00:01
Given the matrix a equal 101, 12100 negative 1, we will show that a is diagonalizeable and we will find an invariable matrix b and a diagonal matrix d, such that p to inverse times a times p equal d.
00:23
Remember that if we can find a matrix p, an invariable matrix p, fulfilling this equation.
00:35
Means that d and a are similar metrics.
00:41
And because d is diagonal, its eigenvalues are placed on the diagonal, on the main diagonal.
00:52
And we know that two similar metrics, two similar matrices get to have the same eigenvalues.
01:02
So the eigenvalues of a are the values that are placed in the main diagonal of d.
01:10
Another thing we know is that a metric is diagonalizable.
01:15
If we can find a base of the space r3 in this case, because we have a 3x3 matrix.
01:23
So if we can find a base formed by eigenvectors of a, one eigenvector for each eigenvalue, we can find that matrix, sorry, that set of three vectors, of eigenvector of a, which is a base of r3, then the matrix is diagonalized.
01:54
And because we're gonna have the eigenvalues of a in the main diagonal of d, so we start by calculating the eigenvalues of a.
02:09
Because with that, we then calculate the eigenvectors and look for a base of this space out of the eigenvectors.
02:19
So the iq values are calculated by finding the roots of the polynomial a minus lambda identity i.
02:33
This i here is a 3x3 identity matrix.
02:38
And this determinant is equal to the determinant of a minus lambda times the identity will be equal to 1 minus lambda 0, 1.
02:53
1, 2 minus lambda, 1, 0, 0, 0, negative 1 minus lambda.
03:06
This determinant can be developed using the third row, because we have 2 zeros there.
03:15
And we get negative 1 minus lambda, which is an even position, because it's 3 -3, and the indices, sum up to 6, which is an even number.
03:29
Directly the coefficient there is one here times the determinant of the sub -matrix 1 minus lambda 0 1 2 minus lambda okay we have that see here okay so that's equal to let's put out a common negative 1 factor out it's negative 1 plus lambda times this determinant here is minus lambda times two minus lambda and then we can put this negative inside this factor for example or or the other one we get we get lambda plus one which is the same as one plus lambda times one minus lambda times lambda minus 2.
04:52
So this is what we call the characteristic polynomial of the matrix a and its roots are the eigenvalues of a and the good thing here is that we have obtained the characteristic polynomial in a factor form so we can tell directly from here which are the roots of this polynomial and so which are the eigenvalues of a first one is negative one which can nullifies this factor the second one orifies this factor here is lambda 2 equal 1 and the third one is 2 so we have these three eigen values of a now for each we calculate an eigen vector so we go for the eigen vectors one associated to each eigenvalue associated to lambda equal negative 1 so the system we get to state is a x equal lambda 1x or x vector different from the vector 0 that is the matrix a is 101 1 1 2 1 and 0 0 negative 1 times vector x let's call it x1 x2 x3 its components and that can be equal to negative 1 which is lambda 1 times the same vector x1 x2 x3 so we get a solve for x1 x2 x3 and we write what this means we have x1 plus x3 equal a negative x1 then x1 plus x2 x2 plus x3 equal negative x2 and the last one is negative x3 equal negative x3 so this equation here down x3 equal x3 is a tautological equation that is doesn't matter the value of x3, it is true.
08:45
So true for any x3, the real number.
08:54
So it doesn't constrain anything, this equation.
08:59
So we stay with the other two, one plus x3 equal negative x1, and the other two, the second one is x1 plus 2x2, plus x3 equal negative x2.
09:16
And then the first one is pass negative x1 to the left we get 2x1 plus x3 equal 0 and passing this negative x2 to the left we get x1 plus x2 plus x3 equal 0.
09:50
So from the first equation to x1 plus x3 equals 0, we can write, for example, x1 in terms of x3, we get negative x3 over 2, for example.
