anonymous
  • anonymous
: Linear Algebra: (Diagonalization) Let A = {{1,1,1},{1,1,1},{0,0,a}}. Find all values a that will make A defective.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
A = \[\left[\begin{matrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & a\end{matrix}\right]\]
anonymous
  • anonymous
you must find all values in which the row <0,0,a> is linearly independent
anonymous
  • anonymous
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.

Looking for something else?

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

More answers

anonymous
  • anonymous
im assuming Defective means it doesnt have a complete set of eigenvectors. is that wrong?
anonymous
  • anonymous
Defective: if an nxn matrix is not diagonizable, i.e. it doesn't have n linearly independent eigenvectors.
anonymous
  • anonymous
correct so find when row is linearly dependent
anonymous
  • anonymous
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.
anonymous
  • anonymous
i understand 0 but not so much 2? how is that?
anonymous
  • anonymous
also We never went over this.. my book doesn't have this
anonymous
  • anonymous
book says a=2, btw.
anonymous
  • anonymous
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.
anonymous
  • anonymous
did the book ask for non trivial?
anonymous
  • anonymous
i think you might be mistaking this with the Null Space, or kernel.
anonymous
  • anonymous
haha it didn't say anything about trivial.
anonymous
  • anonymous
eh? i just looked at the theorem i just pulled up and it says in order for diagonalization,, n linearly independent eigen vectors
anonymous
  • anonymous
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.
anonymous
  • anonymous
Wait, how did you end up with λ(a−λ)(2−λ)=0?
anonymous
  • anonymous
i meant how did you get that as the determinant, to be more precise
anonymous
  • anonymous
scanning and posting in a sec.
anonymous
  • anonymous
okay thanks.
anonymous
  • anonymous
gotcha for some reason i must have got a sign error i got lambda - a
anonymous
  • anonymous
thats fine too. some people do det(lambda I - A)
anonymous
  • anonymous
anonymous
  • anonymous
ahh you were missing a negative lol
anonymous
  • anonymous
lolol
anonymous
  • anonymous
in my defense, the answer will be the same! lol
anonymous
  • anonymous
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
anonymous
  • anonymous
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

Looking for something else?

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