Question from Axler's Linear Algebra Done Right Regarding Isometries and Normal Operators

Question

Exercise 7.C.11 reads: "Suppose $T_1,T_2$ are normal operators on $\mathcal{L}(\mathbb{F}^3)$ and both operators have $2,5,7$ as eigenvalues. Prove that there exists an isometry $S\in\mathcal{L}(\mathbb{F}^3)$ such that $T_1=S^*T_2S$."

I am able to prove this for the case that $\mathbb{F}=\mathbb{C}$: The Complex Spectral Theorem gives two orthonormal bases of eigenvectors of $T_1,T_2$, respectively. The eigenvectors in each basis correspond to the eigenvalues $2,5,7$. Then define $S$ to take one of the bases to the other. Then 7.42(d) implies that $S$ is an isometry and 7.42(h) gives $S^{-1}=S^*$. Some computations show that $T_1=S^*T_2S$, as desired.

But my question is this: How can we prove this for the case that $\mathbb{F}=\mathbb{R}$? It doesn't appear that we can use the Real Spectral Theorem here since $T_1,T_2$ may not be self-adjoint.

Answer 1

For $\Bbb F = \Bbb R$, the thing does not vary much. We prove for the general $\Bbb F$.

Proof. $\blacktriangleleft$ Since $\mathcal T_j \,[ j=1,2]$ has $3$ different eigenvalues and $\dim(\Bbb F^3) =3$, $\mathcal T_j$ is always diagonalizable [Thm 5.44], i.e. $\Bbb F^3$ has an eigenbasis $v_{1,j}, v_{2,j}, v_{3,j}$ and they are corresponding to eigenvalues $2,5,7$ respectively, where $j=1,2$. WLOG we could assume that all $v_{k,j}$ are unit vectors in $\Bbb F^3$ [otherwise scale $v_{k,j}$ by $1/\|v_{k,j}\|$ respectively]. Now since $\mathcal T_j$ is normal, so $(v_{k,j})_{k=1}^3$ is a list of orthogonal vectors [Thm.7.22], hence actually $(v_{k,j})_{k=1}^3$ is an orthonormal basis of $\Bbb F^3$ for $j=1,2$ respectively. Thus the linear operator $\mathcal S$ that takes $ v_{k,1}$ to $v_{k,2}$ actually takes the ONB [OrthoNormal Basis] $(v_{k,1})$ to the ONB $(v_{k,2})$, hence $\mathcal S$ is an isometry [Thm. 7.42]. Now since $$ \mathcal S^* \mathcal T_2 \mathcal S v_{k,1} = \mathcal S^* \mathcal T_2 v_{k,2} = \mathcal S^* (c_k v_{k,2} ) = c_k \mathcal S^* v_{k,2} = c_k v_{k,1} = \mathcal T_1 v_{k,1} $$ where $c_1 =2, c_2 = 5, c_3 =7$ and $k=1,2,3$, we have $\mathcal T_1 = \mathcal S^* \mathcal T_2 \mathcal S$ because $(v_{k,1})_1^3$ is a basis of $\Bbb F^3$. $\blacktriangleright$

Answer 2

If the matrix $T$ is real and the eigenvalue $\lambda$ is real, there will be a real eigenvector, which is a member of the null space of the real matrix $T - \lambda I$.