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I've been reading that the diagonalizable matrices are normal, that is, they commute with their adjoint: ##M^*M=MM^*##, where ##M^*## is the conjugate transpose of ##M##.

So a matrix is diagonalizable if and only if it is normal, see: http://en.wikipedia.org/wiki/Normal_matrix

But from Boyce's section on repeated eigenvalues for systems of first-order differential equations, we have the example $$M=\begin{pmatrix}1&-1\\1&3\end{pmatrix}.$$ Which if we check ##MM^*##, we get $$MM^*=\begin{pmatrix}1&-1\\1&3\end{pmatrix}\begin{pmatrix}1&1\\-1&3\end{pmatrix}

=\begin{pmatrix}2&-2\\-2&8\end{pmatrix}.$$ The last is symmetric, so clearly ##MM^*=M^*M##.

However, Boyce does the problem, finds repeated eigenvalue of ##r=2##, only one eigenvector, for ##MP=PJ##, with ##J## not diagonal, that is, $$MP=

\begin{pmatrix}1&-1\\1&3\end{pmatrix}

\begin{pmatrix}1&0\\-1&-1\end{pmatrix}=

\begin{pmatrix}2&1\\-2&-3\end{pmatrix}=

\begin{pmatrix}1&0\\-1&-1\end{pmatrix}

\begin{pmatrix}2&1\\0&2\end{pmatrix}

=PJ$$

So we've found that ##M## is normal, but it is not diagonalizable. What is going on?

So a matrix is diagonalizable if and only if it is normal, see: http://en.wikipedia.org/wiki/Normal_matrix

But from Boyce's section on repeated eigenvalues for systems of first-order differential equations, we have the example $$M=\begin{pmatrix}1&-1\\1&3\end{pmatrix}.$$ Which if we check ##MM^*##, we get $$MM^*=\begin{pmatrix}1&-1\\1&3\end{pmatrix}\begin{pmatrix}1&1\\-1&3\end{pmatrix}

=\begin{pmatrix}2&-2\\-2&8\end{pmatrix}.$$ The last is symmetric, so clearly ##MM^*=M^*M##.

However, Boyce does the problem, finds repeated eigenvalue of ##r=2##, only one eigenvector, for ##MP=PJ##, with ##J## not diagonal, that is, $$MP=

\begin{pmatrix}1&-1\\1&3\end{pmatrix}

\begin{pmatrix}1&0\\-1&-1\end{pmatrix}=

\begin{pmatrix}2&1\\-2&-3\end{pmatrix}=

\begin{pmatrix}1&0\\-1&-1\end{pmatrix}

\begin{pmatrix}2&1\\0&2\end{pmatrix}

=PJ$$

So we've found that ##M## is normal, but it is not diagonalizable. What is going on?

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