I'm teaching a second course in advanced linear algebra, following the second half of Hoffman-Kunze. I have come across what I believe to be an error, but I want confirmation (or refutation) by experts. Here is the setup:


Let $V$ be an $n$-dimensional inner product space over either $\mathbb C$ or $\mathbb R$. [Here the inner product is always positive definite.] Let $T$ be a *normal* linear operator on $V$. This means $T$ commutes with its adjoint $T^\*$. In Hoffman-Kunze Section 9.6, page 355, they (correctly) prove that if $W$ is any $T$-invariant subspace of $V$, then it is also $T^\*$-invariant, and consequently the orthogonal complement $W^{\perp}$ is also both $T$-invariant and $T^\*$-invariant. Moreover, the restriction of $T$ to both $W$ and $W^{\perp}$ is again normal. [They don't say it this way, but that is implicit in their proof of Theorem 19.]


Next, they say that "one can now easily prove" a \*strengthened cyclic decomposition theorem", which is Theorem 20 at the bottom of page 355. I didn't find this so easy (perhaps I am missing something), but I did indeed find a proof. The relevant fact here is that there is an *orthogonal* direct sum decomposition $V = Z(v\_1, T) \oplus \cdots \oplus Z(v\_r, T)$ where $Z(v, T) = \mathrm{span} \{ v, Tv, T^2v, \cdots \}$ is the $T$-cyclic subspace of $V$ generated by $v$.


It's the "Corollary" at the top of page 356 that I do not believe. It says:


Corollary. If $A$ is a normal matrix over $\mathbb R$ (respectively, $\mathbb C$) then there exists a real orthogonal (respectively, unitary) matrix $P$ such that $P^{-1} A P = P^\* A P$ is in rational canonical form.


This means that one can find an *orthonormal basis* of $V$ in which the matrix for $T$ is in rational canonical form. Since the $Z(v\_i, T)$'s are orthogonal, this means that we can find an *orthonormal* basis of each $Z(v\_i, T)$ with respect to which the matrix of the restriction $T\_i$ of $T$ to $Z(v\_i, T)$ is the companion matrix for the minimal polynomial $p\_i$ of $T\_i$.


But this seems to be equivalent to saying that, if $S$ is a normal operator on an $m$-dimensional inner product space $W$ that admits a cyclic vector $w$, then we can choose that cyclic vector so that $\{ w, Tw, \ldots, T^{m - 1} w \}$ is orthonormal. It is easy to see that this definitely cannot hold if the field is $\mathbb R$, $m > 2$, and $S$ is self-adjoint, for the following reason. Since $S$ is self-adjoint, it is orthogonally diagonalizable by the spectral theorem. Let $\{ w\_1, \ldots, w\_m \}$ be an orthonormal basis of eigenvectors of $S$, with $S w\_i = \lambda\_i w\_i$. Then the $w$ we seek is of the form $w = c\_1 w\_1 + \cdots + c\_m w\_m$ for some scalars $c\_1, \ldots, c\_m \in \mathbb R$. But then we have


$$\langle w, T^2 w \rangle = \sum\_{i=1}^m \lambda\_i^2 c\_i^2 = \langle Tw, Tw \rangle.$$


But we want the left hand side to be zero and the right hand side to be one, which gives a contradiction.


Does my argument look correct? I can't find a mistake. Does anyone else agree that the Corollary on page 356 of Hoffman-Kunze is false?


Thanks!



Error in Hoffman-Kunze (normal operators on finite-dimensional inner product space with a cyclic vector)
I haven't thought carefully about your argument, but I agree that the corollary must be false. Suppose that $A$ is a normal $2 \times 2$ real matrix. The corollary claims that $A$ is orthogonally similar to a matrix $B$ in rational canonical form. Then $B$ is also normal, and is either diagonal (hence $A$ is symmetric), or else
$$
B = \begin{bmatrix} 0 & a \\ 1 & b \end{bmatrix}
$$
for some $a, b$. Normality of $B$ implies that $a = \pm 1$, so $\det A = \det B = \pm 1$. 


So the Corollary entails that every nonsymmetric real normal $2 \times 2$ matrix has determinant $\pm 1$. But $\begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$ is a counterexample.


