1. ## jordan form question

does $\displaystyle A=\left(\begin{array}{cc}a & 1\\0 & a\end{array}\right)$ and
$\displaystyle C=\left(\begin{array}{cc}a & b\\ 0 & a\end{array}\right)$
are similar?
i know a law that if they have the same jordan form then they are similar
i was told that if b=0 they the dont have the same jordan form
if b=0
jordan form of A:
$\displaystyle P(t)=(a-t)^{2}$
if the minimal polinomial is
M(t)=(a-t) then $\displaystyle M(A)=(aI-A)\neq0$
so $\displaystyle M(t)=(a-t)^{2}$
so $\displaystyle j_{A}=\left(\begin{array}{cc}a & 1\\0 & a\end{array}\right)$same form i will get for C
where is my mistake?

2. ## Re: jordan form question

the characteristic polynomial det(A-tI) is not the same as "the Jordan form". two matrices which are NOT similar, can have the same characteristic polynomial.

to determine if the Jordan forms are the same, you HAVE to look at the super-diagonal. it's the non-zero entries which matter.

you can think of a Jordan block as a sum: J = D+N, where D is a "diagonal part" and N is a "nilpotent part".

you can't tell if two matrices are similar just by looking at the characteristic polynomial. if they are similar, their characteristic polynomials

wil be the same, but NOT vice versa (it is a one way implication; you can have same diagonal parts, but different nilpotent parts).

you have to look at ker(A - tI), for each t. for A =

[a 0]
[0 a], ker(A -aI) is span({e1,e2}), dimension = 2. but for B =

[a 1]
[0 a], ker(B-aI) = span({e1}), which is a smaller kernel, dimension = 1.

in this case, we can show explicitly there is no invertible P for which PAP^-1 = B:

PAP^-1 = P(aI)P^-1 = a(PIP^-1) = a(PP^-1) = aI ≠ B, so A and B cannot be similar.

moral of the story: it is not enough to check that the characteristic polynomials are the same, you have to check the dimensions of the eigenspaces

(the dimension of the kernel ker(A-tI)), as well.

you can't just check the algebraic multiplicity of an eigenvalue, you have to check the geometric multiplicity, too. sometimes, they aren't the same.

3. ## Re: jordan form question

i dont say that they are similar because they have the same caractericstic polinomail.
i say
that they are similar if they have the same jordan form

do you argee that if two matrices have the same jordan form then they are similar?

do you agree that if we find the minimal polinomial and caracteristic polinomial then we can know
what is the possioble options for the jordan form
?

4. ## Re: jordan form question

Originally Posted by transgalactic
i dont say that they are similar because they have the same caractericstic polinomail.
i say
that they are similar if they have the same jordan form
but if b=0, they do not have the same Jordan form.

do you argee that if two matrices have the same jordan form then they are similar?
yes.

do you agree that if we find the minimal polinomial and caracteristic polinomial then we can know
what is the possioble options for the jordan form
?
options, yes, but there are many possible jordan forms associated with the same characteristic polynomial.

if you look at your post, you do NOT get the same jordan form for C as you do for A, IF b=0.

that "b" is important, it determines (in this case) whether or not C is diagonalizable. in fact, if you want to create a matrix

that is not diagonalizable, just take a diagonal matrix and sprinkle some 1's on the super-diagonal (like A has).

ok, look: let's try to diagonalize C:

first we find the eigenvalues, which is easy, we have a single eigenvalue of a. now we find the eigenvectors.

[0 b][x]....[0]
[0 0][y] = [0]

leads to: by = 0. there are TWO cases:

case 1) b = 0 ---> in this case ker(C-aI) = span({e1,e2}), dimension 2, we have an eigenbasis.
case 2) b ≠ 0 ---> y = 0, so we get the smaller kernel ker(C-aI) = span({e1}), dimension 1....no eigenbasis.

each of these cases corresponds to a different jordan form. in the first C-aI is of nilpotency 1 (C-aI is actually the 0-matrix), the jordan form is diagonal.

in the second case, C-aI is of nilpotency 2 (and the jordan form is then the same as A).

5. ## Re: jordan form question

i have solved it but without calculating the geometric multiplicity

ok so for $\displaystyle A=\left(\begin{array}{cc}a & 1\\0 & a\end{array}\right)$ the caracteristic polinomial is $\displaystyle P(t)=(a-t)^2$

i have made the calculations and found the minimal polinomial is M(t)=(a-t)^{2} too

so the only form possible is $\displaystyle j_{A}=\left(\begin{array}{cc}a & 1\\0 & a\end{array}\right)$,because the largest block is size 2.

because the power of the minimal polinomial is 2,and the caracteristic polinomial is in power 2 too.

if b=0

$\displaystyle C=\left(\begin{array}{cc}a & b\\0 & a\end{array}\right)is C=\left(\begin{array}{cc}a & 0\\0 & a\end{array}\right)$ the caracteristic polinomial is $\displaystyle P(t)=(a-t)^{2}M(t)=(a-t)$

M(A)=(aI-A)=0

so the largest block is with size 1 and the caracteristic polinomial is in power 2

so this is the only possible form $\displaystyle j_{A}=\left(\begin{array}{cc}a & 0\\0 & a\end{array}\right)$

i have solve it without calculating the geometric multiplicity

did i solved it correctly?

6. ## Re: jordan form question

sure, sometimes the minimal polynomial can help distinguish between a diagonalizable and non-diagonalizable matrix

(and for 2x2 matrices, you either get 2 1x1 jordan blocks, or 1 2x2 jordan block),

because if the minimal polynomial only has linear factors, the matrix is diagonalizable.

7. ## Re: jordan form question

so if the minimal polinomial has only lenear compnents M(t)=(t-a)(t-b)(t-c)..
then it diagonisable

correct.

thanks