigen vectors for given matrix from wolfram :-

http://www.wolframalpha.com/input/?i=eigen+vectors+of+%7B%282%2C2%2C-2%29+%2C+%281%2C3%2C1%29%2C+%281%2C2%2C2%29%7D

Answer 59

good luck !

Answer 60

One more thing

Answer 61

that question was in terms of b. i just needs clarifications on adjustments to find a or c.

Answer 62

that im not sure wat they mean by a, b or c
i think they mean a, b, c are vectors corresponding to each root

Answer 63

Oh, I see. Thanks again.

Answer 64

yw :)