Transcript of 14_01.AVI (Lecture 14, part 1)

By Eckhard M.S. Hitzer, University of Fukui, Prof. Ryosuke Nagaoka (University of the Air, Japan)

Yes, please have a look at the computer graphics. We will do a proof by induction on the number of vectors in a set of k vectors. So over the number of vectors k or the cardinal number.

So here we have a linear combination of the following vectors: The vector x one is a vector in the eigenspace in the wider sense belonging to the eigenvalue alpha one, the vector x two is a vector in the eigenspace in the wider sense belonging to the eigenvalue alpha two, up to x k is a a vector in the eigenspace in the wider sense belonging to the eigenvalue alpha k. Now the theorem is trivial for the case that we just have one vector, because then we have only this first term, and then the coefficient c one must be zero.

Now we make the hypothesis for this induction, that the theorem is already true, or already proved for k minus one vectors, so for a set of just k minus one vectors. And k must be larger or equal than two.

And now we multiply the linear combination from the left on both sides with: A minus the kth eigenvalue alpha k times the unit matrix E, to the power of n. And now especially look at the last term, because we have A minus alpha k times unit matrix E to the power of n, times x k. And x k is precisely a vector in the eigenspace in the wider sense of the eigenvalue alpha k. And therefore this last term will be zero. And we are left with a linear combination of the vectors x one to x k minus one.

And this first term here: A minus alpha k times the unit matrix E to the power of n times the vector x one, is again a vector in the eigenspace in the wider sense belonging to the eigenvalue alpha one. And similarly for the second term, the second term is a vector also in the eigenspace in the wider sense belonging to the eigenvalue alpha two. And also the last term is in this eigenspace in the wider sense belonging to the eigenvalue alpha k minus one. And therefore the induction hypothesis applies. And the coefficients c one, up to c k minus one will be zero. And the linear combination shrinks to just the last term, and therefore also c k will be zero. And therefore we have proved the theorem also for the cardinal number of vectors in the set of k (vectors).

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