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Homework question.


Let $A$ be a $2 \times 2$ matrix with real entries. Suppose that $A$ has an eigenvalue $\lambda$ with the imaginary part of $\lambda \neq 0$. Is there an orthonormal basis of $\mathbb{C}_2$ consisting entirely of eigenvectors of $A$? Explain.


So far I've got the following:

A matrix is self-adjoint iff its eigenvalues are all real. By the spectral theorem, a linear transformation $T$ from $V$ to $V$ is normal iff its eigenvectors span $V$. A linear transformation is normal iff it commutes with its adjoint.

With these three facts, I can state that A is not self-adjoint, and I need to show that it is not normal, assuming that the proposition is false (my belief).

I've found a couple simple counterexamples, but I can't figure out how to make the jump from not self-adjoint to not-normal. (Or any other method of showing it's not normal).

  • 3
    What about $$\begin{pmatrix}0&-1\\1&0\end{pmatrix}$$?2012-05-08
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    There exist normal examples. Take a rotation matrix. (Its adjoint is its inverse.)2012-05-08
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    ...and the example I gave is in fact both a rotation matrix and a skew-symmetric matrix.2012-05-08
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    And the eigenvectors I get are (i,1) and (-i,1), which are orthogonal. Hm, I was wrong then.2012-05-08
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    Is it not true that the adjoint of a matrix is its conjugate transpose? I feel I may be missing something. {{2,3},{-2,1}} doesn't commute with it's conjugate transpose, at least.2012-05-08

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