using python show that your code gives a correct diagonalization of the above matrix a by calcul

Name: using python show that your code gives a correct diagonalization of the above matrix a by calculating pdp1 19417
Uploaded: 2022-05-18T02:13:23-08:00
Duration: 13 min 49 s
Channel: Shu-Ting Huang
Description: using python show that your code gives a correct diagonalization of the above matrix a by calculating pdp1 19417

To diagonalize a matrix ( A ) using Python, we need to follow these steps:

Using python show that your code gives a correct...

Transcript

00:01 In part a, we are going to verify that a is diagonalizable and in part b we are going to compute its diagonalization.

00:11 So first thing to do is to find the eigenvalues.

00:16 So we compute the characteristic polynomial of a, which is a determinant 3 minus lambda to negative 3, negative 3, negative 4 minus lambda, 9, negative negative 2 5 minus lambda we use co -factor expansion from the first row first column then we have 3 minus lambda times determinant this determinant and then the minus negative 3 this determinant and then minus negative 3 this determinant and then plus negative 1 this determinant.

01:24 If we expand it, it will be 3 minus lambda.

01:31 And this part it is lambda square minus lambda minus 2 and this is plus 3 and inside we have negative 2 lambda plus 4 and this term is negative 3 negative plus 6 so it is equal to 3 minus lambda and this is the lambda minus 2 lambda plus 1 plus 1 plus lambda minus 2 and here we have negative 6 here we have 3 so we have a lambda minus 2 and the rest are we expand this here so negative lambda square minus 1 lambda plus 3 lambda plus 2 lambda plus 2 lambda plus 3 minus 6 plus 3 so it is lambda minus 2 and here it is lambda and negative lambda squared plus 2 lambda so it is lambda negative lambda times lambda minus 2 squared so now we have eigen values we need to find eigen vectors so we compute a minus 0i this is just a and then we are going to compute its nor space we swap the first and the third row so negative 1 negative 2 5 and then we add the negative 3 times from the first to the second row so it becomes negative 1 negative 2 5 and here is 0 this is 2, negative 6, and 3 times add to the third row, so 0, negative 4, negative 12.

05:09 So it is 12.

05:18 So from here we have the kernel of a minus 0.

05:26 It is generated by the vector here we put 1 here and here is 3 and 5 minus 6 plus 1 is 0 so here we put negative 1 so this is the eigenvector corresponding to eigenvalue 0.

06:03 Next we compute eigenvector corresponding to ligand value 2.

06:08 So we compute a minus 2i and then compute its normal space.

06:51 Here we use low operation.

06:53 We can eliminate the second and third row.

06:56 So we have the kernel of a minus 2i.

07:01 It is banned by two vectors...