- #1

CGandC

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- 34

- 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?

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?