## windsylph 3 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. windsylph

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

2. Outkast3r09

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

3. joemath314159

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. joemath314159

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

5. windsylph

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

6. Outkast3r09

correct so find when row is linearly dependent

7. joemath314159

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. Outkast3r09

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

9. Outkast3r09

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

10. windsylph

book says a=2, btw.

11. joemath314159

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. Outkast3r09

did the book ask for non trivial?

13. joemath314159

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

14. windsylph

haha it didn't say anything about trivial.

15. Outkast3r09

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

16. joemath314159

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. windsylph

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

18. windsylph

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

19. joemath314159

scanning and posting in a sec.

20. windsylph

okay thanks.

21. Outkast3r09

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

22. joemath314159

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

23. joemath314159

24. Outkast3r09

ahh you were missing a negative lol

25. joemath314159

lolol

26. joemath314159

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

27. joemath314159

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. windsylph

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