00:02
We want to find the largest eigenvalue in magnitude and its corresponding eigenvector by using the power method for the given matrix a equal to 4, 0, 1, negative 2, 0, 1, 2, 0, 1 and the starting initial guess for the eigenvector is vector 1, negative 0 .5, 0 .5.
00:26
So remember here that this method starts with initial guess, x0, a vector, and then any vector in the sequence of vectors xk is defined as matrix a times the previous vector k minus 1.
00:50
So we multiply by the matrix each of the iterates to get the next one.
00:56
So in fact, if the matrix, the central idea is if the vectors, let's say the sequence of vectors xk is approaching the direction of an eigenvector associated to the largest eigenvalue, then what we have here is just the power of, the kth power of a acting on an initial guess x0.
01:27
So that's the main idea.
01:28
And to normalize this process, that is to avoid large magnitude vectors, we normalize at each iteration the vectors so they are two norm equal one vectors or unitary vectors.
01:50
So what we do after we do this calculation, we normalize the vector divided by its norm like this.
02:03
Of course, checking before that the normal vector is not 0, in that case the method has failed to converge.
02:11
So we need to apply this method here with this initial guess x0 equal 1 negative 0 .1 0 .1.
02:20
And in this case we have all eigenvectors of this matrix are different, so we should have eigenvectors that are perpendicular to each other or with inner products equal to 0.
02:37
And so we can find a basis for r3 of eigenvectors.
02:41
In that case, this method is going to converge.
02:45
And this is the case here, we apply the method using the iteration here.
02:51
You can find that the method converges to the eigenvalue.
03:14
Okay, there is something i want to say.
03:23
Converges, i am going to explain a little bit that.
03:27
The eigenvalue with largest magnitude will be 4 .5615528128 or rather approximately equal.
03:55
It is an approximation.
03:57
But in this case, putting an exigence that is the tolerance, meaning two consecutive vectors in this sequence here, defined recursively like this, has the norm of their difference, a small vector, or in other words, the norm of the difference of two consecutive vectors in the sequence is relatively small.
04:21
If we ask that to happen, in this case a difference of less than 10 to the negative 13, we can get this eigenvalue, which is a good approximation to the largest eigenvalue.
04:35
It means that if you calculate all eigenvalues of this matrix, you will see that the one with the largest module or absolute value is this one here.
04:47
And the corresponding eigenvector, which in our case is going to be unitary, that is with two norm equal one, and the eigenvector is 0 .8407275.
05:11
Let me see how many decimals...