Complexification

Vector Space

Suppose $V$ is a real vector space.

The complexification of $V$, denoted by $V_{\mathbb{C}}$, equals $V \times V$. An element of $V_{\mathbb{C}}$ is an ordered pair $(u,v)$, where $u,v \in V$, but we write this as $u + iv$.

• Addition on $V_{\mathbb{C}}$ is defined by

\[(u_1 + iv_1) + (u_2 + iv_2) = (u_1 + u_2) + i(v_1 + v_2)\]

for all $u_1,v_1,u_2,v_2 \in V$.

• Complex scalar multiplication on $V_{\mathbb{C}}$ is defined by

\[(a + bi)(u + iv) = (au - bv) + i(av + bu)\]

for all $a,b \in \mathbb{R}$ and all $u,v \in V$.

With the definitions of addition and scalar multiplication as above, $V_{\mathbb{C}}$ is a complex vector space.

The notation $u + iv$ is suggestive but the two slots are just the two components of an ordered pair. It pays to keep both readings available:

• Addition acts slot by slot: $(u_1,v_1) + (u_2,v_2) = (u_1+u_2,\, v_1+v_2)$.
• Multiplication by a real $a$ also acts slot by slot: $a(u,v) = (au, av)$.
• Multiplication by $i$ mixes the slots: $i(u,v) = (-v, u)$.

If $v_1, \dots, v_n$ is a basis of $V$, then $v_1, \dots, v_n$ is also a basis of the complexification $V_{\mathbb{C}}$.

Here $v_k$ is read as $v_k + i0$, and “basis” means basis over $\mathbb{C}$.

Spanning. Take any $u + iv \in V_{\mathbb{C}}$. Since $v_1,\dots,v_n$ is a basis of $V$, write $u = \sum a_k v_k$ and $v = \sum b_k v_k$ with $a_k, b_k \in \mathbb{R}$. Then

\[u + iv = \sum_{k=1}^n a_k v_k + i\sum_{k=1}^n b_k v_k = \sum_{k=1}^n (a_k + b_k i)\, v_k\]

The real and imaginary parts get absorbed into the complex coefficients.

Independence. Suppose $\sum (a_k + b_k i) v_k = 0$ in $V_{\mathbb{C}}$. Unpacking, that’s the pair $\left(\sum a_k v_k, \sum b_k v_k\right) = (0,0)$. Both components vanish in $V$, so real independence of $v_1,\dots,v_n$ forces every $a_k = 0$ and every $b_k = 0$, hence each complex coefficient is $0$. $\blacksquare$

Nothing is “missing” from the second slot: the scalar $i$ supplies it. Concretely $i(v_k + i0) = 0 + iv_k$, so $(0, v_k)$ already lies in the complex span of $v_k$. This is exactly the step that fails over $\mathbb{R}$ — see the next box.

• $\dim_{\mathbb{C}} V_{\mathbb{C}} = \dim_{\mathbb{R}} V$
• $\dim_{\mathbb{R}} V_{\mathbb{C}} = 2\dim_{\mathbb{R}} V$, with real basis $v_1, \dots, v_n, iv_1, \dots, iv_n$.

The same set $V \times V$ has dimension $n$ or $2n$ depending on which field one counts over.

$V_{\mathbb{C}} = V \oplus iV$, where $V$ is identified with $\{u + i0 : u \in V\}$ and $iV = \{0 + iv : v \in V\}$.

This is a direct sum of real subspaces only: neither summand is closed under multiplication by $i$, which swaps them. An element of $V \times V$ is $0$ exactly when both components are.

Linear Map

Suppose that $V$ and $W$ are real vector spaces and $T \in \mathcal{L}(V,W)$. Define $T_{\mathbb{C}}: V_{\mathbb{C}} \to W_{\mathbb{C}}$ by

\[T_{\mathbb{C}}(u + iv) = Tu + iTv\]

for all $u,v \in V$. In slot form: $T_{\mathbb{C}}(u,v) = (Tu, Tv)$.

