Consider the normal operator $T \in L(C^2)$ whose matrix with respect to the standard basis is

$$ \left( \begin{array}{ccc} 2 & -3 \\ 3 & 2 \end{array} \right)$$

As you can verify, $\frac{(i, 1)}{\sqrt{2}}, \frac{(-i, 1)}{\sqrt{2}}$ is an orthonormal basis of $C^2$ consisting of eigenvectors of T, and with respect to this basis the matrix of T is the diagonal matrix

$$ \left( \begin{array}{ccc} 2 + 3i & 0 \\ 0 & 2 - 3i \end{array} \right)$$

I have a couple questions about this example.

1) I can verify that the orthonormal basis is indeed an orthonormal basis of $C^2$ but where did they come up with the basis in the first place?

2) How did they get the matrix at the end?

This uses the part of the spectral theorem that states that if T is normal in a complex vector space, then T has a diagonal matrix with respect to some orthonormal basis of V. Can someone explain how the answer above was reached? Thank you!


First of all, calculate the characteristic polynomial. The roots are the eigenvalues, in this case we have two distinct eigenvalues. Next, find a non-trivial solution of the linear equation system

$$A-\lambda I=0$$

for each of the eigenvalues. This gives the eigenvectors corresponding to the eigenvalues. In this case, the eigenvalues are orthogonal, so you only have to norm them to get an orthonormalsystem. For symmetric matrices the eigenvectors to distinct eigenvalues are always orthogonal, here they are orthogonal because the given matrix is normal ($A^TA=AA^T$).

Finally the eigenvectors form an invertible matrix T with the property $$T^{-1}AT = D$$

where D is the diagonalmatrix whose diagonal elements are the eigenvalues of A.


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