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!
$\endgroup$ 13 Answers
$\begingroup$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,$$ \frac{1}{6}\left( \begin{array}{ccc} 1 & -8 & -5 \\ -8 & 4 & -8 \\ -5 & -8 & 1 \\ \end{array} \right) \left( \begin{array}{c} -1 \\ 2 \\ -1 \end{array} \right) = \left( \begin{array}{c} -2 \\ 4 \\ -2 \end{array} \right). $$
$\endgroup$ 4 $\begingroup$call the eigenvectors $u_1, u_2$ and $u_3$ the eigenvectors corresponding to the eigenvalues $1, -2, $ and $2.$ then $$A = 1\dfrac{u_1u_1^T}{u_1^Tu_1} - 2\dfrac{u_2u_2^T}{u_2^Tu_2} + 2\dfrac{u_3u_3^T}{u_3^Tu_3}$$
you can verify this by computing $Au_1, \cdots$. this expression for $A$ is called the spectral decomposition of a symmetric matrix.
$\endgroup$ 1 $\begingroup$An $n\times n$ matrix with $n$ independent eigenvectors can be expressed as $A=PDP^{-1}$, where $D$ is the diagonal matrix $\operatorname{diag}(\lambda_1\:\lambda_2\:\cdots\lambda_n)$ and $P$ is the matrix $(\vec{v}_1\:|\:\vec{v}_2\:|\cdots|\:\vec{v}_n)$ where $v_i$ is the corresponding eigenvector to $\lambda_i$.
$$D=\begin{pmatrix}1&0&0\\0&-2&0\\0&0&2\end{pmatrix}$$ $$P=\begin{pmatrix}1&0&-1\\1&1&1\\-1&2&-1\end{pmatrix}$$
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