$T_{\mathbb{C}} \in \mathcal{L}(V_{\mathbb{C}}, W_{\mathbb{C}})$; that is, $T_{\mathbb{C}}$ is linear over $\mathbb{C}$, not merely over $\mathbb{R}$.

Additivity is immediate from slot-by-slot application of the additivity of $T$.

Homogeneity needs the mixing scalar $i$ checked, so take $\alpha = a + bi$ with $a,b \in \mathbb{R}$:

\[T_{\mathbb{C}}\big(\alpha(u + iv)\big) = T_{\mathbb{C}}\big((au - bv) + i(av + bu)\big) = (aTu - bTv) + i(aTv + bTu),\] \[\alpha\, T_{\mathbb{C}}(u+iv) = (a + bi)(Tu + iTv) = (aTu - bTv) + i(aTv + bTu).\]

These agree, using only that $T$ is real-linear. $\blacksquare$

Suppose $v_1, \dots, v_n$ is a basis of $V$ and $w_1, \dots, w_m$ is a basis of $W$. By the basis theorem, these same lists are bases of $V_{\mathbb{C}}$ and $W_{\mathbb{C}}$ over $\mathbb{C}$, so both matrices below are defined:

\[\mathcal{M}\big(T_{\mathbb{C}}, (v_1, \dots, v_n), (w_1, \dots, w_m)\big) = \mathcal{M}\big(T, (v_1, \dots, v_n), (w_1, \dots, w_m)\big).\]

In particular $\mathcal{M}(T_{\mathbb{C}})$ has real entries with respect to such a basis.

Fix $k$. By definition of the matrix of $T$, the entries $A_{1k}, \dots, A_{mk}$ in the $k$-th column of $\mathcal{M}(T)$ are the unique real scalars with

\[Tv_k = \sum_{j=1}^m A_{jk}\, w_j. \tag{1}\]

By definition of the matrix of $T_{\mathbb{C}}$, the entries $B_{1k}, \dots, B_{mk}$ in the $k$-th column of $\mathcal{M}(T_{\mathbb{C}})$ are the unique complex scalars with

\[T_{\mathbb{C}}v_k = \sum_{j=1}^m B_{jk}\, w_j. \tag{2}\]

Here $v_k$ means $v_k + i0$, and the definition of $T_{\mathbb{C}}$ gives

\[T_{\mathbb{C}}(v_k + i0) = Tv_k + i0,\]

so (1) and (2) are two expansions of the same vector of $W_{\mathbb{C}}$ in the basis $w_1, \dots, w_m$. Equation (1) qualifies as an expansion over $\mathbb{C}$, since every real number is a complex number. Expansions in a basis are unique, so

\[B_{jk} = A_{jk} \quad \text{for all } j.\]

As $k$ was arbitrary, the two matrices agree column by column. $\blacksquare$

Polynomial

Lemma. Suppose $V$ is a real vector space, $T \in \mathcal{L}(V)$, and $p \in \mathcal{P}(\mathbb{R})$. Then

\[p(T_{\mathbb{C}}) = (p(T))_{\mathbb{C}}.\]

Complexification preserves sums, products, and multiplication by real scalars. A polynomial in $T$ is built from exactly those three operations, so the whole construction passes through.

Throughout, let $S, T \in \mathcal{L}(V)$ and let $u, v \in V$ be arbitrary; recall the definition $T_{\mathbb{C}}(u + iv) = Tu + i\,Tv$ (and likewise for $S$).

