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windsylph

  • 4 years ago

: Linear Algebra: (Diagonalization) Let A = {{1,1,1},{1,1,1},{0,0,a}}. Find all values a that will make A defective.

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  1. windsylph
    • 4 years ago
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    A = \[\left[\begin{matrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & a\end{matrix}\right]\]

  2. Outkast3r09
    • 4 years ago
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    you must find all values in which the row <0,0,a> is linearly independent

  3. joemath314159
    • 4 years ago
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    the characteristic equation of this guy is:\[-\lambda(a-\lambda)(2-\lambda)=0\]we see that if a= 0 or a = 2, we will have eigenvalues whos algebraic multiplicities are 2, but the geometric multiplicities may not be 2.

  4. joemath314159
    • 4 years ago
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    im assuming Defective means it doesnt have a complete set of eigenvectors. is that wrong?

  5. windsylph
    • 4 years ago
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    Defective: if an nxn matrix is not diagonizable, i.e. it doesn't have n linearly independent eigenvectors.

  6. Outkast3r09
    • 4 years ago
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    correct so find when row is linearly dependent

  7. joemath314159
    • 4 years ago
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    if we have three different eigenvalues, then the matrix is automatically diagonalizable, so we need to make sure one of the eigenvalues is a repeated root of the equation. That means a has to be either 0, or 2.

  8. Outkast3r09
    • 4 years ago
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    i understand 0 but not so much 2? how is that?

  9. Outkast3r09
    • 4 years ago
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    also We never went over this.. my book doesn't have this

  10. windsylph
    • 4 years ago
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    book says a=2, btw.

  11. joemath314159
    • 4 years ago
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    because the equation is:\[-\lambda(a-\lambda)(2-\lambda)=0\]if a is 2 we get:\[-\lambda(2-\lambda)(2-\lambda)=0\iff -\lambda(2-\lambda)^2=0\] the eigenvalue 2 will have algebraic multiplicity 2.

  12. Outkast3r09
    • 4 years ago
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    did the book ask for non trivial?

  13. joemath314159
    • 4 years ago
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    i think you might be mistaking this with the Null Space, or kernel.

  14. windsylph
    • 4 years ago
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    haha it didn't say anything about trivial.

  15. Outkast3r09
    • 4 years ago
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    eh? i just looked at the theorem i just pulled up and it says in order for diagonalization,, n linearly independent eigen vectors

  16. joemath314159
    • 4 years ago
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    basically, you need to get the characteristic equation:\[\det(A-\lambda I)=0\]that turns out to be the\[\lambda(a-\lambda)(2-\lambda)=0\]so if you dont want the matrix to be diagonalisable, then you dont want 3 different eigenvalues. So a has to be either 0, or 2, to match the other values. But if you let a be 0, you will still end up with a diagonalizable matrix. because the algebraic multiplicity will be 2, and the eigenspace will also be 2 dimensional. So your only answer is a = 2.

  17. windsylph
    • 4 years ago
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    Wait, how did you end up with λ(a−λ)(2−λ)=0?

  18. windsylph
    • 4 years ago
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    i meant how did you get that as the determinant, to be more precise

  19. joemath314159
    • 4 years ago
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    scanning and posting in a sec.

  20. windsylph
    • 4 years ago
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    okay thanks.

  21. Outkast3r09
    • 4 years ago
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    gotcha for some reason i must have got a sign error i got lambda - a

  22. joemath314159
    • 4 years ago
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    thats fine too. some people do det(lambda I - A)

  23. joemath314159
    • 4 years ago
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  24. Outkast3r09
    • 4 years ago
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    ahh you were missing a negative lol

  25. joemath314159
    • 4 years ago
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    lolol

  26. joemath314159
    • 4 years ago
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    in my defense, the answer will be the same! lol

  27. joemath314159
    • 4 years ago
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    if im solving for roots of an equation, there is no change in solutions from f(x)= 0 to -f(x)=0, this is why math is great :P

  28. windsylph
    • 4 years ago
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    sorry for making you do all that work, im kind of dizzy right now so I didn't do my determinants correctly. using cofactors and expanding through a-lambda, I just multiplied the two 1-lambda and forgot to subtract 1 from that. sorry for making you do all that after I realized my mistake. thanks so much. :D

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