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I am studying for an upcoming Linear Algebra exam. I am going through the questions from an old exam the instructor gave out, and I have come to this problem:

Give an example of an operator on a complex vector space with characteristic polynomial $(z-2)^3 (z-3)^3$ and with minimal polynomial $(z-2)^3(z-3)^2$.

Now I know that the matrix for this operator must have three $2$'s and three $3$'s down the diagonal, and I know the minimal polynomial divides the characteristic, but I don't know much else. This is in the same chapter as Jordan form, so I think a solution might have to do with Jordan blocks, but I don't have enough intuition about those to get it.

Any help here? :)

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Google "companion matrix". –  DonAntonio Nov 27 '12 at 20:55
    
So the companion matrix gives a matrix having the given polynomial as its minimal polynomial? How do I account for the fact that the characteristic polynomial has an extra factor of $(z-3)$ then? –  Jeff Fuller Nov 27 '12 at 20:59
    
Well, then you have to take a matrix of order one more than the companion one, right? From here you're on your own as I'm not sure whether there's an easy path to that. Perhaps there's some trick or method but I don't know/remember it now. –  DonAntonio Nov 27 '12 at 21:02

2 Answers 2

$$ \left( \begin{array}{cccccc} 2 & 1 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 3 \end{array} \right). $$

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Take a matrix with three Jordan blocks $\pmatrix{A & 0 & 0\cr 0 & B & 0\cr 0 & 0 & C\cr}$. Note that its characteristic polynomial is the product of the characteristic polynomials of $A$, $B$ and $C$. Choose $A$ to have characteristic and minimal polynomial $(z-2)^3$. Choose $B$ to have characteristic and minimal polynomial $(z-3)^2$. What do you suppose $C$ should have?

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So each Jordan block will correspond to one of the eigenvalues, and the number of 1's above the diagonal needs to be chosen so that they disappear when raised to the appropriate power. That makes more sense now. This is confirmed looking at Will's post. –  Jeff Fuller Nov 27 '12 at 21:16
    
About $C$, what minimal polynomial should it have? I am a little confused about that actually. –  Jeff Fuller Nov 27 '12 at 21:17
    
Oh but wait. C should just have $z-3$ as its minimal polynomial, right? Aka C should be just the single value 3 on the diagonal. –  Jeff Fuller Nov 27 '12 at 21:19
    
Robert, could you please take a look at math.stackexchange.com/questions/244038/inhomogeneous-equation ? So far I have written to the author, we agree that the part about the degree is not quite right, but I cannot seem to settle on a fix. One possibility is the degree of the characteristic value in the minimal polynomial instead of the characteristic polynomial, but I'm not sure even of that. Maybe that as an upper bound? –  Will Jagy Nov 27 '12 at 22:59

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