Separating a Complex Eigenvector into Real Vectors
A real matrix acts on real column vectors, but it can also be considered as acting on complex column vectors by the same matrix multiplication. A complex eigenvector therefore contains information that can be separated into real vectors. When its eigenvalue is nonreal, those real vectors are independent, and the real matrix sends every vector in their span back into that span. This gives a real two-dimensional subspace associated with a complex eigenpair, without requiring the matrix itself to have complex entries.
Let \(\mathbf{A}\in\mathbb{R}^{n\times n}\), and suppose \(\lambda=a+ib\in\mathbb{C}\) is an eigenvalue of \(\mathbf{A}\), with complex eigenvector \(\mathbf{z}\ne\mathbf{0}\). Here \(a,b\in\mathbb{R}\), \(i^2=-1\), and \(\mathbf{z}\in\mathbb{C}^n\). Write the real and imaginary parts of \(\mathbf{z}\) as \(\mathbf{x},\mathbf{y}\in\mathbb{R}^n\), so that \(\mathbf{z}=\mathbf{x}+i\mathbf{y}\). The eigenvector equation is \(\mathbf{A}\mathbf{z}=\lambda\mathbf{z}\). Its real and imaginary parts give the key relations.
Proof. Since the entries of \(\mathbf{A}\) are real, multiplying it by the real and imaginary parts separately gives \(\mathbf{A}\mathbf{z}=\mathbf{A}\mathbf{x}+i\mathbf{A}\mathbf{y}\). Multiplying the eigenvalue and eigenvector gives $$ \lambda\mathbf{z}=(a+ib)(\mathbf{x}+i\mathbf{y}) =(a\mathbf{x}-b\mathbf{y})+i(b\mathbf{x}+a\mathbf{y}). $$ The real and imaginary parts of equal complex vectors are equal, coordinate by coordinate. Equating these parts proves both equations. \(\square\)
The minus sign in the first equation comes from \(i^2=-1\). Keeping that sign straight matters: the two equations describe where the matrix sends each of the real vectors, and both will be used to show that their span is preserved.
The Real Span Is Invariant
The span is a subspace because it contains \(\mathbf{0}\), and sums and real scalar multiples of its vectors remain combinations of \(\mathbf{x}\) and \(\mathbf{y}\). More explicitly, the sum of \(s\mathbf{x}+t\mathbf{y}\) and \(r\mathbf{x}+q\mathbf{y}\) is \((s+r)\mathbf{x}+(t+q)\mathbf{y}\), and multiplying the first vector by \(c\in\mathbb{R}\) gives \((cs)\mathbf{x}+(ct)\mathbf{y}\).
Proof. Linear independence means that \(s\mathbf{x}+t\mathbf{y}=\mathbf{0}\) for real \(s,t\) forces \(s=t=0\). Thus the two vectors form a basis for their span, so the span has dimension two. To prove invariance, take any \(\mathbf{w}\in W\). By the definition of span, there are real \(s,t\) such that \(\mathbf{w}=s\mathbf{x}+t\mathbf{y}\). Using the two equations already derived, $$ \begin{aligned} \mathbf{A}\mathbf{w} &=s\mathbf{A}\mathbf{x}+t\mathbf{A}\mathbf{y}\\ &=s(a\mathbf{x}-b\mathbf{y})+t(b\mathbf{x}+a\mathbf{y})\\ &=(as+bt)\mathbf{x}+(-bs+at)\mathbf{y}. \end{aligned} $$ The coefficients \(as+bt\) and \(-bs+at\) are real, so this expression belongs to \(W\). This holds for every \(\mathbf{w}\in W\); hence \(\mathbf{A}(W)\subseteq W\), as required. \(\square\)
The independence condition is important to the dimension claim: it ensures that the span is two-dimensional, rather than a line or just the zero subspace. In fact, when the eigenvalue is nonreal, independence is automatic.
