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

- anonymous

- schrodinger

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- anonymous

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

- anonymous

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

- 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.

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- anonymous

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

- anonymous

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

- anonymous

correct so find when row is linearly dependent

- 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

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

- anonymous

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

- anonymous

book says a=2, btw.

- 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

did the book ask for non trivial?

- anonymous

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

- anonymous

haha it didn't say anything about trivial.

- 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

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

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

- anonymous

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

- anonymous

scanning and posting in a sec.

- anonymous

okay thanks.

- anonymous

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

- anonymous

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

- anonymous

##### 2 Attachments

- anonymous

ahh you were missing a negative lol

- anonymous

lolol

- anonymous

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

- 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

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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