Finding the real-symmetric matrix whose eigenvalues and corresponding eigenvectors are known I am given that $(Λ, x) = \{1, (0,1,1)\} \cup \{2, (1,-1,1)\} \cup \{3, (-2,-1,1)\}$ are eigenpairs for the real symmetric matrix $A,$ how can I construct this matrix?
 A: Hint: symmetric matrices have orthogonal  Eigen vectors, check for orthogonality and try similar transformation.
A: Write $$D = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \\ \end{array} \right]$$
We are going to think of $D$ as the matrix of a linear operator $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ in regard to the matrix $\beta = \{ (0,1/\sqrt{2},1/\sqrt{2}),(1/\sqrt{3},-1/\sqrt{3},1/\sqrt{3}),(-2/\sqrt{6},-1/\sqrt{6},1/\sqrt{6})\}$, whose elements are the normalized eigenvectors, so $\beta$ is orthonormal in the usual inner product of $\mathbb{R}^3$. This is important, because the canonical basis $\alpha = \{(1,0,0),(0,1,0),(0,0,1)\}$ is also orthonormal, so the change of basis matrix is going to be orthogonal, so its inverse will be its transpose.
If $A$ is the matrix of $T$ in the canonical basis, we have that $$A = [I]_{\alpha}^{\beta} D [I]_{\beta}^{\alpha},$$ by the change of basis formula, where $[I]_{a}^{b}$ is the change of basis matrix from the basis $b$ to the basis $a$. If we call $P = [I]_{\alpha}^{\beta}$, then $$A = P D P^T,$$ since both $\alpha$ and $\beta$ are orthonormal.
After we do these calculations, we'll find out that $$A = \frac{1}{3}\left[ \begin{array}{rrr} 8 & 1 & -1 \\ 1 & 5 & -2 \\ -1 & -2 & 5 \\ \end{array} \right]$$
