Spectral decomposition of 4x4 matrix

• CGandC
In summary, the matrix has eigenvalues of 2 and -1, and eigenvectors in the sets e_1' = \{ v_1 = \pmatrix{1 \\1 \\ 0 \\ 3} , v_2 = \pmatrix{1 \\1 \\ 3 \\ 0} \} , e_2' = \{ v_3 = \pmatrix{0 \\1 \\ 0 \\ 1} , v_4 = \pmatrix{1 \\0 \\ 1 \\ 0} \} .
CGandC
Homework Statement
Let ## A = \pmatrix{ -4 & -3 & 3 & 3 \\ -3 & -4 & 3 & 3 \\ -6 & -3 & 5 & 3 \\ -3 & -6 & 3 & 5 } ## over ## \mathbb{R}##.
Write ## A ## in the form ## A = \lambda_2 P_1 + \lambda_1 P_2 ## where ## P_1 , P_2 ## are projections and ## \lambda_i ## are eigenvalues .
Relevant Equations
Spectral decomposition theorem says that if ## T ## is a normal transformation in a finitely generated unitary space ## V ## ( or a self-adjoint linear transformation in a real or complex vector space defined with inner-product ) and let ## \lambda_1 ,..., \lambda_k ## be all the different eigenvalues of ## T ## and let ## V_{\lambda_i} ## be the corresponding eigen-space for eigenvalue ## \lambda_i ##, then denote ## P_i ## as the orthogonal projection transformation on ## V_{\lambda_i} ##. Then:
## T = \lambda_1 P_1 + ... + \lambda_k P_k ##

A linear transformation ## T ## is normal iff ## TT^* = T^* T ##
## A = \pmatrix{ -4 & -3 & 3 & 3 \\ -3 & -4 & 3 & 3 \\ -6 & -3 & 5 & 3 \\ -3 & -6 & 3 & 5 } ## over ## \mathbb{R}##.
Let ## T_A: \mathbb{R}^4 \to \mathbb{R}^4 ## be defined as ## T_A v = Av ##. Thus, ## T_A ## represents ## A ## in the standard basis, meaning ## [ T_A]_{e} = A ##.
I've found the matrix's eigenvalues to be ## \lambda_1 = 2, \lambda_2 = -1 ## , whose eigenvectors are in the sets ## e_1' = \{ v_1 = \pmatrix{1 \\1 \\ 0 \\ 3} , v_2 = \pmatrix{1 \\1 \\ 3 \\ 0} \} ##, ## e_2' = \{ v_3 = \pmatrix{0 \\1 \\ 0 \\ 1} , v_4 = \pmatrix{1 \\0 \\ 1 \\ 0} \} ##, accordingly.
Denote the basis of eigenvectors as ## e' = e_1' \cup e_2' ##.

## [ T_A]_{e ~'} = \pmatrix{ 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1} = S^{-1}AS ## where ## S = \pmatrix{1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \\ 0 & 3 & 0 & 1 \\ 3 & 0 & 1 & 0 }##.

By the spectral decomposition theorem ## T_A ## is normal ( since it is diagonalizable ), hence
## T_A = \lambda_2 E_2 + \lambda_1 E_1 = -E_2 + 2E_1 ## where ## E_1, E_2 ## are projections on the eigenspace of ## \lambda_1 = 2 , \lambda_2 = -1 ## accordingly.
Note that ## [ E_1]_{e ~'} = \pmatrix{1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0}## , ## [ E_2]_{e ~'} = \pmatrix{0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1}##.

And that ## [ E_1]_{e ~} = S^{-1} [ E_1]_{e ~'} S = \pmatrix{-1 & -1 & -\frac{2}{3} & -\frac{1}{3} \\ -1 & -1 & -\frac{1}{3} & -\frac{2}{3} \\ 3 & 3 & 2 & 1 \\ 3 & 3 & 1 & 2}## , ## [ E_2]_{e ~} = S^{-1} [ E_2]_{e ~'} S = \pmatrix{2 & 1 & \frac{2}{3} & \frac{1}{3} \\ 1 & 2 & \frac{1}{3} & \frac{2}{3} \\ -3 & -3 & -1 & -1 \\ -3 & -3 & -1 & -1}##.

So ## [T_A]_{e} = \lambda_2[ E_1]_{e ~} + \lambda_1 [ E_1]_{e ~} = -[ E_2]_{e ~} + 2[ E_1]_{e ~} = 2[ E_1]_{e ~} - [ E_2]_{e ~} = ## ## \pmatrix{-4 & -3 & -2 & -1 \\ -3 & -4 & -1 & -2 \\ 9 & 9 & 5 & 3 \\ 9 & 9 & 3 & 5} ##

The problem is, I should have got that ## [ T_A]_{e} = A ##, but Instead I got something else, do you know why? I can't figure it out and it seems like I didn't make any mistakes, maybe it has to do with me using spectral decomposition in a wrong fashion?

You have $$\lambda_1 \operatorname{diag}(1,1,0,0) + \lambda_2\operatorname{diag}(0,0,1,1) = \Lambda = S^{-1}AS = \lambda_1S^{-1}P_1S + \lambda_2S^{-1}P_2S$$ where $\Lambda = \operatorname{diag}(\lambda_1,\lambda_1,\lambda_2,\lambda_2)$ is the Jordan Normal Form of $A$. It follows that $P_1 = S \operatorname{diag}(1,1,0,0)S^{-1}$ etc, which is not what you have calculated.

CGandC
I've got it now, thank you very much!

What is spectral decomposition of a 4x4 matrix?

Spectral decomposition of a 4x4 matrix is a process in linear algebra where a square matrix is decomposed into its eigenvectors and eigenvalues. This allows for the matrix to be expressed in a simpler form, making it easier to analyze and manipulate.

Why is spectral decomposition important?

Spectral decomposition is important because it allows for a complex matrix to be broken down into simpler components, making it easier to understand and work with. It also provides valuable information about the matrix, such as its eigenvalues and eigenvectors, which can be used in various applications.

How is spectral decomposition calculated for a 4x4 matrix?

The spectral decomposition of a 4x4 matrix is calculated by finding the eigenvalues and eigenvectors of the matrix. The eigenvalues are the values that, when multiplied by the corresponding eigenvectors, result in the original matrix. These values and vectors are then used to construct a diagonal matrix, which is the spectral decomposition of the original matrix.

What are the applications of spectral decomposition of a 4x4 matrix?

Spectral decomposition has various applications in fields such as physics, engineering, and computer science. It is used in solving systems of differential equations, calculating the stability of a system, and in image processing algorithms.

Can a 4x4 matrix always be spectrally decomposed?

Yes, a 4x4 matrix can always be spectrally decomposed as long as it is a square matrix. However, the eigenvalues and eigenvectors may not always be real numbers, which can make the process more complex. In these cases, complex numbers may be used in the spectral decomposition.

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