- #1
ygolo
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It seems to me that http://en.wikipedia.org/wiki/Schur_decomposition" relies on the fact that every linear operator must have at least one eigenvalue...but how do we know this is true?
I have yet to find a linear operator without eigenvalues, so I believe every linear operator does have at least one eigenvalue.
Still how does one prove it?
Since we are looking for solutions to (A-Ia)|V>=|0>, wouldn't it be possible the A-Ia is always nonsingular and that the equation has only trivial solutions?
I have yet to find a linear operator without eigenvalues, so I believe every linear operator does have at least one eigenvalue.
Still how does one prove it?
Since we are looking for solutions to (A-Ia)|V>=|0>, wouldn't it be possible the A-Ia is always nonsingular and that the equation has only trivial solutions?
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