Exercise 9.A.18

Proof. The proof that (a) $\Rightarrow$ (b) below is basically a rewrite of the proof 5.27 with a few tweakings.

Suppose (a) holds. We will prove (b) by induction on $\dim <!--FUNCTION\; APPLICATION-->V$. First, if $\dim <!--FUNCTION\; APPLICATION-->V=1$, then the matrix of $T$ with respect to any basis only has one entry and thus (b) trivially holds. Assume that $\dim <!--FUNCTION\; APPLICATION-->V>1$ and that (a) implies (b) for all vector spaces of lower dimension.

Let $\lambda \in ℝ$ be an eigenvalue of $T$, which exists by (a) and 9.11. Define

Because $\dim <!--FUNCTION\; APPLICATION-->\mathrm{null}<!--FUNCTION\; APPLICATION-->(T-\lambda I)\ge 1$, the Fundamental Theorem of Linear Maps implies that $\dim <!--FUNCTION\; APPLICATION-->U<\dim <!--FUNCTION\; APPLICATION-->V$. Futhermore, $U$ is invariant under $T$ and so $U_{}$

C$isinvariantunder$(T|_U)_C$.Alltheeigenvaluesof$(T|_U)_C$arereal.Toseethis,suppose$$\lambda \in ℂ$ is an eigenvalue of ${(T{|}_U)}_{}$

C$and$u_1 + iu_2$acorrespondingeigenvectorforsome$u_1, u_2 $\in U$. Then

C(u_1 + iu_2) = (T|_U)u_1 + i(T|_U)u_2 = Tu_1 + iTu_2 = T_C(u_1 + iu_2),

which implies that $\lambda$ is an eigenvalue of $T_{}$

C$andthusmustbereal.Bytheinductionhypothesisthereexistsabasis$u_1, …, u_m$of$U$withrespecttowhichthematrixof$T|_U$isuppertriangular.From5.26,wegetthat$

Extend it to a basis $u_1,\dots ,u_m,v_1,\dots ,v_n$ of $V$. For each $k$ we have

The definition of $U$ shows that $(T-\lambda I)v_k\in U$. Hence, the equation above shows that

Using $(4)$, $(5)$ and 5.26, we conclude that the matrix of $T$ with respect to basis $u_1,\dots ,u_m,v_1,\dots ,v_n$ of $V$ is upper triangular.

9.7 and 5.32 show that (b) implies (a). Hence the equivalence between (a) and (b) is established. Now it suffices to show that (a) and (c) are equivalent.

Suppose (a) holds. Let $\lambda \in ℝ$ be an eigenvalue of $T_{}$

C$andlet$d$denoteitsmultiplicity.Let$u_1 + iv_1, …, u_d + iv_d$beabasisof$G($\lambda ,T_{}$

C)$.Foreach$j = 1, …, d$,wehave$

C(u_j + iv_j) = (T_C - $\lambda {)}^{\dim <!--FUNCTION\; APPLICATION-->V}(u_j+i{v}_j)=0.$$$

This implies that $u_1,v_1,\dots ,u_d,v_d\in G(\lambda ,T)$. Since $G(\lambda ,T_{}$

C)$iscontainedinthecomplexspanofthislist,itfollowsthatthedimensionofthiscomplexspanisatleast$d$.9.4(b)nowimpliesthatthedimensionoftherealspanof$u_1, v_1, …, u_d, v_d$isatleast$d$.Hence$$dimG(\lambda ,T)\ge d$. Since the sum of multiplicities of the generalized eigenspaces of $T$ is at most $\dim <!--FUNCTION\; APPLICATION-->V$, it follows that the only way the dimensions will fit is if $\dim <!--FUNCTION\; APPLICATION-->G(\lambda ,T)=d$.

Thus, summing all the generalized eigenspaces of $T$ we get a subspace of $V$ with dimension $\dim <!--FUNCTION\; APPLICATION-->V$, namely $V$. Taking a basis of generalized eigenspace and putting this bases together gives a basis consisting of generalized eigenvectors of $V$. Thus (c) holds.

Now suppose (c) holds. Then $T_{}$

C$hasrealeigenvalues,whosesumofmultiplicitiesalreadyequal$$dimV$. Hence $T_{}$

C$cannothaveanonrealeigenvalue.Thus(a)holds.\square$