## Transcript of 14_02.AVI (Lecture 14, part 2)

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 and here you see a matrix, for which we will look now for the Jordan normal form. And first we calculate this determinant, which you already know quite well, determinant of A minus x times unit matrix E. And we can simplify it in this form: x minus two to the power of two, times x minus three. So we have two eigenvalues: Eigenvalue two and eigenvalue three. And the exponent for this factor x minus two is two.

Now here I write the matrix A minus the first eigenvalue times the unit matrix E, and it has the rank two. So it is rank two equals three minus one, and therefore the dimension of the eigenspace belonging to the eigenvector two will just be one. Now we can calculate the square of this matrix A minus eigenvalue two times E, and we see that if we use this vector zero, one, two, then the square of this matrix A minus eigenvalue two times unit matrix E to the power of two, times p two is zero. Therefore p two is a vector in the eigenspace in the wider sense belonging to the eigenvalue two. And the first column vector, which we want later on to use for constructing the matrix P for getting the Jordan normal form, for this we take A minus two times E (this matrix) times p two. And by the method of construction, because this equation holds, also p one will be an eigenvector of the matrix A with the eigenvalue two.

... Prof. Nagaoka ...

So therefore now we have the two vectors p one and p two, and they form the basis of the eigenspace in the wider sense belonging to the eigenvalue two.

The other eigenvalue is three. And this is the matrix A minus three times unit matrix E, and it has rank two, and therefore the dimension of the eigenspace belonging to the eigenvalue three will be just one. p three is an eigenvector, which belongs to the eigenvalue three. Now these three vectors form the column vectors of the matrix P. And this matrix P and its inverse is used to transform A into its Jordan normal form. And there are two blocks here. The first belongs to the eigenvalue two, and it is a two by two block, therefore the index two here. And this one is just a number three, and so it is just one dimensional, therefore index only one.

... Prof. Nagaoka ...

So please look at this computer graphics. And you see this matrix A, which we also want to transform into its Jordan normal form. The characteristic polynomial has this form of third order and it can be simplified to x minus two to the power of three. So the eigenvalue two has multiplicity three. And we form this matrix A minus two times the unit matrix E. And we see it has rank two, which is three minus one. And therefore the dimension of the eigenspace belonging to the eigenvalue two is simply only one.

... Prof. Nagaoka ...

And now we look at the square of A minus two times the unit matrix E, and we have this matrix here as a result. And if we look at the third power of the matrix, it already becomes zero. The third (eigenvalue of the,) I mean vector in the eigenspace in the wider sense belonging to the eigenvalue two can therefore be chosen freely. For convenience we just choose the vector: one, zero, zero.

... Prof. Nagaoka ...

So as before, the second vector in the eigenspace in the wider sense belonging to the eigenvalue two, can be chosen as p two equal A minus the eigenvalue two times unit matrix E, times vector p three. And the remaining vector, which we are looking for, is again A minus two times unit matrix E, times the second vector p two. And now p one, p two, and p three, they form a basis of the eigenspace in the wider sense belonging to the eigenvalue two. And with these three vectors we can construct the matrix P, column by column, and this matrix P serves us to construct the Jordan normal form of the matrix A.

... Prof. Nagaoka ...

So this is now a three by three matrix, and it belongs to the eigenvalue two. O. k. so this is the Jordan normal form of this matrix.

... Prof. Nagaoka ...

Yes, please look at the computer graphics. And here you see another matrix A of which we want to construct the Jordan normal form. And this is the characteristic polynomial of this matrix. And from it we see that the eigenvalue two also has multiplicity three.

... Prof. Nagaoka ...

And the matrix A minus eigenvalue two times E has just rank one. So it is three minus two. And the eigenspace of the eigenvalue two will therefore have dimension one.

... Prof. Nagaoka ...

And so we take this vector p two, which is in the eigenspace in the wider sense belonging to the eigenvalue two. And the vector p one, constructed by A minus eigenvalue two times unit matrix E, times vector p two is: minus one, minus one, minus one. And it is by construction in the eigenspace belonging to the eigenvalue two.

... Prof. Nagaoka ...

And the vector p three here is also an eigenvector of the matrix A for the eigenvalue two.

... Prof. Nagaoka ...

And so now we have found the three column vectors of the matrix P, which serves us for constructing the Jordan normal form of the matrix A, which you see here. And you see it has a two by two block, which is belonging to the eigenvalue two, and another scalar block, just the eigenvalue two, therefore J one (of) two.

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