I read through similar questions, but I couldn't find an answer to this:
How do you determine the symmetric matrix A if you know:
$\lambda_1 = 1, \ eigenvector_1 = \pmatrix{1& 0&-1}^T;$
$\lambda_2 = -2, \ eigenvector_2 = \pmatrix{1& 1& 1}^T;$
$\lambda_3 = 2, \ eigenvector_3 = \pmatrix{-1& 2& -1}^T;$
I tried to solve it as an equation system for each line, but it didn't work somehow.
I tried to find the inverse of the eigenvectors, but it brought a wrong matrix.
Do you know how to solve it?
Thanks!
Best Answer
Writing the matrix down in the basis defined by the eigenvalues is trivial. It's just $$ M=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 2 \end{array} \right). $$ Now, all we need is the change of basis matrix to change to the standard coordinate basis, namely: $$ S = \left( \begin{array}{ccc} 1 & 1 & -1 \\ 0 & 1 & 2 \\ -1 & 1 & -1 \\ \end{array} \right). $$ This is just the matrix whose columns are the eigenvectors. We can change to the standard coordinate bases by computing $SMS^{-1}$. We get $$ SMS^{-1} = \frac{1}{6}\left( \begin{array}{ccc} 1 & -8 & -5 \\ -8 & 4 & -8 \\ -5 & -8 & 1 \\ \end{array} \right). $$ You can check that this matrix has the desired eigensystem. For example, $$ \frac{1}{6}\left( \begin{array}{ccc} 1 & -8 & -5 \\ -8 & 4 & -8 \\ -5 & -8 & 1 \\ \end{array} \right) \left( \begin{array}{c} -1 \\ 2 \\ -1 \end{array} \right) = \left( \begin{array}{c} -2 \\ 4 \\ -2 \end{array} \right). $$