I am self-studying out of Hoffman and Kunze's Linear Algebra. In the section on cyclic decomposition and the rational form, exercise 3 asks:

Let $T$ be the linear operator on $R^3$ which is represented in the standard ordered basis by the matrix


Let $W$ be the null space of $T - 2I$. Prove that $W$ has no complementary $T$-invariant subspace.

A hint is given:

Hint: Let $\beta = \epsilon_1$ and observe that $(T - 2I)\beta$ is in $W$. Prove there is no $\alpha$ in $W$ with $(T - 2I)\beta = (T - 2I)\alpha$.

Following the suggestions, $(T - 2I)\beta = (0,1,0)$. Now, $W$ must be one-dimensional by rank-nullity, and $null(T-2I) = span\{(0,1,0)\}$; since $\beta \in W$, there is no $\alpha \in W$ such that $(T - 2I)\alpha = (T - 2I)\beta$, since $\alpha \in W \implies (T-2I)\alpha = 0 \neq (0,1,0) = (T - 2I)\beta$.

However, I do not see how to proceed from this observation to the conclusion that $W$ has no complementary, $T$-invariant subspace. In particular, I do not see how to use the cyclic decomposition theorem; indeed, it seems that if I could apply it, it would, in fact, yield a complementary $T$-admissible subspace.

I observe that $\beta$ is in the null space of $(T-2I)^2$, but do not see how to make use of this either, unless I am to find the primary decomposition, but this does not seem to help and seems out of place for an exercise in the section on cyclic decomposition.

What is the correct direction here?

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    $\begingroup$ can you please remind me what a complementary subspace is? $\endgroup$ – abel Jan 3 '15 at 16:53
  • $\begingroup$ It's a subspace whose intersection with the given subspace is $\{0\}$. But it is $(T-2I)$ which sends $\beta$ into $W$, not $T$, since $\beta \in null(T-2I)^2 \supset null(T-2I) = W$. $\endgroup$ – user169845 Jan 3 '15 at 16:56
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    $\begingroup$ why is not $span\{(1,0,0)^T, (0,0,1)^T\}$ a complementary subspace of $W = span\{(0,1,0)^T\}?$ $\endgroup$ – abel Jan 3 '15 at 17:08
  • $\begingroup$ It is, but it is not a $T$-invariant one. $T(1,0,0) = (2,1,0) \not \in span\{(1,0,0),(0,0,1)\}$... oh. Then why this business with $T-2I$ and $\beta$? $\endgroup$ – user169845 Jan 3 '15 at 17:11
  • $\begingroup$ what has invariance got to do with the subspace being complementary? are you adding more constraint that $T$ restricted to the complementary subspace should not have $2$ for an eigenvalue? $\endgroup$ – abel Jan 3 '15 at 17:18

If $U$ is a $T$-stable complementary subspace to $W=\ker(T-2I)$ then one can write $\epsilon_1=u+w$ with $u\in U$ and $w\in W$. Then $\epsilon_2=(T-2I)(\epsilon_1)=(T-2I)(u+w)=(T-2I)(u)$, which is in$~U$ by $T$-stability of the latter, but also $\epsilon_2\in W$. This is in contradiction with $U\cap W=\{0\}$.

This tries to use (part of) the given hint. However one can also reason as follows. If $U$ were a $T$-stable complementary subspace of $\ker(T-2I)$, then the restriction of $T-2I$ to $U$ is well defined and injective (since its kernel is $\ker(T-2I)\cap U=\{0\}$). But then writing an arbitrary vector $v=u+w$ one sees that $(T-2I)^2(v)=0$ implies $u=0$ and therefore already $(T-2I)(v)=0$, which contradicts the fact that there are vectors in the kernel of $(T-2I)^2$ that are not in the kernel of $T-2I$ (for instance $\epsilon_1$ is such a vector).

  • $\begingroup$ That makes complete sense. Thank you. $\endgroup$ – user169845 Jan 3 '15 at 19:47

At abel's guidance in comments on the original question, the answer is simple. Observe that $W = span\{(0,1,0)\}$. Then, a complementary subspace of $W$ would have to be spanned by $(1,0,0)$ and $(0,0,1)$. However, $T(1,0,0) = (2,1,0)$, which is not spanned by those vectors, and so the complementary subspace of $W$ is not $T$-invariant.

I'm not sure what the hint was there for.

This is incorrect as per the below comment. See Marc's answer.

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    $\begingroup$ This is not correct. There are (infinitely) many complementary subspaces of $W$, you cannot assume the complementary subsapce has to be the one spanned by $(1,0,0)$ and $(0,0,1)$. $\endgroup$ – Marc van Leeuwen Jan 3 '15 at 19:25

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