Diagonalisation with a degenerate eigenvalue

Every entry of $J^2$ is $1+1+1=3$, so $J^2=3J$. If $J\mathbf v=\mu\mathbf v$ with $\mathbf v\neq0$, then $\mu^2\mathbf v=3\mu\mathbf v$, hence $\mu=0$ or $\mu=3$. The vector $(1,1,1)$ satisfies $J(1,1,1)=3(1,1,1)$, while every vector whose components sum to zero satisfies $J\mathbf v=0$. The eigenvalues of $A=\mathbf{1}+J$ are therefore $4$ and $1$.

To calculate the characteristic polynomial, subtract the second column from the first and the third column from the second, which does not change the determinant:

$$\begin{aligned} \det(A-\lambda\mathbf{1})=\begin{vmatrix}2-\lambda&1&1\\1&2-\lambda&1\\1&1&2-\lambda\end{vmatrix} =\begin{vmatrix}1-\lambda&0&1\\-(1-\lambda)&1-\lambda&1\\0&-(1-\lambda)&2-\lambda\end{vmatrix}. \end{aligned}$$

Factoring $(1-\lambda)$ out of each of the first two columns leaves

$$\begin{aligned} (1-\lambda)^2\begin{vmatrix}1&0&1\\-1&1&1\\0&-1&2-\lambda\end{vmatrix}=(1-\lambda)^2\bigl[(2-\lambda+1)+1\bigr]=(1-\lambda)^2(4-\lambda). \end{aligned}$$

The eigenvalue $1$ has multiplicity two, while the eigenvalue $4$ is simple. We verify that their sum $1+1+4=6$ equals the trace of $A$.

The eigenspace $E_4$ is the line spanned by $\ket{v_1}=\frac{1}{\sqrt3}(1,1,1)$. The eigenspace $E_1$ is the kernel of $A-\mathbf{1}=J$, that is, the plane with equation $x+y+z=0$, of dimension $2$, equal to the multiplicity of the eigenvalue. To obtain an orthonormal basis, begin with $(1,-1,0)$, which lies in the plane, then seek a vector in the plane orthogonal to it, of the form $(1,1,c)$ with $1+1+c=0$, namely $(1,1,-2)$. We obtain

$$\ket{v_2}=\frac{1}{\sqrt2}\begin{pmatrix}1\\-1\\0\end{pmatrix},\qquad \ket{v_3}=\frac{1}{\sqrt6}\begin{pmatrix}1\\1\\-2\end{pmatrix}.$$

This basis of $E_1$ is not unique: any pair $\cos\alpha\ket{v_2}+\sin\alpha\ket{v_3}$, $-\sin\alpha\ket{v_2}+\cos\alpha\ket{v_3}$ is also suitable, as is, more generally, any orthonormal basis of the plane. Only the eigenspace is determined by the matrix. We verify that $E_1$ is orthogonal to $E_4$, as predicted by the spectral theorem.

We have $P_4=\ket{v_1}\bra{v_1}=\frac13J$. To find $P_1$, we could calculate $\ket{v_2}\bra{v_2}+\ket{v_3}\bra{v_3}$, but it is simpler to use the completeness relation $P_1+P_4=\mathbf{1}$:

$$\begin{aligned} P_1=\mathbf{1}-\frac13J=\frac13\begin{pmatrix}2&-1&-1\\-1&2&-1\\-1&-1&2\end{pmatrix}. \end{aligned}$$

The projector onto a subspace depends only on the subspace, not on the orthonormal basis chosen to calculate it: $P_1$ is the same for every $\alpha$. The spectral decomposition is

$$A=4P_4+1\cdot P_1=\frac43J+\mathbf{1}-\frac13J=\mathbf{1}+J,$$

which is indeed the original matrix. We also verify that $P_4P_1=\frac13J-\frac19J^2=\frac13J-\frac13J=0$: the two projectors are orthogonal.

Take the three orthonormal eigenvectors as the columns of $O$:

$$\begin{aligned} O=\begin{pmatrix}\frac{1}{\sqrt3}&\frac{1}{\sqrt2}&\frac{1}{\sqrt6}\\[2pt] \frac{1}{\sqrt3}&-\frac{1}{\sqrt2}&\frac{1}{\sqrt6}\\[2pt] \frac{1}{\sqrt3}&0&-\frac{2}{\sqrt6}\end{pmatrix}, \qquad O^{\mathsf T}AO=\begin{pmatrix}4&0&0\\0&1&0\\0&0&1\end{pmatrix}. \end{aligned}$$

Since its columns are orthonormal, $O$ is orthogonal, and $O^{\mathsf T}AO$ is the matrix of $A$ in the eigenvector basis, by the change-of-basis formula. The order of the eigenvalues on the diagonal is the order of the columns of $O$.