Help finding a transition matrix between the Jordan form and a general form

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 9K views
Quantumpencil
Messages
96
Reaction score
0

Homework Statement



First year-linear algebra (Proof based... and this is my first exposure to proofs so I'm like... lol). This question is pretty computational though.

Find J, The Jordan Canonical form of a Given Matrix A, and an invertible Matrix Q such that J = Q(A)(Q^-1)

Homework Equations


The matrix is a 3x3 matrix with entries
(0 1 -1)
(-4 4 -2) = A
(-2 1 1)

The Attempt at a Solution



I think I've got the first part, but I really want to understand this stuff thoroughly so I'd like to have my "justifications" checked.

First we find the eigenvalues of A, by calculating det ([tex]\lambda[/tex]I-A) and factoring the characteristic polynomial. In this case the Characteristic polynomial is [tex]\lambda^{3}[/tex]-5[tex]\lambda^{2}[/tex]+8[tex]\lambda[/tex]-4. Factoring gives the eigenvalues 1, and 2 multiplicity 2.

Since 1 has multiplicity one, the corresponding eigenspace cannot have dimension greater than one, therefore there is a single eigenvector of 1 which spans the entire space.

However, 2 has multiplicity of 2, so \exists some v_{1}, v_{2} in Ker (A-2I)^{2} and some u_{1}, u_{2} in Ker (A-2I) such that (A-2I)v_{1}=u_{1}, and similarly for v_{2}. Which implies there are vectors in Ker (A-2I) such that Av_{1}= u_{1} + 2v_{1}, similarly for v_{2}

(1 0 0)
(0 2 1) = J
(0 0 2)

(In this particular case, I think there exist two linearly independent vectors in Ker (A-2I), so perhaps generalized eigenvectors are not necessary? I'd like confirmation, as in this case the matrix would be diagonal)

Now after finding J-form, I need to find a matrix Q which satisfies the relation in the problem statement. I tried adjoining the eigenvectors I calculated {(1,2,1),(1,0,2),(0,1,1)} but that was ineffectual. Then I tinkered around with row operations and found one matrix which produced the desired effect, but that's not helpful. I think I might need to find a particular basis, but I'm not sure what properties my basis needs to satisfy (My class is using Axler's LA done right, and it doesn't really have much in the way of algorithms) / how to go about "choosing" in order to construct the transition matrix. I'd prefer if possible, a "tip" in the right direction rather than an outright solution.

Thanks!
 
Last edited:
Physics news on Phys.org
Ok, I just re-worked my calculations.

For Ker (T-I) I got the condition v_{1}=2v_{2}=v_{3}, implying Ker (T-I) is spanned by (1, 2, 1) = the eigenvector.

For Ker (T-2I) the the set of spanning vectors to be (1,0,2), and (0,1,1). So why isn't the U_{1}[tex]\oplus[/tex]U_{2}[tex]\oplus[/tex]U_{3} the transition matrix in question? (The matrix Q which satisfies that relation...)

Or am I missing something still?
 
Blah, typo, meant (1,0,-2) for the second eigenvector.

But their direct sum still doesn't satisfy the condition for the Matrix of Q.

What am I missing?
 
Last edited:
The direct sum of what?

If you let

[tex]Q = \begin{pmatrix} 1 & 1 & 0 \\ 2 & 0 & 1 \\ 1 & -2 & 1 \end{pmatrix},[/tex]

then

[tex]Q^{-1}AQ = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}.[/tex]
 
Really? That's the matrix I calculated... I guess I must've just been messing up the matrice multiplication or something.

Thanks.