Proof. Suppose instead that \(\mathbf{x}\) and \(\mathbf{y}\) are dependent. Since \(\mathbf{z}=\mathbf{x}+i\mathbf{y}\ne\mathbf{0}\), at least one of \(\mathbf{x},\mathbf{y}\) is nonzero. Choose a nonzero vector \(\mathbf{u}\) among them. Dependence implies that both vectors are real multiples of \(\mathbf{u}\), so \(\mathbf{x}=\alpha\mathbf{u}\) and \(\mathbf{y}=\beta\mathbf{u}\) for some \(\alpha,\beta\in\mathbb{R}\). Therefore \(\mathbf{z}=(\alpha+i\beta)\mathbf{u}\). The scalar \(\alpha+i\beta\) is nonzero, since otherwise \(\mathbf{z}=\mathbf{0}\). Substituting this expression into \(\mathbf{A}\mathbf{z}=\lambda\mathbf{z}\) and dividing by that nonzero scalar gives \(\mathbf{A}\mathbf{u}=\lambda\mathbf{u}\). The left side is real. The imaginary part of the right side is \(b\mathbf{u}\), so equality forces \(b\mathbf{u}=\mathbf{0}\). Since \(b\ne0\), this gives \(\mathbf{u}=\mathbf{0}\), contradicting the choice of \(\mathbf{u}\). Thus \(\mathbf{x},\mathbf{y}\) are independent. \(\square\)
Worked Examples
Worked Example: A Planar Rotation
Let $$ \mathbf{A}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}, \qquad \lambda=i, \qquad \mathbf{z}=\begin{bmatrix}1\\-i\end{bmatrix}. $$ Verify the eigenvector equation and identify the invariant real span.
First multiply the matrix by the proposed eigenvector: $$ \mathbf{A}\mathbf{z} =\begin{bmatrix}0&-1\\1&0\end{bmatrix} \begin{bmatrix}1\\-i\end{bmatrix} =\begin{bmatrix}i\\1\end{bmatrix}. $$ Multiplication by the eigenvalue gives $$ \lambda\mathbf{z} =i\begin{bmatrix}1\\-i\end{bmatrix} =\begin{bmatrix}i\\1\end{bmatrix}. $$ Thus \(\mathbf{A}\mathbf{z}=\lambda\mathbf{z}\), and \(\mathbf{z}\ne\mathbf{0}\). Its real and imaginary parts are $$ \mathbf{x}=\begin{bmatrix}1\\0\end{bmatrix}, \qquad \mathbf{y}=\begin{bmatrix}0\\-1\end{bmatrix}. $$ If \(s\mathbf{x}+t\mathbf{y}=\mathbf{0}\), then \((s,-t)^{\top}=(0,0)^{\top}\), so \(s=t=0\). They are independent and span \(\mathbb{R}^2\). The component equations also check directly: \(\mathbf{A}\mathbf{x}=(0,1)^{\top}=-\mathbf{y}\), while \(\mathbf{A}\mathbf{y}=(1,0)^{\top}=\mathbf{x}\). Both images lie in the span. In this case the invariant subspace is the whole real plane.
Worked Example: An Invariant Plane Inside R³
Consider $$ \mathbf{A}=\begin{bmatrix}3&-1&0\\1&3&0\\0&0&4\end{bmatrix}, \qquad \lambda=3+i, \qquad \mathbf{z}=\begin{bmatrix}1\\-i\\0\end{bmatrix}. $$ Check that the real and imaginary parts span an invariant plane.
Direct multiplication gives $$ \mathbf{A}\mathbf{z} =\begin{bmatrix}3+i\\1-3i\\0\end{bmatrix}, \qquad \lambda\mathbf{z} =\begin{bmatrix}3+i\\(3+i)(-i)\\0\end{bmatrix} =\begin{bmatrix}3+i\\1-3i\\0\end{bmatrix}. $$ So \(\mathbf{z}\) is indeed an eigenvector for \(\lambda\). Here \(a=3\), \(b=1\), and $$ \mathbf{x}=\begin{bmatrix}1\\0\\0\end{bmatrix}, \qquad \mathbf{y}=\begin{bmatrix}0\\-1\\0\end{bmatrix}. $$ These vectors are independent: \(s\mathbf{x}+t\mathbf{y}=(s,-t,0)^{\top}=\mathbf{0}\) implies \(s=t=0\). Their span consists of vectors whose third coordinate is zero. To check the action explicitly, $$ \mathbf{A}\mathbf{x}=\begin{bmatrix}3\\1\\0\end{bmatrix} =3\mathbf{x}-\mathbf{y}, \qquad \mathbf{A}\mathbf{y}=\begin{bmatrix}1\\-3\\0\end{bmatrix} =\mathbf{x}+3\mathbf{y}. $$ Thus each image remains in the span. The third coordinate of a general vector in the span stays zero after applying \(\mathbf{A}\), which also confirms the invariance.
Worked Example: Scaling an Eigenvector Changes Its Parts
Let $$ \mathbf{A}=\begin{bmatrix}1&-2\\2&1\end{bmatrix}, \qquad \lambda=1+2i, \qquad \mathbf{z}=\begin{bmatrix}1+i\\1-i\end{bmatrix}. $$ Verify the eigenpair and the resulting real span.
