Why Is (A-3I)^2 Used Instead of (A-3I)^3 in Finding Eigenvectors?

[franky2727]

my question is take A= {(5,0,-1),(2,3,-1),(4,0,1)} find all eigenvalues and eigenvectors

by using the characteristic equation i get -(lamda-3)³

however its the next bit i don't understand, in the answers (A-3I)(x,y,z)=(0,0,0) is used which I'm perfectly ok with and then (A-3I)² is used and not (A-3I)³, my point is how am i supost to know which to use and when? i have a similar question that i am trying to apply this to with (lamda-1)²(lamda-4) and am not sure which values to take?

 

[Hurkyl]

I assume you meant to say "generalized eigenvectors"?

What happens if you 'use' (A-3I)^3? Have you tried it?

my point is how am i supost to know which to use and when?

Presumably, you know some basic structural theorems about generalized eigenspaces. Apply them!

(e.g. you apparently knew not to bother with (A-3I)^4. Why is that? What would happen if you did 'use' (A-3I)^4?)

 

[HallsofIvy]

Try each in turn. Once you realize that 3 is a triple root of the characteristic equation, it follows that there may 1, 2, or 3 independent eigenvectors. The first thing you should do is try (A- 3I)(x,y,z)= (0,0,0) as you say, or A(x,y,z)= 3(x,y,z) (which is, in my mind more fundamental- the basic definition of "eigenvector". A(x,y,z)= 3(x,y,z) gives 5x- z= 3x, 2x+ 2y- z= 3y, and 4x+ z= 3z. The first equation is, of course, the same as z= 2x and so is the third. Putting z= 2x into the second equation gives 2y= 3y which gives y= 0. Any vector of the form <x, 0, 2x>= x<1, 0, 2> is an eigenvector. Since we do not get two independent eigenvectors, we look for a vector that does NOT satisfy (A- 3I)x= 0 but DOES satisfy (A- 3I)²x= 0. We can rewrite that as (A-3I)[(A-3I)x]= 0. Since any vector satisfying (A- 3I)v= 0 must be a multiple of <1, 0, 2>, we look for x so that (A- 3I)(x,y,z)= <1, 0, 2>. That is the same as 2x- z= 1, 2x+y-z= 0, 4x- 2z= 2. Again the first equation says z= 2x-1. Putting that into the third, 4x-2(2x-1)= 4x-4x+2= 2 is automatically satisfied. Putting z= 2x-1 into the third, 2x+ y-(2x-1)= y+ 1= 0 so y= -1. Taking x= 1, such a vector is <1, -1, 1>.

NOW to get a third independent vector, you could try (A- 3I)^3x= 0 which would lead to (A-3I)(x,y,z)= <1,-1,1>.

 

[franky2727]

i think i understand you, do you mean that using these different multiplications we are simply trying to get the same amount of eigenvectors as the dimension and it doesn't really matter which ones we use as long as we end up with 3 answers, the only reason we are using ^1 then ^2 is logical order?

 

[franky2727]

and no i didnt mean to say generalised eigenvectors the question isn't about generalised eigenvectors its simply eigenvectors and eigenvalues

 

[Hurkyl]

and no i didnt mean to say generalised eigenvectors the question isn't about generalised eigenvectors its simply eigenvectors and eigenvalues

Then why is (A-3I)^2 involved? And why would you think there exists a set of three linearly independent eigenvectors?