10:26
And putting that in the second equation from x1 plus 3, x2, plus x3 equals 0, and x3, and x3, x1 equal negative x3 over 2 we get so we replace this expression instead of x2 instead of x1 here and we get negative x3 half which is x1 plus 3 x2 plus x3 equal 0 multiplied by 2 both sides we get negative x3 plus 6 x2 plus 2 plus 2 x3 equals 0 and simplify the x3 here and we get x3 plus 6 x2 equals 0 and again x2 can be written in terms of x3 we get negative x3 over 6 so x3 is free is any value and x1 and x2 are written in terms of x3.
12:22
For example, if we take, taking x3 equals 6, we get x1 equal, we use this expression here, negative x3 over 2 will be negative 6 over 2 and that is negative 3.
12:57
And using this expression here we get x2 equal negative 6 over 6 that is negative 1 so the vector x1 is negative 3 x2 is negative 1 x 3 is an eigenvector very important remember the icon vector must be different from zero and this is the case here for this choice it's an eigenvector associated to the eigenvalue and the one equal negative one so we have our first vector right here now we go for the second one that is an eigenvector associated two second igen value lambda 2 equal 1 so we start as here associated to lambda 1 now associated associated to lambda 2 equal 1 so we state the equation a x equal lambda 2x 4 x a vector known 0 this means that 1 0 1 1 2 1 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 times a vector x1 x2 is 3 is equal to 1 times the same vector x1 x2 x3 and we do the multiplication and state the equality of the vectors and we get the system x1 plus x3 equal x1 x1 plus x3 equal x1 x1 plus x3 equal x2 and negative x3 equal x2 so from here we get 2x3 equal 0 that is x3 must be 0 okay now the system reduces 2 because x3 is 0 we put it in the first 2 equation and we get x1 equal 1 x1 and x1 plus 2 x2 equal x2 equal x2 because we have put there x3 equals 0 that's a tautological equation that is true for any value x1 the real number so it does not constrate anything and from here we get x1 plus x2 equals 0 that is for example x2 equal negative x1 so x1 is any value we we can choose and x2 in terms of x1 x3 is equal to 0 that is we put any x1 then using that value we choose for x1 we put it here and we get x2 and x3 got to be always equal to 3 to 0 so, for example, it is obvious that x1 can be 0, because if we choose x1 equals 0, the vector will be 0 is not an eigenvector.
18:54
So, for example, we can take x2, x1 equal 1, or x1 equal negative 1, for example, we get x2 equal 1.
19:12
And so an eigenvector, remember that the eigenvectors are not unique.
19:23
Once we have one, we can find infinitely many of them.
19:32
In this case, we are calculating an eigenvector, in particular, taking this value for x1.
19:39
So an eigenvector associated to the second eigenvalue, than the 2 equal 1, is, the vector x1 equal negative 1, x2 equal 1, and x3 0.
20:02
And we have then the second iguen vector.
20:07
And we can see that because we have a third component equal to 0, these two vectors are linearly independent.
20:15
We can never reproduce this vector by multiplying this other vector by a constant.
20:23
Because we have a 0 in the third component and any scalar times that 0 will be, which is not the third component of this vector so they are linearly independent so we go correct in the correct direction so we need another aga vector now we are for lambda 3 equal 2 same equation a x equal lambda 2x 4 x and then non -zero vector so this equation will be 1 .01 1 2 1 2 1 and 0 0 negative 1 times x1 x2 x3 equal in this case is 2 times x1 x2 x3 and this state the system x1 plus x3 equal 2x1 x1 plus 2 x2 plus x3 equal 2x2 and negative x3 equal x3 so from negative x3 equal x3 3 3 3 3 3 we get 2 x3 equal 0 and so x3 get to be 0 and we reduce the system to the following reduces to putting x3 equals 0 in the first equation we get x1 equal 2 x1 and the second is x1 plus 2 x2 equal to x2 the first equation means x1 equals 0 because as we put this turn to the right we get x1 equals 0 and this term here cancel out with this 1 and we get x1 equals 0 so is in reality only one equation so both x3 and x1 going to be zero but x2 is any value okay sorry in this case x2 in the real numbers can be any value for example taking x2 equal it's important that x2 is not zero so taking x2 equal so taking x2 equal so let's say one, we get an eigenvector associated to the eigenvalue, lambda 3 equal 2...