00:01
This problem we are going to be working with matla.
00:05
We have four parts.
00:08
Part a, we enter the matrix f, big f equals 01, 0 ,0 in medlab.
00:14
We find an invertible matrix p and a diagonal matrix d, such that p times d, then p into the negative 1 that is inverse of p equal f.
00:25
In part b, we use matlaf to compare the power 10 of f of matrix d.
00:33
F and p times power 10 of t times inverse of p and in part c we let f equal the vector the column vector 1 -1 and compute the matrix f times vector f then square of the metric f times vector f and we do that up to the fifth power of f times vector f we are going to describe the pattern of the resulting vectors from these operations and in part c we consider the fibonacci sequence 1 -1, 2, 3, 5, 8, 13, and so on, where each term is calculated as the sum of the previous two terms.
01:17
So we start with f equal small f equal f0, f1, and then matrix f times vector f equal f1 as the first component, and f0 plus f1 of second component.
01:35
And that is just f1 and f2, that is the third and the second and the third terms in the few and hess sequence.
01:46
In general, we have that the power n of matrix f times vector f is the nth term of c .e if you're anage sequence, or terms of index n rather.
01:58
And as first component and the second components is the term of index n plus one in feronage sequence.
02:07
Here we are setting f1 as 0 and f1 as 1 as we have here, of course.
02:16
And with that we want to compute f30, that is the terms in f5naci sequence with index 30.
02:27
That will be the 31st term because we are indexing the fnaci terms starting at f0.
02:36
Good, so let's start with part a.
02:41
We enter the matrix f equals 01 is the first row and 11 is the second row.
02:53
There is a way to find the matrix b and matrix d we're talking about here.
02:58
First, we got to say that finding this matrix b with this property, what means mathematically is that f is similar to a diagonal matrix.
03:09
Or in other words, matrix f is diagonalizable.
03:16
And we can find directly matrix p and matrix d by using the equal value common of matlab.
03:25
That is, we do the following common in badlap.
03:31
We get a vector of two components as output.
03:38
Those are the first component.
03:39
The first output is matrix p and the second output is matrix d equal common is e i g for eigenvalue of f so when we use this function in madlat with and with only one output it will give us only the eigenvalues of the matrix in a vector but when we use two outputs like this we will have the first output will be a matrix whose columns are the, are eigenvectors as sitting to the eigenvalues, and those eigenvectors are unitary, that is there are of norm, to norm equal one.
04:27
And d will be a diagonal matrix, whose diagonal elements are the eigenvalues of matrix f.
04:35
So i'm going to put the values here for reference only.
04:39
P is about only for you to know which are the magnitude values.
04:48
So it's negative 0 .8507 rounding values and here is 0 .527 and here is 0 .527 and the second row is 0 .527.
05:15
Same value here so it's symmetric and 0 .5807 because it's same value in the diagonal with change in sign so this is about matrix b and matrix d is about negative 0 .6 -180 and 0 here 0 here and this diagonal matrix and then 1 .6180.
05:54
These are the eigenvalues, these two values, this one here and this one here, are the eigenvalues of the matrix f.
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And this column here, let me use another column, this green.
06:13
So this column here, this column corresponds to an eigenvector associated to the first eigenvalue in the diagonal, this one, and the second column here corresponds to an eigenvector associated to this eigenvalue.
06:31
And we also have that these two eigenvectors forming the columns of matrix v are of two norm equal one, that is they are unitary vectors.
06:44
Okay, we have all this given directly by madlab in this comment.
06:48
It's easy to verify that f is equal.
06:56
Of course, there are some rounding happening here because these calculations might not do in this function are real value of floating point operations.
07:09
So there are some rounding happening around here.
07:13
That's why you're not going to obtain exactly this metric to us, but essentially that matrix we take into account the rounding.
07:23
That is happening.
07:24
So f is about p times d times inverse of p and that can be verified in my lab by calculating the norm, the matrix to norm of the difference between f and p times d times inverse of p.
07:51
Exactly as i'm writing that over here.
07:55
This is a common in my lab.
07:56
And if you do that, you will find that that norm is about 10 to the negative, let me see, 1 .6, let me say, times 10 to the negative 14, meaning they are essentially the same metrics.
08:19
We have that because of rounding, as i said.
08:23
1 .6 so we have that and that's part a now in part p what we're going to do so let me emphasize here is the comment you get to run in metlab in order to obtain p and d and then verify that f is essentially equal to p times d times inverse of p by calculating the two norm two matrix norm of the difference between the matrix f and the product p times d times inverse of p given a small number.
09:11
Okay, so part b, we're going to use madap to compare f to the 10th power and p times t to the tenth power times p into the negative one power.
09:29
And that got to be the same.
09:32
This is they are essentially the same matrix.
09:36
And if you do that, you calculate, these two matrices.
09:43
Here you are going to have a floating point numbers, of course, and here you're going to have integer numbers.
09:48
But if you calculate it again the norm of f to the 10th power minus p times t to the 10th power times inverse of p, you're going to know what is the magnitude to the difference, and if you do that, you get the following number about 1 .6 times 10 to the 9014 again, as we did here.
10:22
So they are essentially equal.
10:25
And the reason that's true, i'm going to say quickly here, for example, if you want to calculate this and p d square p inverse, they got to be the same, but that's because f square so sorry first remember in part a we prove that f is equal to p d p inverse so f is equal to p d p inverse and so f square which is we know f times f will be p d p inverse times itself because f is equal to that and and then we can associate this priority's way, pd times p inverse p, associating the internal matrices here, times d p inverse.
11:23
That's gotta be the same.
11:24
But p inverse times f is the identity matrix, as we know.
11:33
And then we associate again, the identity matrix times any matrix is the other matrix.
11:38
So we get p d times d p inverse, and that is nothing by.
11:44
P d square p inverse.
11:47
If we do that again for the third power and use induction, we prove that any power of f can be calculated as p times the same power of t times inverse of p.
12:02
And the good thing of having that a property is that the power of t, being d diagonal matrix, can be calculated very easily by raising to that power, diagonal elements of d.
12:17
So d raised to any power is again a diagonal matrix where the diagonal elements are the diagonal elements of d raised to that power, which is a very good property to have.
12:31
So we can verify this in maladap to this and we verify it by saying that the two norm of the matrix two norm of the difference between this power of f and p d to the 10th p inverse is a that is relatively small 10 to and 84...