## Transcript of 13_02.AVI (Lecture 13, part 2)

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

Here you see a matrix A. And we can calculate the characteristic polynomial of this matrix. It is x minus three to the power of two. So three is an eigenvalue of multiplicity two, and the matrix A can not be diagonalized.

But let us take the matrix S, which is A minus the eigenvalue three times E. You can calculate, that it has this form. And if you take the square of S, which is a good exercise for you, we find that it is a zero matrix. So the matrix S is actually nilpotent. In order to find the Jordan normal form of the matrix S, we now need first a vector e one, under the condition, that S times e one is not zero. As you can easily verify, this vector e one is one possible vector, to fulfill this condition. The other vector we need should be a vector, which has the condition S times e two to be zero. And if we take this vector e two, which is actually S times e one, as you can verify as well, and I recommend it as an exercise. So the vector S times e one, we take e two to be this vector: minus one half, minus one half.

Now we take the vectors together to form the matrix P. And we use this matrix to transform the matrix S. P inverse times S times P, as you can verify, is: zero, zero, one, zero. And this is the Jordan normal form of the matrix S. S equals A minus three E, and now we put A on the left side. And we transform this equation also with the matrix P. So we get P inverse A P, equals P inverse S P, plus three E. And as you can see we have here the Jordan normal form of S plus three times the unit matrix, which is the Jordan normal form of A.

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Soli Deo Gloria. Created by Eckhard Hitzer (Fukui).