anonymous
  • anonymous
Let A is a 4x4 matrix such that A(0,1,0,1)^t=(0,1,0,1)^t and A(1,0,1,0)^t=(-1,0,-1,0)^t. Also, suppose A has trace 4 and det 9/4. Find all eigenvalues of A and their multiplicites and explain why (1,1,1,1) is not an eigenvector of A.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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phi
  • phi
the two equations A(0,1,0,1)^t=(0,1,0,1)^t and A(1,0,1,0)^t=(-1,0,-1,0)^t. tell you 2 of the eigenvalues (and vectors): 1 and -1 the trace is the sum of the eigenvalues the set is the product of the eigenvalues
phi
  • phi
**the determinant is the product of the eigenvalues
anonymous
  • anonymous
All i'm confused about it the information at the beginning

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anonymous
  • anonymous
of the question
anonymous
  • anonymous
how do you already know two eigen values?
phi
  • phi
A(0,1,0,1)^t=(0,1,0,1)^t where t means transpose is a way to write \[A \left[\begin{matrix}0 \\ 1\\0 \\1\end{matrix}\right]= 1 \cdot \left[\begin{matrix}0 \\ 1\\0 \\1\end{matrix}\right]\]
anonymous
  • anonymous
yea i understood that before.
phi
  • phi
which should look familiar: \[ A v = \lambda v\]
anonymous
  • anonymous
ohhhhhh
anonymous
  • anonymous
yep. so the other would be -1. get ya
anonymous
  • anonymous
yep now its easy to get lamda3 and lamda4
phi
  • phi
yes. for the last part, [1 1 1 1] is a linear combination of the two eigenvectors they give you.
anonymous
  • anonymous
yea. it looked obvious by looking at it from the beginning. thanks for starting me off!
anonymous
  • anonymous
alright. so i found that lamda=1,-1,9/2,-1/2
anonymous
  • anonymous
then does this mean their multiplicities are just 1?
phi
  • phi
yes, all eigenvalues are distinct, so you have 4 eigenvalues, with multiplicity of 1 for each (a repeated value would give a multiplicity > 1)
anonymous
  • anonymous
yea. how would you represent in mathematical terms? or is what you wrote sufficient?
phi
  • phi
what I wrote should be sufficient
anonymous
  • anonymous
so with b) (1,1,1,1)=(0,1,0,1)+(1,0,1,0) hence this is a linear combination of the two basis hence not an eigen vector?
phi
  • phi
yes
anonymous
  • anonymous
is it right to say that (0,1,0,1) and (1,0,1,0) are eigenbasis's?
phi
  • phi
yes
anonymous
  • anonymous
but each of them are eigen vectors?
phi
  • phi
yes, the 4 distinct eigenvectors (because you have 4 distinct eigenvalues) make up a basis in 4d
anonymous
  • anonymous
legend.
anonymous
  • anonymous
thanks phi
Loser66
  • Loser66
thanks phi

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