Sums.

\[(S+T)_{\mathbb{C}}(u+iv) = (S+T)u + i(S+T)v = (Su + i\,Sv) + (Tu + i\,Tv) = S_{\mathbb{C}}(u+iv) + T_{\mathbb{C}}(u+iv).\]

Products.

\[(ST)_{\mathbb{C}}(u+iv) = (ST)u + i\,(ST)v = S(Tu) + i\,S(Tv) = S_{\mathbb{C}}(Tu + i\,Tv) = S_{\mathbb{C}}T_{\mathbb{C}}(u+iv).\]

Induction on $k$ then gives \((T^k)_{\mathbb{C}} = (T_{\mathbb{C}})^k\), the base case being \((T^0)_{\mathbb{C}} = I_{\mathbb{C}} = I\) on $V_{\mathbb{C}}$.

Real scalars. For $a \in \mathbb{R}$:

\[(aT)_{\mathbb{C}}(u+iv) = a\,Tu + i\,a\,Tv = a(Tu + i\,Tv) = a\,T_{\mathbb{C}}(u+iv).\]

Assembly. Write $p(z) = \sum_{k=0}^m a_k z^k$ with every $a_k \in \mathbb{R}$. Then

\[p(T_{\mathbb{C}}) = \sum_k a_k (T_{\mathbb{C}})^k = \sum_k a_k (T^k)_{\mathbb{C}} = \sum_k (a_k T^k)_{\mathbb{C}} = \Big(\sum_k a_k T^k\Big)_{\mathbb{C}} = (p(T))_{\mathbb{C}}. \qquad \blacksquare\]

The hypothesis $p \in \mathcal{P}(\mathbb{R})$ is not a technicality — without it the right-hand side is undefined. Evaluating $p(T)$ means forming $\sum a_k T^k$ inside $\mathcal{L}(V)$, a vector space over $\mathbb{R}$; there is no operator “$iT$” on a real space. In the proof, real-ness is used exactly once, at the step \(a_k (T^k)_{\mathbb{C}} = (a_k T^k)_{\mathbb{C}}.\)

For a general $p \in \mathcal{P}(\mathbb{C})$, split $p = g + ih$ with $g, h$ real. The Lemma applies to each part, giving

\[p(T_{\mathbb{C}}) = (g(T))_{\mathbb{C}} + i\,(h(T))_{\mathbb{C}},\]

and this is as far as one can go: the sum is not $(\cdot)_{\mathbb{C}}$ of anything. Test it on $u + i0$. The first term returns $g(T)u + i0$, but the second returns $0 + i\,h(T)u$, which has first component $0$. A complexification always sends $u + i0$ to something of the form $(\text{vector}) + i0$, so unless $h(T) = 0$ the sum is not one.

Theorem. Suppose $\mathbf{F} = \mathbb{R}$, $V$ is finite-dimensional, and $T \in \mathcal{L}(V)$. Then the minimal polynomial of $T_{\mathbb{C}}$ equals the minimal polynomial of $T$.

Write $p = p_T$ and $q = p_{T_{\mathbb{C}}}$, and set $n = \deg p$, $m = \deg q$.

Note the asymmetry in what we may assume. Since $V$ is a vector space over $\mathbb{R}$, the definition of the minimal polynomial forces $p \in \mathcal{P}(\mathbb{R})$. But $V_{\mathbb{C}}$ is a vector space over $\mathbb{C}$, so $q$ is drawn from the strictly larger pool $\mathcal{P}(\mathbb{C})$ and may a priori have complex coefficients. The content of the theorem is that this larger pool buys nothing.

Step 1 ($q \mid p$, hence $m \le n$).

$p$ has real coefficients, so the Lemma applies: \(p(T_{\mathbb{C}}) = (p(T))_{\mathbb{C}} = 0_{\mathbb{C}} = 0.\) Thus $p$ annihilates $T_{\mathbb{C}}$, and since the minimal polynomial divides every annihilating polynomial, $q \mid p$. In particular $m \le n$.

Step 2 (split $q$ into real and imaginary parts).

