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?

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

and what is a, b and c?

- anonymous

##### 1 Attachment

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- anonymous

Thx, wio. Let me try that.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

\[
A-\lambda_i I
\]

- anonymous

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

- anonymous

Yes, why?

- anonymous

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

- anonymous

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

- anonymous

So, are you suggesting the question was wrong?

- anonymous

That is a possibility.

- anonymous

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

- anonymous

Okay. Here is the full thing.

##### 1 Attachment

- anonymous

Hello wio. Still there?

- ganeshie8

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

- ganeshie8

@d3Xter you able to find the eigenvectors ?

- anonymous

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

- ganeshie8

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

- ganeshie8

you knw how to find the corresponding eigenvector right ?

- anonymous

Okay.

- anonymous

How do I go about that?

- ganeshie8

\[ \left( \begin{array}{ccc}
2-\lambda & 2 & -2 \\
1 & 3-\lambda & 1 \\
1 & 2 & 2-\lambda
\end{array} \right)
\left( \begin{array}{c}
x \\
y \\
z \end{array} \right)
= 0
\]

- ganeshie8

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

- ganeshie8

\[
\left( \begin{array}{ccc}
2-2 & 2 & -2 \\
1 & 3-2 & 1 \\
1 & 2 & 2-2
\end{array} \right)
\left( \begin{array}{c}
x \\
y \\
z \end{array} \right)
= 0
\]
\[
\left( \begin{array}{ccc}
0 & 2 & -2 \\
1 & 1 & 1 \\
1 & 2 & 0
\end{array} \right)
\left( \begin{array}{c}
x \\
y \\
z \end{array} \right)
= 0
\]

- ganeshie8

reduce the matrix

- ganeshie8

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

- ganeshie8

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

- ganeshie8

corresponding eigen vector for \(\lambda = 2\)
\[
\left( \begin{array}{ccc}
-2z \\
z \\
z
\end{array} \right)
\]

- ganeshie8

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

- anonymous

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

- anonymous

The first row.

- ganeshie8

R1-R2

- anonymous

Okay, understood, but why did you do that?

- ganeshie8

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

- ganeshie8

so, in the x row, ive simply eliminated y

- anonymous

hmm, that makes sense.

- ganeshie8

cool/

- anonymous

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

- anonymous

Why the z-s?

- ganeshie8

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

- ganeshie8

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

- anonymous

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

- ganeshie8

x = -2z
dint get ur q ?

- anonymous

oops, that was an oversight

- ganeshie8

hey im not getting u hmm

- anonymous

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

- ganeshie8

we have the vector in terms of z like below :-
-2z
z
z

- ganeshie8

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

- ganeshie8

put z =1
-2
1
1
is the eigen vector.

- ganeshie8

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

- anonymous

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

- ganeshie8

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

- ganeshie8

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

- ganeshie8

good luck !

- anonymous

One more thing

- anonymous

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

- ganeshie8

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

- anonymous

Oh, I see. Thanks again.

- ganeshie8

yw :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.