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anonymous
 5 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.
anonymous
 5 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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A = \[\left[\begin{matrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & a\end{matrix}\right]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you must find all values in which the row <0,0,a> is linearly independent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im assuming Defective means it doesnt have a complete set of eigenvectors. is that wrong?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Defective: if an nxn matrix is not diagonizable, i.e. it doesn't have n linearly independent eigenvectors.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0correct so find when row is linearly dependent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i understand 0 but not so much 2? how is that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also We never went over this.. my book doesn't have this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did the book ask for non trivial?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think you might be mistaking this with the Null Space, or kernel.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha it didn't say anything about trivial.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0eh? i just looked at the theorem i just pulled up and it says in order for diagonalization,, n linearly independent eigen vectors

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0basically, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wait, how did you end up with λ(a−λ)(2−λ)=0?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i meant how did you get that as the determinant, to be more precise

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0scanning and posting in a sec.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0gotcha for some reason i must have got a sign error i got lambda  a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats fine too. some people do det(lambda I  A)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahh you were missing a negative lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in my defense, the answer will be the same! lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry 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 alambda, I just multiplied the two 1lambda 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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