Non-diagonalisable matrices and normal matrices

The characteristic polynomial of $N$ is $\lambda^2$: its only eigenvalue is $0$. The associated eigenspace is the kernel of $N$, consisting of the vectors $(x,y)$ such that $N(x,y)=(y,0)=0$, that is, the line $y=0$. It has dimension $1$, whereas two independent eigenvectors would be required to diagonalise $N$. Alternatively, if $N$ were diagonalisable, then, since its only eigenvalue is $0$, it would be similar to the zero matrix, and hence would itself be zero. Similarly, $T=\mathbf{1}+N$ has $1$ as its only eigenvalue, and the eigenspace $\ker(T-\mathbf{1})=\ker N$ has dimension $1$: $T$ is not diagonalisable. We verify that $N$ and $T$ are not normal: for example, $NN^\dagger=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $N^\dagger N=\begin{pmatrix}0&0\\0&1\end{pmatrix}$.

The eigenvalues of this triangular matrix are its diagonal entries, $1$ and $2$. They are distinct, so $B$ is diagonalisable. For $\lambda=1$, we find the eigenvector $(1,0)$; for $\lambda=2$, the equation $-x+y=0$ gives $(1,1)$. These two vectors are not orthogonal: their inner product is $1$. Since each eigenspace is a line, there is no other possible choice: $B$ is not diagonalisable in an orthonormal basis. We verify that it is not normal:

$$\begin{aligned} BB^\dagger=\begin{pmatrix}1&1\\0&2\end{pmatrix}\begin{pmatrix}1&0\\1&2\end{pmatrix}=\begin{pmatrix}2&2\\2&4\end{pmatrix}, \qquad B^\dagger B=\begin{pmatrix}1&0\\1&2\end{pmatrix}\begin{pmatrix}1&1\\0&2\end{pmatrix}=\begin{pmatrix}1&1\\1&5\end{pmatrix}. \end{aligned}$$

If $A$ is diagonalisable in an orthonormal basis, there exist a unitary matrix $U$ and a diagonal matrix $D$ such that $A=UDU^\dagger$. Then $A^\dagger=UD^\dagger U^\dagger=UD^*U^\dagger$, and

$$AA^\dagger=UDU^\dagger UD^*U^\dagger=UDD^*U^\dagger, \qquad A^\dagger A=UD^*DU^\dagger .$$

The diagonal matrices $D$ and $D^*$ commute, so $AA^\dagger=A^\dagger A$. For the two requested examples, the theorem is not even needed: if $A^\dagger=A$, then $AA^\dagger=A^2=A^\dagger A$; if $A^\dagger=A^{-1}$, then $AA^\dagger=\mathbf{1}=A^\dagger A$.

The matrix is not Hermitian, since $C^\dagger=\begin{pmatrix}1&1\\-1&1\end{pmatrix}\neq C$. We calculate

$$\begin{aligned} CC^\dagger=\begin{pmatrix}1&-1\\1&1\end{pmatrix}\begin{pmatrix}1&1\\-1&1\end{pmatrix}=\begin{pmatrix}2&0\\0&2\end{pmatrix}=C^\dagger C, \end{aligned}$$

so $C$ is normal, but $CC^\dagger=2\mathbf{1}\neq\mathbf{1}$: it is not unitary. In fact, $C=\sqrt2\,R(\pi/4)$, where $R$ is the plane rotation matrix. Its eigenvectors are therefore those of rotations, $\frac{1}{\sqrt2}(1,-i)$ and $\frac{1}{\sqrt2}(1,i)$, which are orthogonal. For example, $C(1,-i)=(1+i,\,1-i)=(1+i)(1,-i)$, since $(1+i)(-i)=1-i$. The eigenvalues are $1\pm i$, which are non-real and have modulus $\sqrt2$. In quantum mechanics, observables are represented by Hermitian matrices, which additionally have real eigenvalues; non-Hermitian normal matrices, such as unitary operators, are used instead to represent transformations.