The Lemma cannot be applied to $q$ directly, since $q$ need not be real. So write each coefficient $c_k = a_k + i b_k$ with $a_k, b_k \in \mathbb{R}$ and regroup:

\[q(z) = \sum_{k=0}^m c_k z^k = g(z) + i\,h(z), \qquad g(z) = \sum_k a_k z^k, \quad h(z) = \sum_k b_k z^k,\]

so that $g, h \in \mathcal{P}(\mathbb{R})$. Now the Lemma does apply, to each piece separately:

\[0 = q(T_{\mathbb{C}}) = g(T_{\mathbb{C}}) + i\,h(T_{\mathbb{C}}) = (g(T))_{\mathbb{C}} + i\,(h(T))_{\mathbb{C}}.\]

Step 3 ($g(T) = 0$ and $h(T) = 0$).

Apply the operator identity above to an arbitrary $u + i0 \in V_{\mathbb{C}}$. Since \((g(T))_{\mathbb{C}}(u + i0) = g(T)u + i0\) and \(i\,(h(T))_{\mathbb{C}}(u + i0) = 0 + i\,h(T)u,\) we get

\[0 = g(T)u + i\,h(T)u, \qquad \text{i.e. the ordered pair} \quad \big(g(T)u,\; h(T)u\big) = (0,0).\]

An element of $V_{\mathbb{C}} = V \times V$ is zero exactly when both components are, so $g(T)u = 0$ and $h(T)u = 0$. As $u \in V$ was arbitrary, $g(T) = 0$ and $h(T) = 0$.

Step 4 (locate the degrees of $g$ and $h$).

By construction $\deg g \le m$ and $\deg h \le m$, since both were assembled from the coefficients $c_0, \dots, c_m$ of $q$. Everything sharper follows from the single top coefficient. Since $q$ is monic of degree $m$,

\[c_m = 1 = 1 + i \cdot 0,\]

and the decomposition of a complex number into real and imaginary parts is unique, so $a_m = 1$ and $b_m = 0$. Hence:

• the degree-$m$ coefficient of $g$ is $a_m = 1 \neq 0$, so $\deg g = m$ and $g$ is monic;
• the degree-$m$ coefficient of $h$ is $b_m = 0$, so $\deg h \le m - 1 < m$ (this includes $h = 0$, under the convention $\deg 0 = -\infty$).

The asymmetry is the point: $a_m \ne 0$ pins $g$ at degree $m$, while $b_m = 0$ pushes $h$ below it. The strictness is used in Step 5.

Step 5 (conclusion).

By Steps 3 and 4, $g$ is a monic polynomial in $\mathcal{P}(\mathbb{R})$ with $g(T) = 0$ — that is, $g$ is a competitor to $p$ in the very contest that defines $p$. Minimality of $p$ therefore gives $n \le \deg g = m$. Combined with Step 1,

\[n = m = \deg g.\]

So $g$ is a monic real annihilator of $T$ of minimal degree, and the minimal polynomial is the unique such polynomial, whence $g = p$.

Finally $h = 0$: otherwise, dividing $h$ by its leading coefficient would give a monic real annihilator of $T$ of degree $\deg h < m = n$, contradicting minimality of $p$.

Therefore $q = g + i\,h = g = p. \qquad \blacksquare$

Corollary. The minimal polynomial of $T_{\mathbb{C}}$ has real coefficients.

Eigenvalue

Both results below rest on the standard characterization: for $\lambda \in \mathbf{F}$, $\lambda$ is an eigenvalue of an operator iff $\lambda$ is a zero of its minimal polynomial.

Suppose $ \mathbf{F} = \mathbb{R} $ and $ \lambda \in \mathbb{R} $. Then

$\lambda$ is an eigenvalue of $T \iff \lambda$ is an eigenvalue of the complexification $ T_{\mathbb{C}}. $

Let $p$ be the minimal polynomial of $T$. By the theorem above, $p$ is also the minimal polynomial of $T_\mathbb{C}$. Then for $\lambda \in \mathbb{R}$:

\[\lambda \text{ is an eigenvalue of } T \iff p(\lambda) = 0 \iff \lambda \text{ is an eigenvalue of } T_\mathbb{C},\]

the first equivalence by the characterization applied to $T$ over $\mathbb{R}$, the second by the same characterization applied to $T_\mathbb{C}$ over $\mathbb{C}$ — legitimate because the two operators have the same polynomial $p$. $\blacksquare$

Directly. ($\Rightarrow$) If $Tv = \lambda v$ with $v \neq 0$, then

\[T_\mathbb{C}(v + i0) = Tv + i0 = \lambda v + i0 = \lambda(v + i0),\]

the last step because $\lambda$ is real and so scales slots; and $v + i0 \neq 0$.

