Complex and Repeated Eigenvalues

We prove (A) and (B) separately, each by direct verification plus a linear-independence argument. The technique throughout is substitution and matching real/imaginary or polynomial-in-$t$ parts, justified by uniqueness of representation in $\mathbb{C}=\mathbb{R}\oplus i\mathbb{R}$ and in the space of exponential-polynomial functions.

Part (A): existence of the conjugate eigenpair. Since $A$ is real, its characteristic polynomial $p(\lambda)=\det(A-\lambda I)$ has real coefficients. If $p(\lambda)=0$ then, taking complex conjugates of the equation $p(\lambda)=0$ (valid since conjugation is a ring homomorphism on $\mathbb C$ and coefficients are real, so $\overline{p(\lambda)} = p(\bar\lambda)$), we get $p(\bar\lambda)=0$. Hence $\bar\lambda$ is also an eigenvalue. If $\mathbf v$ satisfies $(A-\lambda I)\mathbf v = \mathbf 0$, take the entrywise conjugate of this equation. Since $A$ is real, $\bar A = A$, so

which shows $\bar{\mathbf v}$ is an eigenvector for $\bar\lambda$. This establishes the first claim of (A).

Part (A): $\mathbf z(t)=e^{\lambda t}\mathbf v$ solves the system. Differentiate directly (valid since $\mathbf v$ is a constant complex vector and $e^{\lambda t}$ is differentiable in $t$ with $\frac{d}{dt}e^{\lambda t}=\lambda e^{\lambda t}$, a fact provable from the power series definition of the complex exponential, term-by-term differentiable by uniform convergence on compact $t$-intervals):

using $(A-\lambda I)\mathbf v=\mathbf 0 \iff A\mathbf v = \lambda\mathbf v$, and linearity of $A$ (a constant real matrix) to pull it past the scalar $e^{\lambda t}$. So $\mathbf z$ solves the complex-extended system $\mathbf z' = A\mathbf z$.

Part (A): real and imaginary parts also solve the real system. Write $\mathbf z(t) = \mathbf u(t) + i\mathbf w(t)$ with $\mathbf u,\mathbf w$ real-valued. Since $A$ is a real matrix, $A\mathbf z = A\mathbf u + iA\mathbf w$ has real part $A\mathbf u$ and imaginary part $A\mathbf w$. Differentiation is applied entrywise (each entry is a differentiable complex function of the real variable $t$, and differentiation commutes with taking real/imaginary parts because $\operatorname{Re}$ and $\operatorname{Im}$ are $\mathbb R$-linear continuous operations), so

Matching real and imaginary parts of $\mathbf u' + i\mathbf w' = A\mathbf u + iA\mathbf w$ (using that a complex-valued function of a real variable is zero iff both its real and imaginary parts are zero, applied to $\mathbf z' - A\mathbf z=0$) gives

So both $\mathbf u$ and $\mathbf w$ are (real) solutions of $\mathbf x'=A\mathbf x$.

Part (A): the explicit formulas. Using Euler's formula $e^{(a+bi)t} = e^{at}(\cos bt + i\sin bt)$ (itself a consequence of matching power series, a standard fact assumed known) and $\mathbf v = \mathbf p + i\mathbf q$,

Reading off real and imaginary parts gives exactly $\mathbf u(t)$ and $\mathbf w(t)$ as claimed.

Part (A): linear independence. First, $\mathbf p,\mathbf q$ must be linearly independent over $\mathbb{R}$ whenever $b\neq 0$: if $\mathbf q = c\mathbf p$ for real $c$ (the case $\mathbf p=\mathbf 0$ forces $\mathbf v=i\mathbf q$, an entirely imaginary eigenvector, handled symmetrically below), then $\mathbf v = (1+ic)\mathbf p$, so $A\mathbf p = \operatorname{Re}\left(\frac{\lambda \mathbf v}{1+ic}\right)$; a short computation shows $A\mathbf p = a'\mathbf p$ for a real scalar $a'$ forcing $\mathbf p$ to be a real eigenvector, and consequently $\lambda$ would have to be real (a real matrix acting on a real eigenvector produces a real eigenvalue via $A\mathbf{p}=a'\mathbf{p}$ compared against $A\mathbf v=\lambda\mathbf v=\lambda(1+ic)\mathbf p$, forcing $\lambda = a'\in\mathbb R$), contradicting $b\neq 0$. Hence $\mathbf p,\mathbf q$ are $\mathbb R$-independent. Now evaluate at $t=0$: $\mathbf u(0)=\mathbf p$, $\mathbf w(0)=\mathbf q$. If $c_1\mathbf u(t)+c_2\mathbf w(t)\equiv 0$ for all $t$ for constants $c_1,c_2\in\mathbb R$, evaluating at $t=0$ gives $c_1\mathbf p + c_2\mathbf q = \mathbf 0$, and independence of $\mathbf p,\mathbf q$ forces $c_1=c_2=0$. Hence $\mathbf u,\mathbf w$ are linearly independent solutions (their values at $t=0$ alone are independent vectors, which suffices to show the functions themselves are independent, since a nontrivial linear combination vanishing identically would vanish at $t=0$ too).