Multiplying by the matrix gives $$ \mathbf{A}\mathbf{z} =\begin{bmatrix}(1+i)-2(1-i)\\2(1+i)+(1-i)\end{bmatrix} =\begin{bmatrix}-1+3i\\3+i\end{bmatrix}. $$ Multiplying by the eigenvalue gives $$ \lambda\mathbf{z} =\begin{bmatrix}(1+2i)(1+i)\\(1+2i)(1-i)\end{bmatrix} =\begin{bmatrix}-1+3i\\3+i\end{bmatrix}. $$ Hence the eigenvector equation holds. In this choice of eigenvector, $$ \mathbf{x}=\begin{bmatrix}1\\1\end{bmatrix}, \qquad \mathbf{y}=\begin{bmatrix}1\\-1\end{bmatrix}. $$ If \(s\mathbf{x}+t\mathbf{y}=\mathbf{0}\), then \(s+t=0\) and \(s-t=0\). Adding these equations gives \(2s=0\), and subtracting them gives \(2t=0\), so the vectors are independent. The component equations are verified by $$ \mathbf{A}\mathbf{x} =\begin{bmatrix}-1\\3\end{bmatrix} =\mathbf{x}-2\mathbf{y}, \qquad \mathbf{A}\mathbf{y} =\begin{bmatrix}3\\1\end{bmatrix} =2\mathbf{x}+\mathbf{y}. $$ The two images are in the real span, as the theorem predicts.
Common Sign Error
A frequent mistake is to write \(\mathbf{A}\mathbf{x}=a\mathbf{x}+b\mathbf{y}\) and \(\mathbf{A}\mathbf{y}=a\mathbf{y}-b\mathbf{x}\). This reverses the signs and does not follow from the complex product. For the rotation example, that incorrect rule would give \(\mathbf{A}\mathbf{x}=\mathbf{y}\). But direct multiplication gives \(\mathbf{A}\mathbf{x}=(0,1)^{\top}=-\mathbf{y}\), so the proposed equation is wrong. Expanding \((a+ib)(\mathbf{x}+i\mathbf{y})\) resolves the issue: $$ (a+ib)(\mathbf{x}+i\mathbf{y}) =(a\mathbf{x}-b\mathbf{y})+i(b\mathbf{x}+a\mathbf{y}). $$ The corrected real equation is \(\mathbf{A}\mathbf{x}=a\mathbf{x}-b\mathbf{y}\), and the corrected imaginary equation is \(\mathbf{A}\mathbf{y}=b\mathbf{x}+a\mathbf{y}\).
Why This Construction Matters
A real matrix can have nonreal eigenvalues when considered over \(\mathbb{C}\), but the corresponding eigenvector need not be a real vector. Taking its real and imaginary parts converts the eigenvector equation into statements entirely about real vectors. For a nonreal eigenvalue, those parts are independent, and together they span a real two-dimensional subspace preserved by the matrix. The result identifies a real part of the action of \(\mathbf{A}\) even when no real eigenvector is available for that eigenvalue.
The conclusion is specifically about the span of these two vectors. It does not say that each vector is an eigenvector over \(\mathbb{R}\); the equations generally show that the image of one part involves both parts. Keeping that distinction in view helps connect complex eigenpairs to real subspaces without treating a complex eigenvector as though it were real.
Check Your Understanding
Use the component equations and verify any numerical claims by direct substitution.
- If \(\lambda=4-3i\) and \(\mathbf{z}=\mathbf{x}+i\mathbf{y}\) is an eigenvector of a real matrix, write the equations satisfied by \(\mathbf{A}\mathbf{x}\) and \(\mathbf{A}\mathbf{y}\).
- For a nonreal eigenvalue, why must the real and imaginary parts of a nonzero eigenvector be independent over \(\mathbb{R}\)?
- For \(\mathbf{A}=\begin{bmatrix}2&-1\\1&2\end{bmatrix}\), verify that \(\lambda=2+i\) has eigenvector \(\mathbf{z}=(1,-i)^{\top}\), and identify its real and imaginary parts.
- Explain why showing that \(\mathbf{A}\mathbf{x}\) and \(\mathbf{A}\mathbf{y}\) both lie in \(\operatorname{span}\{\mathbf{x},\mathbf{y}\}\) proves that the whole span is invariant.
- Give the incorrect sign in the common error and correct it by expanding \((a+ib)(\mathbf{x}+i\mathbf{y})\).