($\Leftarrow$) If $T_\mathbb{C}(u + iv) = \lambda(u+iv)$ with $u + iv \neq 0$, comparing slots gives $Tu = \lambda u$ and $Tv = \lambda v$. At least one of $u,v$ is nonzero, so $\lambda$ is an eigenvalue of $T$. $\blacksquare$

The characterization requires $\lambda \in \mathbf{F}$: the scalar must belong to the field of the space. Since $\mathbb{R} \subseteq \mathbb{C}$, a real $\lambda$ is eligible on both sides, which is why the hypothesis $\lambda \in \mathbb{R}$ appears. For non-real $\lambda$ the right-hand side still makes sense but the left does not — there is no equation $Tv = \lambda v$ to write down, since $\lambda v$ is undefined for $v \in V$. So the non-real case needs a different statement entirely.

Suppose $ \mathbf{F} = \mathbb{R} $ and $ \lambda \in \mathbb{C} $. Then

$\lambda$ is an eigenvalue of $T_{\mathbb{C}} \iff \bar{\lambda} $ is an eigenvalue of $ T_{\mathbb{C}} $.

Via the minimal polynomial. Let $p$ be the minimal polynomial of $T$, which is also that of $T_\mathbb{C}$; crucially $p \in \mathcal{P}(\mathbb{R})$. Then

\[\lambda \text{ eigenvalue of } T_\mathbb{C} \iff p(\lambda) = 0 \iff \overline{p(\lambda)} = 0 \iff p(\bar\lambda) = 0 \iff \bar\lambda \text{ eigenvalue of } T_\mathbb{C}.\]

The only nontrivial step is $\overline{p(\lambda)} = p(\bar\lambda)$, and that is exactly where real coefficients enter. Conjugation preserves sums and products, so with $p(z) = \sum a_k z^k$,

\[\overline{p(\lambda)} = \sum \overline{a_k}\,\overline{\lambda}^k = \sum a_k \bar\lambda^{\,k} = p(\bar\lambda),\]

using $\overline{a_k} = a_k$. Had $p$ carried a genuinely complex coefficient this would fail, and the spectrum would have no reason to be conjugate-symmetric. $\blacksquare$

Via conjugation. Define $C : V_{\mathbb{C}} \to V_{\mathbb{C}}$ by

\[C(u + iv) = u - iv.\]

Three properties:

• $C$ is an involution, hence a bijection: $C\big(C(u+iv)\big) = C(u-iv) = u+iv$, so $C^{-1} = C$. In particular $C$ sends nonzero vectors to nonzero vectors.
• $C$ is conjugate-linear: $C(w_1 + w_2) = Cw_1 + Cw_2$ and $C(\alpha w) = \bar\alpha\, Cw$ for $\alpha \in \mathbb{C}$. For the second, with $\alpha = a+bi$ and $w = u+iv$, both sides equal $(au - bv) + i\big(-(av+bu)\big)$ by the scalar-multiplication rule. (So $C$ is real-linear but not complex-linear.)
• $C$ commutes with $T_{\mathbb{C}}$:

\[T_{\mathbb{C}}\big(C(u+iv)\big) = T_{\mathbb{C}}(u - iv) = Tu - iTv = C(Tu + iTv) = C\big(T_{\mathbb{C}}(u+iv)\big).\]

Now suppose $T_{\mathbb{C}}w = \lambda w$ with $w \neq 0$. Apply $C$ to both sides: on the left the commuting relation gives $C(T_{\mathbb{C}}w) = T_{\mathbb{C}}(Cw)$, and on the right conjugate-linearity gives $C(\lambda w) = \bar\lambda\,Cw$. Hence

\[T_{\mathbb{C}}(Cw) = \bar\lambda\,(Cw),\]

with $Cw \neq 0$ since $C$ is a bijection. So $\bar\lambda$ is an eigenvalue of $T_{\mathbb{C}}$.

