Ket-bra operators

The matrix of $\ket u\bra v$ is the product of the column vector $\ket u$ and the row vector $\bra v=\frac{1}{\sqrt2}(0,1,1)$:

$$\begin{aligned} A=\frac12\begin{pmatrix}1\\ i\\0\end{pmatrix}\begin{pmatrix}0&1&1\end{pmatrix}=\frac12\begin{pmatrix}0&1&1\\0&i&i\\0&0&0\end{pmatrix}. \end{aligned}$$

From $\hat A\ket w=\braket vw\ket u$, every vector in the image is proportional to $\ket u$: the image is the line spanned by $\ket u$, and the rank is $1$. The kernel is the set of vectors orthogonal to $\ket v$, namely the plane with equation $w_2+w_3=0$, of dimension $2$, in accordance with the rank—nullity theorem.

For all $\ket w$, $\ket z$, we have $\braket{w}{\hat A z}=\braket vz\braket wu$. The operator $\ket v\bra u$ satisfies $\braket{(\ket v\bra u)w}{z}=\braket uw^*\braket vz=\braket wu\braket vz$, which is the same quantity. Thus

$$\begin{aligned} \hat A^\dagger=\ket v\bra u, \qquad A^\dagger=\frac12\begin{pmatrix}0&0&0\\1&-i&0\\1&-i&0\end{pmatrix}, \end{aligned}$$

which is indeed the conjugate transpose of $A$. More generally, $(\ket a\bra b)^\dagger=\ket b\bra a$. The operator $\ket a\bra b$ is therefore self-adjoint if and only if $\ket a\bra b=\ket b\bra a$. Applying both sides to $\ket a$ gives $\braket ba\ket a=\norm a^2\ket b$, so $\ket b$ is proportional to $\ket a$: $\ket b=\lambda\ket a$. The equality then becomes $\lambda^*\ket a\bra a=\lambda\ket a\bra a$, hence $\lambda$ is real. A ket-bra is self-adjoint if and only if the two vectors are proportional with a real coefficient; this is the case for projectors $\ket a\bra a$.

Using associativity of the product,

$$\hat A^2=\ket u\braket vu\bra v=\braket vu\,\hat A .$$

Now $\braket vu=\frac12(0\cdot1+1\cdot i+1\cdot0)=\frac i2$, so $\hat A^2=\frac i2\hat A$. For the trace, we use an arbitrary orthonormal basis $(\ket{e_k})$:

$$\tr\hat A=\sum_k\braket{e_k}{u}\braket{v}{e_k}=\sum_k\braket{v}{e_k}\braket{e_k}{u}=\braket vu=\frac i2,$$

by the closure relation $\sum_k\ket{e_k}\bra{e_k}=\mathbf{1}$. In the matrix, the sum of the diagonal entries is indeed $\frac12(0+i+0)=\frac i2$, and direct calculation also verifies that $A^2=\frac i2A$.

The kernel, of dimension $2$, is the eigenspace associated with the eigenvalue $0$. Moreover, $\hat A\ket u=\braket vu\ket u=\frac i2\ket u$: $\ket u$ is an eigenvector with eigenvalue $\frac i2$. Since $\frac i2\neq0$, there are $2+1=3$ linearly independent eigenvectors, and $\hat A$ is diagonalisable. Its eigenvectors are not orthogonal, however: for example, $\ket u$ is not orthogonal to the vector $(1,0,0)$ in the kernel, since their inner product is $\frac{1}{\sqrt2}$.

This is consistent with the spectral theorem: an orthonormal basis of eigenvectors exists only for a normal operator, $\hat A\hat A^\dagger=\hat A^\dagger\hat A$. Now $\hat A\hat A^\dagger=\ket u\braket vv\bra u=\ket u\bra u$, whereas $\hat A^\dagger\hat A=\ket v\braket uu\bra v=\ket v\bra v$: these two projectors are different, and $\hat A$ is not normal.