00:01
We will find the ica values and corresponding igv vectors of each of the following matrices.
00:08
Part 8, a equal 1 -2 -1 -0 -3 -1 and 0 -3 -1.
00:17
And in part b, matrix b equal 1 -9 -1 -0 -1, 0 -2 and 0 -0 -1.
00:25
So start by part a, matrix a.
00:29
We know that the polynomial characteristic polynomial which we find in this case because the matrix can do it that way the characteristic polynomial of a is the determinant of the matrix a minus lambda i lambda is variable and it's going to give us the eigenvalues of the matrix so we formed this so we got to subtract lambda from the diagonal element of a and calculate the determinant of the resulting matrix.
01:19
This is determinant of the matrix 1 minus lambda 2 1 0 0 3 minus lambda 1 and 0 negative 3 negative 1 minus lambda so we need the determinant of this matrix let's give a name to this polynomial p of lambda is so we can develop the determinant of this matrix here by the first column.
01:52
So we get 1 minus lambda times the determinant of the 2x2 matrix obtained by removing the first column and the first row of these 3x3 matrix here.
02:07
So we get 3 minus lambda 1, negative 3, and negative 1 minus 1 minus lambda.
02:17
So we get 1 minus lambda times and this determinant here is, as usual, product of 3 minus lambda times negative 1 minus lambda minus negative 3 times 1.
02:35
So the characteristic polynomial p of lambda is 1 minus lambda times this negative sign here in the second factor can be put out and put in the.
02:50
To the first factor so we get lambda minus 3 times lambda plus 1 and the part here is plus 3.
03:03
So we can develop these expression inside this parenthesis and we get lambda square minus 3 lambda plus lambda negative 3 plus 3.
03:22
So these two values cancel out and we get and the characteristic polynomial of a is 1 minus lambda times lambda square minus 2 lambda.
03:42
And that is the same as lambda common factor here times 1 minus lambda times lambda minus 2.
03:52
And so we have the characteristic polynomial factor factor out already.
03:57
And we can see easily that the roots of polynomial characteristic polynomial which is the same as the eigenvalues of a r lambda equals zero which is the value that nullifies the factor of lambda lambda equals one which nullifies this factor and lambda equal two which notifies and this third factor.
04:38
So we have these three eigenvalues, which are the roots of the characteristic polynomial of a.
04:47
Okay, so now we calculate the eigenvectors associated to, or corresponding to, better, responding to lambda equal, start by, let's see here, and then see here, and that's zero.
05:28
And so we got to establish the equation defining the eigenvectors and eigenvec values that is a x equal lambda x for x different from zero so in this case if x is a general vector x1 x2 x3 with three and 0 negative 3 negative 1 times a generating vector x1 x2 a 3 which will be an eigenvector associated to lambda equal 0 if we take if the vector is a solution to this equation and it's not not equal to 0 so he's 0 times because lambda's equal 0 in this case x1 x2 or 3 sorry, and that is 0 -0 .0.
06:42
So this system here implies that the first equation will be x1 plus 2, x2 plus 3x3 equals 0.
06:55
Then the second one is 3x2 plus x3 equals 0, and the third negative x3 x2 minus 2, sorry minus x3 equals 0.
07:18
Then we can see that the third equation and multiplying by negative one both sides is the same as the second equation.
07:25
So we have only two equations.
07:28
X1 plus 2x2 plus 3x3.
07:32
Sorry, i put 3, but it's not 3 because we have, let me show you 1 here.
07:40
So it's another 3.
07:41
It's a 1.
07:42
So here, sorry, it's 1 and here it's 1.
07:46
Okay, we equal 0 and the second one is x3 equal negative to x2 for example.
07:56
I have put x3 in terms of x2 and we can do the same for x1.
08:04
The first equation you get x1 equal negative 2 x2 minus x3.
08:11
That is negative 2 x2 minus x3 will be plus.
08:19
To x2 x3 is negative okay so here i think i have a problem in some equations okay i have another error here sorry the second equation is x3 here i'm looking at here this equation which is the same as this one here we are solving for x3 so it's negative 3 x2.
09:04
So here, let's get rid of this a little bit, is 3 here.
09:11
As we are solving this equation for x3.
09:15
We get negative 3x2.
09:19
Okay, so that's what we put in here.
09:23
And so the last term will be negative x3 is plus 3x2.
09:31
So it that gives us x2.
09:33
So x1 and x2 are the same.
09:36
Are equal and x3 is given in terms of x2 this way and x1 is any value there is no restriction in sorry x2 is any value there is no restriction on that so any eigenvector corresponding to lambda equals 0 is of the form with here here is any value in the second component x2, let's say alpha, and then the first component x1 is equal to second component against 2 is alpha, and the third x3 here is negative 3 times alpha.
10:48
For alpha, alpha, a real number different from 0.
10:57
So we can say that a vector generating these or expanding the eigenspaced, corresponds to the eigenvalue 0 is a vector 1 -1 -negative 3 for example and divided that vector by its norm, its 2 norm, we obtain a unitary vector expanding the eigenspace correspond to the eigenvalal and that equals 0.
11:26
Okay, we get that and this is, let's say, the general answer you can find a particular eigenvector if you want by putting alpha equal a non -negative value and non -zero value.
11:41
For example, alpha equal 1, we will give us 1 -1 -negative 3.
11:47
Okay, so we go now for the eigen vectors corresponding to the eigen value lambda equal 1.
12:06
And that, similar to what we did before, that is we state equation a x equal lambda x for lambda equal 1, and x will be a non -zero vector.
12:19
That is a vector with not all its component equal to zero.
12:29
So if we write that in this case is 1 to 1, 0, 3 ,1, 0 ,0, 0, negative, 1 is 0, negative 1 is 0 ,000, is 0 ,000, is 3 negative 1 is the matrix times a generic vector we are looking for, x1 x2 is 3 equal 1 times the vector x1 x2 is 3 that is equal x1 x2 x3 and that becomes a system of equations which is x1 plus 2 x2 plus x3 equal x1 3 x2 plus x3 equal x2 plus x2 plus x2 equal x2 and negative 3 x2 minus x3 equal x3.
13:27
What we get from here, we get the first equation, we cancel out x1 both sides of the equation, so we get 2x2 plus x3 equals 0.
13:47
From the second we get 3x2 minus x2 is 2x2 plus x3 equals 0, so it's the same equation as the first 1...