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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!

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    $\begingroup$ Related: math.stackexchange.com/questions/437253/… $\endgroup$ – darij grinberg Jul 24 '18 at 19:57
  • $\begingroup$ Yes, I have seen that list before, but thanks for linking to it. $\endgroup$ – Spiro Karigiannis Jul 24 '18 at 23:37
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    $\begingroup$ My point was that you should probably make a post in that m.se thread as well, at least once this is cleared up. But now I see you've done exactly that :) $\endgroup$ – darij grinberg Jul 25 '18 at 0:54
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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.

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  • $\begingroup$ Yes, that looks right to me, and it also take care of the case $m = \dim W = 2$, which my counterexample didn't cover. Thanks. $\endgroup$ – Spiro Karigiannis Jul 24 '18 at 23:43

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