Question

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

Answer 1

A = \[\left[\begin{matrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & a\end{matrix}\right]\]

Answer 2

you must find all values in which the row <0,0,a> is linearly independent

Answer 3

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.

Answer 4

im assuming Defective means it doesnt have a complete set of eigenvectors. is that wrong?

Answer 5

Defective: if an nxn matrix is not diagonizable, i.e. it doesn't have n linearly independent eigenvectors.

Answer 6

correct so find when row is linearly dependent

Answer 7

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.

Answer 8

i understand 0 but not so much 2? how is that?

Answer 9

also We never went over this.. my book doesn't have this

Answer 10

book says a=2, btw.

Answer 11

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.

Answer 12

did the book ask for non trivial?

Answer 13

i think you might be mistaking this with the Null Space, or kernel.

Answer 14

haha it didn't say anything about trivial.

Answer 15

eh? i just looked at the theorem i just pulled up and it says in order for diagonalization,, n linearly independent eigen vectors

Answer 16

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.

Answer 17

Wait, how did you end up with λ(a−λ)(2−λ)=0?

Answer 18

i meant how did you get that as the determinant, to be more precise

Answer 19

scanning and posting in a sec.

Answer 20

okay thanks.

Answer 21

gotcha for some reason i must have got a sign error i got lambda - a

Answer 22

thats fine too. some people do det(lambda I - A)

Answer 23

ahh you were missing a negative lol

Answer 24

lolol

Answer 25

in my defense, the answer will be the same! lol

Answer 26

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

Answer 27

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