00:01
We want to determine whether or not the given matrix is diagonalizable.
00:08
If it is, we want to find a diagonalizing matrix p and a diagonal matrix d such that p inverse times a times p equals d.
00:18
And using the diagonalization of a, we will compute a raised to the fifth power.
00:26
The matrix a is 2x2, 5 negative 4 and 3 negative 2.
00:34
So what we're going to do first is to calculate the eigenvalues of a.
00:41
That is because if a matrix is diagonalizable, it means we know that there is a matrix p in variable matrix b and a diagonal matrix d such that we have this equation.
01:02
This equation means that a and d are similar, and we know similar matrices have the same eigenvalues.
01:09
So the eigenvalues of a are the eigenvalues of d, but d is diagonal.
01:16
So its eigenvalues are its main diagonal.
01:19
So the main diagonal of d contains all the eigenvalues of a.
01:28
That's why we calculate the eigenvalues of a and with that we know which is the matrix d.
01:35
Okay, so for that we construct the characteristic polynomial, which is a, which is a matrix d.
01:35
Okay, so for that we construct the characteristic polynomial, is the determinant of a minus lambda i and the variable of the characteristic polynomial is lambda that is determinant of the matrix 5 minus lambda negative 4 3 negative 2 minus lambda and that determinant is equal to 5 minus lambda times negative 2 minus lambda plus 12 that is lambda minus 5 times lambda plus 2 plus 12 and we develop this product here and that is lambda square minus 3 lambda minus 10 plus 12 and finally we get that the characteristic polynomial of a is lambda square minus 3 lambda plus 2 and this is easily factor out as lambda minus 2 times lambda minus 1 so the eigenvalues of a are lambda 1 equal 2 and lambda 2 equal 1 so so d is 1 .002.
03:29
It's a diagonal matrix with the eigenvalues of a on its main diagonal.
03:35
And to construct the matrix b, we know we have to find eigenvectors associated to the eigenvalues, and the columns of p will be those eigen vectors.
03:46
So we calculate the eigen vectors of the eigen values of a.
03:55
So we say for lambda 1 equal here i'm going to change it a bit notation because i said lambda 1 is 2 and the other is 1 i'll say that lambda 1 is 1 and then the 2 is 2.
04:14
Okay so with that we want to solve the equation a x equal lambda 1x for a non -zero vector x.
04:26
So the equation would be 5 .0 .0 .3, negative 2 times an eigenvector we are looking for x1, x2, equal 1 times the eigenvector.
04:47
And this means that 5x1 minus 4x2 equal x1 and 3x1 minus 2x2 equal x1 and 3x1 minus 2x2 equal x2.
05:02
We simplify both equations and we get 4 x1 minus 4 x2 equals 0 and 3x1 minus 3x2 equals 0.
05:17
And both equations are the same when we simplify the first 1 divided by 4 both sides and the second one divided by 3 both sides.
05:27
The conclusion is the same.
05:30
X1 equals x2.
05:33
We can say that all eigenvectors associated to the eigenvalue lambda 1 equal 1 are of the form alpha alpha for alpha a real number different from 0.
06:01
In fact, this means that any eigenvector associated to lambda 1 equal 1 is of the form alpha times 1 -1, alpha a real number different from 0.
06:33
That is 1 -1 is a generator of that eigenspace, of the vector space of eigenvectors associated to lambda 1 equal 1.
06:46
So 1 -1 is a candidate we can take.
06:52
Okay, so you're going to take.
06:54
So the first column column of p can be and be the vector one one transpose okay so we now calculate eigenvectors number two eigenvectors associated to lambda 2 equal 2 the equation is the same but now considering lambda 2 okay so this means 5 negative 4 3 negative 2 times the eigenvector x1 x2 equal to x1 x2 and this means two equations 5x1 minus 4 x2 equal 2 x1 and 3x1 minus 2 x2 3x1 minus 4 x2 equals 0 exactly the same equation.
08:31
So the conclusion is that x2, for example, is 3 over 4 x1.
08:41
We can do the other way around...