The converse needs no new work: apply this to $\bar\lambda$ in place of $\lambda$ and use $\overline{\bar\lambda} = \lambda$. $\blacksquare$

Lemma If $U$ is a complex subspace of $V_{\mathbb{C}}$, then $\dim C(U) = \dim U$.

Let $x_1, \dots, x_d$ be a basis of $U$; we claim $Cx_1, \dots, Cx_d$ is a basis of $C(U)$.

Spanning. Any element of $C(U)$ is $Cx$ with $x = \sum \alpha_k x_k \in U$, and conjugate-linearity gives $Cx = \sum \overline{\alpha_k}\, Cx_k$.

Independence. Suppose $\sum \beta_k\, Cx_k = 0$. Writing $\beta_k = \overline{\overline{\beta_k}}$ and pulling the scalars back through $C$, this says

\[C\Big(\sum \overline{\beta_k}\, x_k\Big) = 0.\]

Since $C$ is injective and $C0 = 0$, we get $\sum \overline{\beta_k}\, x_k = 0$, so each $\overline{\beta_k} = 0$ by independence of $x_1, \dots, x_d$, hence each $\beta_k = 0$. $\blacksquare$

The second proof gives more than the statement: $C$ restricts to a bijection

\[C\big(E(\lambda, T_{\mathbb{C}})\big) = E(\bar\lambda, T_{\mathbb{C}}),\]

and a conjugate-linear bijection carries a complex basis to a complex basis. Hence

\[\dim E(\lambda, T_{\mathbb{C}}) = \dim E(\bar\lambda, T_{\mathbb{C}}).\]

The same argument upgrades to generalized eigenspaces. From conjugate-linearity and commuting, $C(T_{\mathbb{C}} - \lambda I) = (T_{\mathbb{C}} - \bar\lambda I)C$, so by induction \(C(T_{\mathbb{C}} - \lambda I)^k = (T_{\mathbb{C}} - \bar\lambda I)^k C\) for every $k$. Therefore

\[C\big(G(\lambda, T_{\mathbb{C}})\big) = G(\bar\lambda, T_{\mathbb{C}}), \qquad \dim G(\lambda, T_{\mathbb{C}}) = \dim G(\bar\lambda, T_{\mathbb{C}}).\]

By definition \(G(\lambda, T_{\mathbb{C}}) = \operatorname{null}(T_{\mathbb{C}} - \lambda I)^{n}\) with $n = \dim_{\mathbb{C}} V_{\mathbb{C}}$. If $w \in \operatorname{null}(T_{\mathbb{C}} - \lambda I)^{n}$, then

\[(T_{\mathbb{C}} - \bar\lambda I)^{n}(Cw) = C\big((T_{\mathbb{C}} - \lambda I)^{n} w\big) = C(0) = 0,\]

so $Cw \in \operatorname{null}(T_{\mathbb{C}} - \bar\lambda I)^{n}$. That is,

\[C\big(G(\lambda, T_{\mathbb{C}})\big) \subseteq G(\bar\lambda, T_{\mathbb{C}}).\]

The same argument with $\bar\lambda$ in place of $\lambda$ gives

\[C\big(G(\bar\lambda, T_{\mathbb{C}})\big) \subseteq G(\lambda, T_{\mathbb{C}}).\]

