Let $S$ be an operator on $\mathbb{R}^4$ having eigenvectors $$((1 ,1, 1, 1)^T, (1 ,1, -1, -1)^T, (1 ,-1, 1, -1)^T, (1, -1, -1, 1)^T)$$ with corresponding eigenvalues $2$, $3$, $4$, $5$. Let $T$ be the operator on $\mathbb{R}^4$ having eigenvectors $$((1, 1, 1, 1)^T ,(1, 1, -1, -1)^T ,(1, -1, 1, -1)^T, (1, 0 ,0 ,0)^T)$$ with corresponding eigenvalues $2$, $3$, $4$, $5$. Prove that there exists an invertible operator $ā$ such that $ā^{-1}Sā=T$, but it is not possible for $ā$ to be an isometry.
example of a base change operator for two matrices with given eigenvectors which is not an isometry
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linear-algebra