anonymous
  • anonymous
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?
Linear Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
and what is a, b and c?
anonymous
  • anonymous
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anonymous
  • anonymous
To find eigen values: \[ \det(A-\lambda I)=0 \]Solve for \(\lambda\).

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anonymous
  • anonymous
Basically: 1) subtract \(\lambda\) from the diagonals 2) find the determinate, giving you a polynomial 3) find the roots of that polynomial
anonymous
  • anonymous
Thx, wio. Let me try that.
anonymous
  • anonymous
If you have repeated or imaginary eigen values, it gets really hairy. Hopefully they're all unique and real.
anonymous
  • anonymous
Okay, I have -2, -1, -2 on the diagonal. Do i subtract λ diagonally and proceed or?
anonymous
  • 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
  • anonymous
I think? Actually I'm pretty sure it's just finding the null space of: \[ (A-\lambda_i) \]
anonymous
  • anonymous
Which may be the same thing as that equation above...
anonymous
  • anonymous
\[ A-\lambda_i I \]
anonymous
  • anonymous
I'm gettting confused here. Did you see the pic I attached?
anonymous
  • anonymous
Yes, why?
anonymous
  • anonymous
I understand I need to subtract λ diagonally, but why is it given?
anonymous
  • anonymous
I don't know. It isn't even an Eigen value for the picture you uploaded.
anonymous
  • anonymous
So, are you suggesting the question was wrong?
anonymous
  • anonymous
That is a possibility.
anonymous
  • anonymous
I think we might be interpreting it wrong, or you could have written it down wrong.
anonymous
  • anonymous
Okay. Here is the full thing.
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anonymous
  • anonymous
Hello wio. Still there?
ganeshie8
  • 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
  • ganeshie8
@d3Xter you able to find the eigenvectors ?
anonymous
  • anonymous
Well, using \[ \lambda I _{3}-A.\] I got \[\lambda ^{3}-7\lambda ^{2}+14\lambda-8\]
ganeshie8
  • 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
  • ganeshie8
you knw how to find the corresponding eigenvector right ?
anonymous
  • anonymous
Okay.
anonymous
  • anonymous
How do I go about that?
ganeshie8
  • 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
  • ganeshie8
put \(-\lambda = 2\) and solve above
ganeshie8
  • 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
  • ganeshie8
reduce the matrix
ganeshie8
  • 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
  • ganeshie8
that gives you x+2z = 0 y-z = 0 z is free
ganeshie8
  • ganeshie8
corresponding eigen vector for \(\lambda = 2\) \[ \left( \begin{array}{ccc} -2z \\ z \\ z \end{array} \right) \]
ganeshie8
  • ganeshie8
\[ \left( \begin{array}{ccc} -2 \\ 1 \\ 1 \end{array} \right) \]
anonymous
  • 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
  • anonymous
The first row.
ganeshie8
  • ganeshie8
R1-R2
anonymous
  • anonymous
Okay, understood, but why did you do that?
ganeshie8
  • ganeshie8
i wanted to express x and y in terms of free variable z
ganeshie8
  • ganeshie8
so, in the x row, ive simply eliminated y
anonymous
  • anonymous
hmm, that makes sense.
ganeshie8
  • ganeshie8
cool/
anonymous
  • anonymous
How about this? \left( \begin{array}{ccc} -2z \\ z \\ z \end{array} \right)
anonymous
  • anonymous
Why the z-s?
ganeshie8
  • ganeshie8
cuz our last row is all 0s. z is the free variable here
ganeshie8
  • ganeshie8
you will get infinite multiples for that vector by substituting a value for z
anonymous
  • anonymous
So, that makes it one since x=2z, right?
ganeshie8
  • ganeshie8
x = -2z dint get ur q ?
anonymous
  • anonymous
oops, that was an oversight
ganeshie8
  • ganeshie8
hey im not getting u hmm
anonymous
  • anonymous
Since x = -2z, z=1, right?
ganeshie8
  • ganeshie8
we have the vector in terms of z like below :- -2z z z
ganeshie8
  • ganeshie8
put z = 1, 2, 3, 4.... u wil get different vectors all corresponding to eigen value of 2
ganeshie8
  • ganeshie8
put z =1 -2 1 1 is the eigen vector.
ganeshie8
  • ganeshie8
Also every non-zero multiple of this vector is an eigen vector corresponding to \(\lambda = 2 \)
anonymous
  • anonymous
Thanks a lot. This really got me crazy for the past week.
ganeshie8
  • ganeshie8
np :) you may use wolframalpha when you're stuck
ganeshie8
  • 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
  • ganeshie8
good luck !
anonymous
  • anonymous
One more thing
anonymous
  • anonymous
that question was in terms of b. i just needs clarifications on adjustments to find a or c.
ganeshie8
  • 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
  • anonymous
Oh, I see. Thanks again.
ganeshie8
  • ganeshie8
yw :)

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