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Let $M$ be a $n$ by $n$ matrix over a field $F$.

When $F$ is $\mathbb{C}$, $M$ always has a Schur decomposition, i.e. it is always unitarily similar to a triangular matrix, i.e. $M = U T U^H$ where $U$ is some unitary matrix and $T$ is a triangular matrix.

  1. I was wondering for an arbitrary field $F$, what are some conditions for $M$ to admit Schur decomposition?
  2. Consider a generalization of Schur decomposition, $M = P T P^{-1}$ where $P$ is some invertible matrix and $T$ is a triangular matrix. I was wondering what some conditions are for $M$ to admit such an decomposition?

    Note that $M$ admit such an decomposition when $F$ is $\mathbb{C}$, since it always has Schur decomposition.

Thanks!

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You just wrote above that for $\mathbb C$ you have such a decomposition with a unitary matrix, so of course, you still have the decomposition if you don't ask for a unitary matrix, but just for a invertible one. –  Phira Nov 27 '12 at 13:18
    
@Phira: That's right! Thanks! I will edit my post. –  Tim Nov 27 '12 at 13:20
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2 Answers 2

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If the characterisic polynomial factors in linear factors then the Jordan decomposition works as your triangular matrix.

If you have a similar triangular matrix then the characteristic polynomial of $M$ is the characteristic polynomial of $T$ which clearly factors into linear factors.

So, the criterion is exactly the same as for Jordan decomposition.

The similar triangular matrix is just a lazy variant of Jordan decomposition.

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+1. Thanks! Is it correct that the necessary and sufficient condition for existence of Schur decomposition and those for Jordan (normal form) decomposition are the same? –  Tim Nov 27 '12 at 17:29
    
@Tim Yes, because generalized eigenvectors belonging to different eigenvalues are automatically orthogonal, and the vectors belonging to the same eigenvalue can be made orthogonal by using Gram-Schmidt. –  Phira Nov 27 '12 at 18:31
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This is a thought. If you look at the construction of schur decomposition, at every step, one uses a new eigenvector to triangularize further and further (see here). So as long as the matrix has $n$ eigenvalues (distinct or repeated), I don't see any problem in extending schur decomposition to any field.

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+1, thanks! (1) "one uses a new eigenvector to triangularize further and further", what does "new" mean? I guess it should be weaker than linearly independent with previous eigenvectors? (2) "the matrix has n eigenvalues (distinct or repeated)", do you mean eigenvalues repeated as many times as their algebraic multiplities? Is it true that any n by n matrix over any field always has n eigenvalues (algebraic multiplities counted), if and only if its characteristic polynomial can factor into linear factors over the field F? –  Tim Nov 27 '12 at 13:49
    
Every n-degree polynomial with coefficients from a algebraically closed field will have $n$ roots. See wikipedia entry en.wikipedia.org/wiki/Algebraically_closed_field This is also equivalent to saying it should factor into linear factors (it is stated in the previous link). –  dineshdileep Nov 27 '12 at 14:26
    
Again in the construction of schur decomposition, the succesive eigenvectors comes from a different lower dimensional matrix. See the construction here en.wikipedia.org/wiki/Schur_decomposition . It is not like we are using the set of all eigenvectors of the orginal matrix we start with. –  dineshdileep Nov 27 '12 at 14:33
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