Case 1. Then, (double), (simple). By a well known theorem, On the other hand:
That is, is diagonalizable.
Case 2. Then, (triple). So,
In this case, is not diagonalizable.
Hi,
I have this problem I'm stuck on:
A is a 3x3 matrix
A=
a,b,c are real numbers
1) determine for what values of a,b,c the matrix A is diagnoizable. (advice: distinguish between the cases a=c and a=\=c)
My attempt at a solution has been finding det(tI-A)=0 , the characteristic polynomial is (t-c)(t-a)^2 and therefore the eigenvalues are a and c.
for t=c we get the matrix let it be C. for Cx=0 we get the system of linear equations:
(c-a)x_{1}-x_{3}=0
(c-a)x_{2} -bx_{3}=0
likewise, for t=a we get
-1x_{3}=0
-bx_{3}=0
(a-c)x_{3}=0
Now, when I tried to find the eigenspaces that correspond to a and c to find out their dimensions I got pretty lost.
I have no idea what to do. help would be very welcomed!
we can try to proceed with your original approach, and "see what happens". i often do this "just out of curiosity".
it's clear that you found the characteristic equation properly, it's (t - c)(t - a)^{2}.
now let's find some eigenvectors. first let's tackle c:
it's clear that x_{3} is "free", and provided a ≠ c:
which means that any eigenvector (belonging to c) is a scalar multiple of (1,b,c-a) (we have a one-dimensional eigenspace).
(note that if we DID have a = c, we we get instead is: x_{3} = 0. more on this later).
next we look at a:
gives:
x_{3} = 0
bx_{3} = 0 (and thus x_{3} = 0, this just duplicates the first equation)
(c-a)x_{3} = 0 (another duplication).
thus we are free to choose x_{1} and x_{2} independently, so we can pick (1,0,0) and (0,1,0) as linearly independent eigenvectors.
so if {(1,0,0),(0,1,0),(1,b,c-a)} are linearly independent, A is diagonalizable (we have an eigenbasis). it should be clear that this will be true if and only if c -a ≠ 0, that is: if a ≠ c.
on the other hand, if a = c, we only get an eigenspace of dimension 2: namely {(x,y,0) in R^{3}}, so there is no eigenbasis, and A is not diagonalizable.