Spectral decomposition of 4x4 matrix

[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?

 

[pasmith]

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!