# example of a base change operator for two matrices with given eigenvectors which is not an isometry

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 $∏$ such that $∏^{-1}S∏=T$, but it is not possible for $∏$ to be an isometry.

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An isometry is a transformation that preserves length. Linear transformations that preserve length are always given by orthonormal matrices. So it is enough to prove $∏$ isn't orthonormal (orthogonal is not orthonormal) if it exists. Define $D=diag(2,3,4,5)$. $diag(.)$ denotes a diagonal matrix with its arguments as its diagonal entries. Since your matrix have distinct (non-zero) eigenvalues, it is also diagonalizable with respect to this operator.
Define \begin{align} A=\begin{bmatrix} 1 &1&1&1 \\ 1 & 1 & -1& -1 \\ 1 & -1 & 1 & -1 \\ 1 & -1 & -1 & 1 \end{bmatrix} \end{align} This is the set of eigenvectors of your operator $S$
Define \begin{align} B=\begin{bmatrix} 1 &1&1&1 \\ 1 & 1 & -1& 0 \\ 1 & -1 & 1 & 0 \\ 1 & -1 & -1 & 0 \end{bmatrix} \end{align} This is the set of eigenvectors of your operator $T$
Now it should be clear that $S=ADA^{-1}$ and $T=BDB^{-1}$. \begin{align} T &=BDB^{-1} \\ &= B(A^{-1}A)D(A^{-1}A)B^{-1} \\ &=(BA^{-1})ADA^{-1}(AB^{-1}) \\&=(BA^{-1})S(AB^{-1}) \\ &=(BA^{-1})S(BA^{-1})^{-1} \end{align} This implies $∏$ indeed exists and is given by $∏=(BA^{-1})$. To prove $∏$ isn't an isometry, it is enough to prove it isn't orthonormal. It isn't because it is a product of two non-orthogonal matrices which are not inverses of each other. (In this case, this is sufficient).