Spectral Decomposition of Linear Operator T

[MathIdiot]

1. Let T be the linear operator on R^n that has the given matrix A relative to the standard basis. Find the spectral decomposition of T.

A=

7, 3, 3, 2
0, 1, 2,-4
-8,-4,-5,0
2, 1, 2, 3


3. eigen values are 1 (mulitplicity 1), -1 (mult. 1), 3 (mult. 2). And associated eigen vectors:

(1,-2,0,0)
(1,-6,4,-1)
(1,0,-1,-1/2), (0,1,-1/2,-3/4), respectively.

So, T = P1 - P2 + 3P3 (P1, P2, P3 being projection matrices)

I really need some sort of algorithm with perhaps this as an example, because I will have to solve more like it. Thanks so much!

 

[gumi_kr]

Try typing 'finding eigenvalues' in google and you will find your answer there. It is very common thing and you shall find it in any advanced 'linear algebra' book. (ctrl+f + eigenvalue)

The general idea of 'spectral decomposition' is to find eigenvalues and vectors associated with it (choose any of them), change base to eigenvectors and have a diagonal matrix.

 

[Architeuthis Dux]

I would first like to congratulate you on successfully finding the spectral decomposition of the given linear operator T. This is an important technique in linear algebra and is used to break down a linear operator into its eigenvalues and eigenvectors.

Now, let's discuss the spectral decomposition in more detail. The spectral decomposition of a linear operator T is a way of representing T as a sum of projection matrices, each corresponding to a different eigenvalue of T. In this case, we have three eigenvalues: 1, -1, and 3.

To find the spectral decomposition, we first need to find the eigenvalues and eigenvectors of T. This can be done by solving the characteristic equation det(A - λI) = 0, where A is the given matrix and λ is the eigenvalue. Once we have the eigenvalues, we can find the corresponding eigenvectors by solving the system (A - λI)x = 0.

In this case, you have already found the eigenvalues and their corresponding eigenvectors. Now, to find the spectral decomposition, we simply need to construct the projection matrices P1, P2, and P3. This can be done by using the formula P = vv^T, where v is the eigenvector.

So, for the eigenvalue 1, we have the eigenvector (1,-2,0,0). Using the formula, we get P1 = (1,-2,0,0)(1,-2,0,0)^T =
(1, -2, 0, 0)(1, -2, 0, 0) =
(1, -2, 0, 0)
(-2, 4, 0, 0)
(0, 0, 0, 0)
(0, 0, 0, 0)

Similarly, for the eigenvalue -1, we have the eigenvector (1,-6,4,-1). Using the formula, we get P2 = (1,-6,4,-1)(1,-6,4,-1)^T =
(1, -6, 4, -1)
(-6, 36, -24, 6)
(4, -24, 16, -4)
(-1, 6, -4, 1)

For