 Solution

The vector $(0,1,-1)$ is an eigenvector with eigenvalue one. The other two orthogonal directions give eigenvalues two and five, leading toIts columns are an orthonormal eigenbasis, so $V^{\mathsf{T}}V=I$ and $V^{\mathsf{T}}AV=D$. Multiplying by $V$ and $V^{\mathsf{T}}$ gives $\boxed{A=VD{V}^{\mathsf{T}}}$. Eigenvector signs and the joint order of columns and diagonal entries may be changed consistently.