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Exponentiation of a matrix is normal

Give an example of a matrix A that is not normal but e^A is. Exponentiation is to be taken with respect to the usual definition.

Solution

Let P= \begin{pmatrix} 2 &1 \\ 1 & 1 \end{pmatrix} which is clearly invertible and P^{-1} = \begin{pmatrix} 1 &-1 \\ -1 & 2 \end{pmatrix}. Let D = \begin{pmatrix} 2 \pi i &0 \\ 0 & -2\pi i \end{pmatrix}. Now consider

    \[A = P D P^{-1} = \begin{pmatrix} 2 &1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 2 \pi i &0 \\ 0 & -2\pi i \end{pmatrix} \begin{pmatrix} 1 &-1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 6 \pi i & -8 \pi i \\ 4\pi i  & - 6 \pi i \end{pmatrix}\]

It follows easily that the conjugate transpose matrix of A is

    \[A^* = \begin{pmatrix} -6 \pi i & -4\pi i\\ 8 \pi i & 6  \pi i \end{pmatrix}\]

Also,

    \[A A^* = \begin{pmatrix} 6 \pi i & -8 \pi i \\ 4\pi i  & - 6 \pi i \end{pmatrix} \begin{pmatrix} -6 \pi i & -4\pi i\\ 8 \pi i & 6  \pi i \end{pmatrix} = \begin{pmatrix} 100 \pi^2 & 72 \pi^2 i \\ 72 \pi^2 & 52 \pi^2 i \end{pmatrix}\]

and

    \[A^* A =\begin{pmatrix} -6 \pi i & -4\pi i\\ 8 \pi i & 6  \pi i \end{pmatrix} \begin{pmatrix} 6 \pi i & -8 \pi i \\ 4\pi i  & - 6 \pi i \end{pmatrix} = \begin{pmatrix} 52 \pi^2 & -72 \pi^2 \\ -72 \pi^2 & 100 \pi^2 \end{pmatrix}\]

Hence A A^* \neq A^* A and therefore A is not normal. Furthermore,

    \begin{align*} \exp A &= \exp \left ( P D P^{-1} \right ) \\ &=P \exp (D) P^{-1} \\ &= P \; \mathbb{I}_{2 \times 2} \; P^{-1} \\ &= P P^{-1} \\ &= \mathbb{I}_{2 \times 2} \end{align*}

which is normal.

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