[Math] Spectral decomposition of a normal matrix

linear algebraspectral-theory

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?

Best Answer

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!

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