00:02
Hello there.
00:04
Okay, so for this exercise we have the following linear transformation and we need to find the eigen space and the eigen values for this linear transformation.
00:15
So it might seem to be complicated, but the best way to do this is to consider the matrix representation of this linear transformation.
00:25
So the matrix representation can be easily obtained by considering the coefficients of each variable at each position of the vector that we obtain here.
00:40
So this matrix is going to be 2, minus 1, minus 1, 1, 0, minus 1, and minus 1, 1, 2.
00:53
So now that we have a more familiar setting, we can work using this matrix to find the eigen values and the eigen space.
01:01
Case.
01:04
Okay, so how to, so the first thing we need to obtain is the eigenvalues and we know that for that we need to consider the determinant of lambda, in this case the identity 3 by 3, minus here the matrix that we're interested to obtain the eigen values and we need to equate this to 0.
01:30
This will return us the characteristic equation and and from that, so basically here we have these determinant of lambda minus 2, 1 minus 1 lambda, 1 and minus 1 and lambda minus 2.
01:55
So this determinant gives us the characteristic equation that in this case is equals to lambda so it is cube minus 4 lambda square plus 5 lambda and minus 2.
02:13
This should be equal to 0.
02:16
Ok, so we should have 3 roots for this 3 degree polynomial.
02:23
And happen that we can decompose this polynomial as lambda minus 2 times lambda minus 1 square.
02:33
Basically, what we have here is a double root and the root lambda equals to 2.
02:40
So we have the first eigenvalue that is 2 and we have our second eigenvalue 1, but in this case this has multiplicity equals to 2.
02:59
Great.
02:59
So we have the eigenvalues and now we need to obtain the eigenvectors.
03:05
So the second thing that we need to do is to find the eigen vectors...