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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?

  • 1
    The $P_i$ are the orthogonal projections onto the Eigenspaces, so you need to find the eigenspaces of the eigenvectors. Find a nonzero eigenvector for each eigenvalue; because $A$ is normal, the eigenvectors will be mutually orthogonal. Then you just need to normalize them; the $P_i$ are given by the orthogonal projections onto the subspaces spanned by these vectors.2011-04-19
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    Shouldn't $P_i$ be the *projection* onto the $i$-th eigenspace? Also, shouldn't they sum to the identity instead of to 0?2011-04-19
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    @Arturo: I'm stuck after finding the eigenspaces. I got $$V_{\lambda_1}=sp{(\frac{1}{\sqrt 2}, 0, \frac{1}{\sqrt 2})}, V_{\lambda_2}=sp{(\frac{-1}{\sqrt 3}, \frac{1}{\sqrt 3}, \frac{1}{\sqrt 3})}, V_{\lambda_3}=sp{(\frac{-1}{\sqrt 6}, \frac{-2}{\sqrt 6}, \frac{1}{\sqrt 6})}$$ And I know that if $v=v_1+v_2+v_3, v_i \in V_{\lambda_i}$ then $P_i = v_i$. But how do I actually find $[P_i]$? I'm confused.2011-04-20
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    The $P_i$ are the orthogonal projections onto the spans of the eigenspaces. The eigenvectors you have are already an orthonormal basis, and the matrices of $P_i$ relative to that basis are very easy: for example, if you order your basis as $\beta=[v_{\lambda_1},v_{\lambda_2},v_{\lambda_3}]$, then $$P_1 = \left(\begin{array}{ccc}1&0&0\\0&0&0\\0&0&0\end{array}\right).$$ To write $[P_i]_{\beta}$ in terms of the standard orthonormal basis of $\mathbb{C}^3$, just take the change of basis matrix $M$, and compute $M[P_i]_{\beta}M^{-1}$ (cont)2011-04-20
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    Here, $M$ changes from the $\beta$ basis to the standard basis, so the columns of $M$ are the vectors of $\beta$.2011-04-20
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    Just curious. Are you avoiding the usual eigendecomposition on purpose to exploit the normality structure?2011-09-16

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