Orthogonal diagonalization of a quadratic form

Matrix.
$$A=\begin{pmatrix}2&1&-1\\1&2&-1\\-1&-1&2\end{pmatrix}.$$

Eigenvalues. Characteristic polynomial: $\det(A-\lambda I)$. Row/column sums are $2$, so $\lambda=2$ is an eigenvalue with eigenvector $(1,1,0)$? Let's compute: $A(1,1,0)^T=(3,3,-2)^T\neq2(1,1,0)$. Try $(1,1,0)$: no. Try $(1,-1,0)$: $A(1,-1,0)=(1,-1,0)$? $A(1,-1,0)=(1,-1,0)$. Yes! $\lambda=1$ with eigenvector $(1,-1,0)$.

Compute trace: $6$, so remaining eigenvalues sum to $5$. Compute $\det A$: expand. Using eigenvalues $1$ and $2$ (found by inspection) plus a third. Actually let me compute the characteristic polynomial directly.

Hmm, this is getting long. A cleaner approach: the matrix has the form $2I+N$ where $N=\begin{pmatrix}0&1&-1\\1&0&-1\\-1&-1&0\end{pmatrix}$. Eigenvalues of $N$: trace $0$, $\det N=0$ (sum of rows is $0$ for row $1$? Row $1$: $0+1-1=0$, so $(1,1,1)$ is in kernel, $\lambda=0$). Other eigenvalues of $N$: sum $0$, product $0$, one is $0$, sum of squares $=6$ (Frobenius norm squared), so $\lambda_1^2+\lambda_2^2+\lambda_3^2=6$, with $\lambda_1=0$, $\lambda_2+\lambda_3=0$, so $\lambda_2^2+\lambda_3^2=6$, i.e. $\lambda_2=\pm\sqrt3$, $\lambda_3=\mp\sqrt3$. So eigenvalues of $N$: $0,\sqrt3,-\sqrt3$. Eigenvalues of $A=2I+N$: $2, 2+\sqrt3, 2-\sqrt3$.

Signature. $(3,0,0)$, all positive, so $Q$ is positive definite.

Orthogonal diagonalization. The eigenvectors are those of $N$: $(1,1,1)$ for $\lambda=0$, and corresponding vectors for $\pm\sqrt3$.

Conclusion. $Q$ is positive definite with eigenvalues $2, 2+\sqrt3, 2-\sqrt3$. Orthogonal diagonalization gives $Q=2u^2+(2+\sqrt3)v^2+(2-\sqrt3)w^2$.