Spectral decomposition of a normal matrix I'd like to find the spectral decomposition of $A$:
$$A = \begin{pmatrix}
2-i & -1 & 0\\ 
-1 & 1-i & 1\\ 
0 & 1 & 2-i
\end{pmatrix}$$
i.e. $A=\sum_{i}\lambda_i P_i$ where $P_i$ are the coordinate matrices (in the standard basis) of the corresponding orthogonal transformations in the spectral decomposition of $T_A$ and $\lambda_i$ are the eigenvalues.
I started off by showing that $A$ is normal, piece of cake.  
Then found the eigenvalues of $A$, those are: $\lambda_1 = 2-i, \lambda_2 = 3-i, \lambda_3 = -i$.
I tried using these known facts from the spectral theorem:


*

*$A=(2-i)P_1+(3-i)P_2-iP_3$

*$I=P_1+P_2+P_3$

*$\forall i\neq j, P_i P_j=0$

*$P^*_i=P_i$


The only example I have in my book uses these but I couldn't get it to work here. The terms don't cancel out it seems.
What else can I try?
 A: Using the primary decomposition theorem (PDT): Find the minimal polynomial of $A$. Clearly that would be $m_A(x)=(x-\lambda_1)(x-\lambda_2)(x-\lambda_3)$. Define $f_i(x)=\frac{m_A(x)}{(x-\lambda_i)}$. Observe that $f_1,...,f_3$ are co-prime (i.e. $gcd(f_1,f_2,f_3)=1$). Hence you can find polynomials $g_1,g_2,g_3$ such that $g_1f_1+g_2g_2+g_3f_3=1$. \
Finally, define $P_i=g_i(A)f_i(A)$.
Check why does it work!
A: The following result is useful
"A matrix is normal if and only if it is unitarily similar to a diagonal matrix, and therefore any matrix A satisfying the equation $A^{*}A=AA^{*}$ is diagonalizable". That is
$$ \mathbf{A} = \mathbf{U} \mathbf{\Lambda} \mathbf{U}^* . $$
The next step is to find the eigenvectors $v_i$. Once that done, then you can get the matrices $P_i$ such that
$$ P_i = v_iv^{*}_i, $$ 
which gives,
$$ A=(2-i)P_1+(3-i)P_2-iP_3 = (2-i)v_1v^{*}_1+ (3-i)v_2v^{*}_2-i v_3v^{*}_3.  $$ 
Note: You should check that $P_i,i=1,2,3$ satisfy the properties you listed above.
Eigenvectors: I computed the eigenvectors by Maple which corresponds to the eigenvalues that you already computed $2-i,3-i,-i$ respectively
$$v_1= \left[ \begin {array}{c} 1\\0
\\1 \end {array} \right ], v_2=\left[ \begin {array}{c} -1\\1
\\1 \end {array} \right ], v_3=\left[ \begin {array}{c} -1\\-2
\\1 \end {array} \right ]   $$ 
A: it is also possible to compute the Projection matrices without finding each eigen vector (corresponding to each eigen-value) of the matrix A.  The trick is to use the Caley-Hamilton Theorem.  First find the partial fraction decomposition of 1/charpoly(A,x) where charpoly(A,x) is the characteristics polynomial of the matrix A factored in the complex field as linear factorr.