Part (B): existence of the generalized eigenvector $\mathbf v_2$. Since $\lambda$ has algebraic multiplicity $2$, the matrix $A-\lambda I$ is singular with $1$-dimensional kernel (by hypothesis, geometric multiplicity $1$), hence has rank $n-1$. We must show $\mathbf v_1 \in \operatorname{range}(A-\lambda I)$, i.e., that the linear equation $(A-\lambda I)\mathbf x = \mathbf v_1$ is solvable. Consider the restriction of $A - \lambda I$ to $\mathbb{R}^n$, and let $N = \ker(A-\lambda I)$, $R=\operatorname{range}(A-\lambda I)$, with $\dim N = 1$, $\dim R = n-1$ by rank–nullity. We use the algebraic fact (a standard consequence of the primary decomposition / Jordan form theorem, established independently of the present theorem and not circular here since it is pure linear algebra about a single matrix, not about solving ODEs) that for an eigenvalue of algebraic multiplicity $2$ and geometric multiplicity $1$, the generalized eigenspace

has dimension exactly $2$. Granting this, pick any $\mathbf v_2 \in \ker\big((A-\lambda I)^2\big)\setminus \ker(A-\lambda I)$ — such a vector exists because the kernel of the square strictly contains the kernel of $(A - \lambda I)$ (dimension $2 > 1$). Set $\mathbf v_1' := (A-\lambda I)\mathbf v_2$. Then

so $\mathbf v_1' \in N = \ker(A-\lambda I)$, a $1$-dimensional space spanned by $\mathbf v_1$; also $\mathbf v_1'\neq \mathbf 0$ since $\mathbf v_2\notin \ker(A-\lambda I)$. Hence $\mathbf v_1' = c\,\mathbf v_1$ for some nonzero real scalar $c$. Rescaling $\mathbf v_2 \mapsto \mathbf v_2/c$ gives $(A-\lambda I)\mathbf v_2 = \mathbf v_1$ exactly, as required. This proves existence.

Part (B): verifying $\mathbf x_2(t)$ solves the system. Compute the derivative of $\mathbf x_2(t) = e^{\lambda t}(t\mathbf v_1+\mathbf v_2)$ using the product rule (justified since each entry is a product of the differentiable scalar function $e^{\lambda t}$ and a differentiable, in fact affine, function of $t$):

On the other hand,

using $A\mathbf v_1 = \lambda\mathbf v_1$ (eigenvector equation) and linearity of $A$. We need $\mathbf x_2' = A\mathbf x_2$, i.e., comparing the two displayed expressions (both multiplied by the common nonzero factor $e^{\lambda t}$),

which is exactly the defining equation for $\mathbf v_2$ established above. Hence $\mathbf x_2$ solves $\mathbf x'=A\mathbf x$. (And $\mathbf x_1(t)=e^{\lambda t}\mathbf v_1$ solves the system by the same computation as in Part (A) with $b=0$.)

Part (B): linear independence of $\mathbf x_1,\mathbf x_2$. Suppose $c_1\mathbf x_1(t) + c_2 \mathbf x_2(t) \equiv \mathbf 0$ for all $t\in\mathbb R$, with $c_1,c_2\in\mathbb R$. Dividing by the nonvanishing scalar $e^{\lambda t}$,

i.e.

Differentiate this identity in $t$ (it holds identically, so its derivative is also identically zero): $c_2 \mathbf v_1 = \mathbf 0$. Since $\mathbf v_1\neq \mathbf 0$ (it is an eigenvector), $c_2=0$. Substituting back, $c_1\mathbf v_1 = \mathbf 0$ forces $c_1 = 0$ as well. Hence $\{\mathbf x_1,\mathbf x_2\}$ is linearly independent. $\blacksquare$