'Operator' or 'linear operator'?
The term 'operator' is used in the statement and the proof. Wouldn't it be more correct (and clearer) to say 'linear operator'? The Wikilink goes to linear operator, even though the text says 'operator'. --220.127.116.11 (talk) 16:30, 29 February 2008 (UTC)
- In my experience, when one says "operator", they almost always mean "linear operator" (I can't think of a single counterexample to this, though there probably are a few), and thus the word "linear" is often omitted for the sake of brevity. Clarity would be preferable over brevity in this situation though in my opinion, so I have no objection to changing it to "linear operator". JokeySmurf (talk) 16:00, 4 May 2009 (UTC)
Which definition is correct?
The definition for Schur decomposition at Mathworld is completely different
I can find other sources that use the definition on this page and also for the Mathworld one...so which one is correct? Can we at least have a section explaining WHY there are two definitions?
- The definitions may look different but they are closely related. The definition here is A = QUQ−1, with Q unitary and U triangular. Multiply this equation from the left by Q−1 and from the right by Q to get Q−1AQ = U. Finally, note that Q−1 equals Q* (the conjugate transpose of Q) because Q is a unitary matrix and you get the definition from MathWorld. -- Jitse Niesen (talk) 17:36, 23 April 2009 (UTC)
Real Schur Decomposition
The article should mention real Schur decomposition, i.e. where the matrix T is quasi-upper-triangular, with 2x2 blocks on the diagonal, which contain conjugated pairs of complex eigenvalues.
Comment that used to be at the top of the page
"The article does not describe what happens if the set of null-vectors for the given matrix do not form a base (cf. defective matrices)." (This was unsigned)
I assume that what was meant was "...if the set of eigenvectors do not form a basis." The response is: the Schur decomposition always works (over the complex numbers, or if all of the eigenvalues are real), even if the eigenvectors do not form a basis. If they do form a basis, then the matrix can be diagonalized, although not by a unitary matrix (unless the matrix is normal). -- Spireguy (talk) 22:11, 18 February 2011 (UTC)
Problems with the proof
I see two problems with the first version of the proof presented in the article. (1) The distinction between a matrix and an operator is not made clear, and two matrices are said to be equal when what is really meant is that they are two matrices for the same operator in different bases (hence they are similar). (2) The details of how the recursive argument work to put together the matrix Q are more complicated than the current presentation implies, since at each stage you are working with progressively smaller subspaces of the original matrix. So it would be good to spell that out. I doubt I'll have time to do it myself. -- Spireguy (talk) 22:11, 18 February 2011 (UTC)