Let $ A \in L ( \mathbb{C}^4) $ be a normal operator with characteristic polynomial $ k_{A} = (\lambda - 1)^2 * (\lambda - 2)^2$. Is then the matrix for the operator just a diagonal matrix with eigenvalues on the diagonal?
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Not necessarily. The operator is certainly - as it is normal, then diagonalizable - similar to a diagonal matrix with the eigenvalues on the diagonal.
Joca Ramiro
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But how do I find the operators matrix then? – user15269 Feb 14 '19 at 18:33
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You can't determine the operator from the characteristic polynomial. – Joca Ramiro Feb 14 '19 at 18:37
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See this question: https://math.stackexchange.com/questions/1658106/determining-a-matrix-from-its-characteristic-polynomial – Joca Ramiro Feb 14 '19 at 18:38
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The operator $A$ is unitarily diagonizable, as are all normal operators (see this wikipedia page), though it may not itself be diagonal in the initial form it is presented; since the characteristic polynomial of $A$,
$k_A = (x - 1)^2(x - 2)^2, \tag 1$
has two zeroes $1$ and $2$, each of multiplicity $2$, it follows that there is a basis of $\Bbb C^4$ in which the matrix of $A$ takes the form
$[A] = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{bmatrix}. \tag 2$
Robert Lewis
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