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windsylph
: Linear Algebra: (Diagonalization) Let A = {{1,1,1},{1,1,1},{0,0,a}}. Find all values a that will make A defective.
A = \[\left[\begin{matrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & a\end{matrix}\right]\]
you must find all values in which the row <0,0,a> is linearly independent
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.
im assuming Defective means it doesnt have a complete set of eigenvectors. is that wrong?
Defective: if an nxn matrix is not diagonizable, i.e. it doesn't have n linearly independent eigenvectors.
correct so find when row is linearly dependent
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.
i understand 0 but not so much 2? how is that?
also We never went over this.. my book doesn't have this
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.
did the book ask for non trivial?
i think you might be mistaking this with the Null Space, or kernel.
haha it didn't say anything about trivial.
eh? i just looked at the theorem i just pulled up and it says in order for diagonalization,, n linearly independent eigen vectors
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.
Wait, how did you end up with λ(a−λ)(2−λ)=0?
i meant how did you get that as the determinant, to be more precise
scanning and posting in a sec.
gotcha for some reason i must have got a sign error i got lambda - a
thats fine too. some people do det(lambda I - A)
ahh you were missing a negative lol
in my defense, the answer will be the same! lol
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
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