Apply $C$ to both sides of this second inclusion. Since $C \circ C = I$ and $\overline{\bar\lambda} = \lambda$, it becomes

\[G(\bar\lambda, T_{\mathbb{C}}) \subseteq C\big(G(\lambda, T_{\mathbb{C}})\big),\]

which is the reverse of the first inclusion. Hence

\[C\big(G(\lambda, T_{\mathbb{C}})\big) = G(\bar\lambda, T_{\mathbb{C}}).\]

A generalized eigenspace is the null space of a complex-linear operator, so it is a complex subspace, and the box above on $\dim C(U)$ applies to it:

\[\dim G(\lambda, T_{\mathbb{C}}) = \dim G(\bar\lambda, T_{\mathbb{C}}).\]

Non-real eigenvalues of a real operator come in conjugate pairs with equal multiplicities, geometric and algebraic.

flowchart LR
    subgraph CX["Eigenvalues of T_C  (scalars in C)"]
        direction TB
        A["real λ"]
        B["non-real λ"]
        C["conjugate λ-bar"]
        B <== "conjugation" ==> C
    end

    subgraph RE["Eigenvalues of T  (scalars in R)"]
        direction TB
        D["λ"]
        E["nothing"]
    end

    A <== "same scalar" ==> D
    B -.-> E
    C -.-> E

    style E fill:none,stroke-dasharray:4 4

Application: odd dimension forces a real eigenvalue

Suppose $V$ is a real vector space with $\dim V$ odd and $T \in \mathcal{L}(V)$. Then $T$ has an eigenvalue.

Let $n = \dim_{\mathbb{R}} V$, so $\dim_{\mathbb{C}} V_{\mathbb{C}} = n$ is odd, and let $\lambda_1, \dots, \lambda_r$ be the distinct eigenvalues of $T_{\mathbb{C}}$ (there is at least one, since $V_{\mathbb{C}}$ is a nonzero complex vector space).

Decompose $V_{\mathbb{C}}$ into generalized eigenspaces:

\[V_{\mathbb{C}} = G(\lambda_1, T_{\mathbb{C}}) \oplus \cdots \oplus G(\lambda_r, T_{\mathbb{C}}), \qquad \sum_{j=1}^r d_j = n, \quad d_j := \dim G(\lambda_j, T_{\mathbb{C}}).\]

Split the eigenvalues into the real ones and the non-real ones. By the previous box, conjugation permutes the eigenvalues and preserves the $d_j$; and it has no fixed points among the non-real ones, since $\lambda = \bar\lambda$ forces $\lambda$ real. So the non-real eigenvalues fall into genuine two-element pairs ${\lambda, \bar\lambda}$, each contributing $d + d = 2d$ to the total. Hence

\[n = \underbrace{\sum_{\lambda_j \in \mathbb{R}} d_j}_{=:N} \;+\; \underbrace{\sum_{\text{pairs}} 2d}_{\text{even}}.\]

Since $n$ is odd, $N$ is odd, so $N \neq 0$ and $T_{\mathbb{C}}$ has at least one real eigenvalue $\lambda$. By the first eigenvalue result, that $\lambda$ is an eigenvalue of $T$. $\blacksquare$

The parity is used only to rule out $N = 0$: the pairing makes the non-real contribution even, so an odd total cannot be accounted for by non-real eigenvalues alone. For even $n$ the argument collapses, as it must — rotation by $90°$ on $\mathbb{R}^2$ has eigenvalues $i, -i$, one conjugate pair, and no real eigenvalue.

The count must be by multiplicity. It is tempting to argue “an odd-dimensional space cannot have its eigenvalues occur solely in conjugate pairs”, but the number of distinct eigenvalues has no bearing on parity of the dimension — an operator on a $3$-dimensional space may have exactly two distinct eigenvalues. Ordinary eigenspaces are also not enough, since they need not fill $V_{\mathbb{C}}$ when $T_{\mathbb{C}}$ is not diagonalizable. Generalized eigenspaces are what make the sum come out to exactly $n$.