## anonymous 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.

1. anonymous

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

2. anonymous

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

3. 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.

4. anonymous

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

5. anonymous

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

6. anonymous

correct so find when row is linearly dependent

7. 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.

8. anonymous

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

9. anonymous

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

10. anonymous

book says a=2, btw.

11. 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.

12. anonymous

did the book ask for non trivial?

13. anonymous

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

14. anonymous

haha it didn't say anything about trivial.

15. anonymous

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

16. 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.

17. anonymous

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

18. anonymous

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

19. anonymous

scanning and posting in a sec.

20. anonymous

okay thanks.

21. anonymous

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

22. anonymous

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

23. anonymous

24. anonymous

ahh you were missing a negative lol

25. anonymous

lolol

26. anonymous

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

27. 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

28. 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