An operator that is diagonalizable but not normal. I'm looking for an example of operator $T \in L(\mathbb{C}^2)$ that is diagonalizable but not normal.
But, I think any diagonalizable $T \in L(\mathbb{C}^2)$ is normal because
$$M(T)= \begin{bmatrix}
a & 0\\ 
 0& b
\end{bmatrix}$$ with respect to some basis.
So, $$M(T^\ast )=\begin{bmatrix}
\bar{a} & 0\\ 
 0& \bar{b}
\end{bmatrix}$$ with respect to the same basis.
Thus, $$M(T)M(T^\ast )=\begin{bmatrix}
\left \| a \right \|^2 & 0 \\ 
 0& \left \| b \right \|^2
\end{bmatrix}=M(T^\ast )M(T)$$
Hence, if $T \in L(\mathbb{C}^2)$ is diagonalizable, then $T$ is noraml.
This is my reasoning, but it has to be wrong.
Which part in my reasoning doesn't make sense?
And, what is a good example of $T$ that is diagonalizable but not normal?
 A: Normal would imply orthogonal diagonalization. However, if $A$ is a $2\times 2$ matrix complex matrix with non-orthogonal 1-dimensional subspaces, then $A$ cannot be normal. For example, suppose
$$
              A\hat{a}=\hat{a},\;\;\; A\hat{b}=-\hat{b}
$$
where
$$
             \hat{a}\cdot\hat{b}\ne 0.
$$
For example,
$$
                 \hat{a}=\left[\begin{array}{1}1 \\ 1\end{array}\right],\;\;\; \hat{b}=\left[\begin{array}{1}1 \\ 0 \end{array}\right].
$$
Suppose $A=[a_{ij}]$. Then we want
$$
                   A\left[\begin{array}{c}1 \\ 1\end{array}\right]=\left[\begin{array}{c}1 \\ 1\end{array}\right] \\
               A\left[\begin{array}{c}1 \\ 0\end{array}\right]=\left[\begin{array}{c}-1 \\ 0\end{array}\right]
$$
Or,
$$
               a_{11}+a_{12}=1 \\
               a_{21}+a_{22}=1 \\
               a_{11} = -1 \\
               a_{21} = 0
$$
The resulting matrix $A$ is
$$
               \left[\begin{array}{cc}-1 & 2 \\ 0 & 1\end{array}\right]
$$
You can check that $A$ is not normal:
$$
            AA^{T} = \left[\begin{array}{cc}-1 & 2 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc} -1 & 0 \\ 2 & 1\end{array}\right]
   = \left[\begin{array}{cc}5 & 2 \\ 2 & 1\end{array}\right] \\
        A^{T}A = \left[\begin{array}{cc} -1 & 0 \\ 2 & 1\end{array}\right]\left[\begin{array}{cc}-1 & 2 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & -2 \\ -2 & 5\end{array}\right] \ne AA^{T}
$$
But $A$ is diagonalizable.
