From a Complex Eigenpair to Real Vector Equations
A real matrix can act on complex vectors by using the same real entries and performing the usual complex arithmetic. When a nonreal eigenvalue and its eigenvector are written in terms of real and imaginary parts, the eigenvalue equation splits into two linked equations involving real vectors only. Those equations describe the matrix’s action on the plane spanned by the two parts. The order in which the parts are used as a basis determines the signs in the resulting two-by-two matrix.
Let \(\mathbf{A}\in\mathbb{R}^{n\times n}\). A complex eigenpair consists of a scalar \(\lambda\in\mathbb{C}\) and a nonzero vector \(\mathbf{z}\in\mathbb{C}^n\) such that \(\mathbf{A}\mathbf{z}=\lambda\mathbf{z}\). Write \(\lambda=a+ib\) and \(\mathbf{z}=\mathbf{x}+i\mathbf{y}\), where \(a,b\in\mathbb{R}\) and \(\mathbf{x},\mathbf{y}\in\mathbb{R}^n\). In this tutorial the eigenvalue is nonreal, so \(b\ne0\). The real and imaginary parts \(\mathbf{x}\) and \(\mathbf{y}\) are real column vectors.
Proof. Since \(\mathbf{A}\) has real entries, multiplying it by \(\mathbf{x}+i\mathbf{y}\) gives \(\mathbf{A}\mathbf{x}+i\mathbf{A}\mathbf{y}\). Multiplying the scalar and vector on the other side gives \((a+ib)(\mathbf{x}+i\mathbf{y})=(a\mathbf{x}-b\mathbf{y})+i(b\mathbf{x}+a\mathbf{y})\). Equality of two complex vectors means equality of their real parts and of their imaginary parts. This proves the two displayed equations. To prove independence, suppose \(c\mathbf{x}+d\mathbf{y}=\mathbf{0}\) for real \(c,d\). If \(d=0\), then \(c\mathbf{x}=\mathbf{0}\). If \(c\ne0\), this implies \(\mathbf{x}=\mathbf{0}\). The equation \(\mathbf{A}\mathbf{x}=a\mathbf{x}-b\mathbf{y}\) would then give \(b\mathbf{y}=\mathbf{0}\), so \(\mathbf{y}=\mathbf{0}\) because \(b\ne0\). That contradicts \(\mathbf{z}\ne\mathbf{0}\). Thus \(c=0\). If \(d\ne0\), the assumed relation gives \(\mathbf{y}=t\mathbf{x}\) for \(t=-c/d\). The case \(\mathbf{x}=\mathbf{0}\) would also force \(\mathbf{y}=\mathbf{0}\), as just shown, so \(\mathbf{x}\ne\mathbf{0}\). The two real equations now give \(\mathbf{A}\mathbf{x}=(a-bt)\mathbf{x}\) and \(\mathbf{A}\mathbf{y}=(b+at)\mathbf{x}\). But \(\mathbf{y}=t\mathbf{x}\) also implies \(\mathbf{A}\mathbf{y}=t\mathbf{A}\mathbf{x}=(at-bt^2)\mathbf{x}\). Equating coefficients of the nonzero vector \(\mathbf{x}\) yields \(b+at=at-bt^2\), or \(b(1+t^2)=0\). This is impossible for real \(t\) and nonzero \(b\). Therefore \(d\ne0\) is impossible as well, and the only relation has \(c=d=0\). \(\square\)
The Matrix in the Ordered Basis
Set \(W=\operatorname{span}\{\mathbf{x},\mathbf{y}\}\), and use the ordered basis \(\mathcal{B}=(\mathbf{x},\mathbf{y})\) for this real two-dimensional space. The theorem’s equations show that applying \(\mathbf{A}\) to either basis vector gives a linear combination of \(\mathbf{x}\) and \(\mathbf{y}\), so the result remains in \(W\). In the matrix of this action, the first column gives the coordinates of \(\mathbf{A}\mathbf{x}\), and the second gives the coordinates of \(\mathbf{A}\mathbf{y}\). In particular, the entries of each column are ordered according to \((\mathbf{x},\mathbf{y})\).
The first column is \((a,-b)^{\top}\) because \(\mathbf{A}\mathbf{x}=a\mathbf{x}-b\mathbf{y}\). The second is \((b,a)^{\top}\) because \(\mathbf{A}\mathbf{y}=b\mathbf{x}+a\mathbf{y}\). Keeping these columns in the same order as the basis vectors is essential: the two equations give the block’s columns, not its rows.
Worked Example: A Complex Eigenpair of a Real Two-by-Two Matrix
Consider \(\mathbf{A}=\begin{bmatrix}2&1\\-1&2\end{bmatrix}\), \(\mathbf{x}=\begin{bmatrix}1\\0\end{bmatrix}\), \(\mathbf{y}=\begin{bmatrix}0\\1\end{bmatrix}\), and \(\lambda=2+i\). Verify the eigenpair and find the matrix in the ordered basis \((\mathbf{x},\mathbf{y})\).
First form \(\mathbf{z}=\mathbf{x}+i\mathbf{y}=\begin{bmatrix}1\\i\end{bmatrix}\). Multiplication by \(\mathbf{A}\) gives \(\mathbf{A}\mathbf{z}=\begin{bmatrix}2+i\\-1+2i\end{bmatrix}\). On the other hand, \((2+i)\mathbf{z}=\begin{bmatrix}2+i\\(2+i)i\end{bmatrix} =\begin{bmatrix}2+i\\-1+2i\end{bmatrix}\). Thus \(\mathbf{A}\mathbf{z}=\lambda\mathbf{z}\), with \(\mathbf{z}\ne\mathbf{0}\).
Here \(a=2\) and \(b=1\). The real equations are \(\mathbf{A}\mathbf{x}=2\mathbf{x}-\mathbf{y}\) and \(\mathbf{A}\mathbf{y}=\mathbf{x}+2\mathbf{y}\). Indeed, direct multiplication gives \(\mathbf{A}\mathbf{x}=\begin{bmatrix}2\\-1\end{bmatrix} =2\begin{bmatrix}1\\0\end{bmatrix}-\begin{bmatrix}0\\1\end{bmatrix}\) and \(\mathbf{A}\mathbf{y}=\begin{bmatrix}1\\2\end{bmatrix} =\begin{bmatrix}1\\0\end{bmatrix}+2\begin{bmatrix}0\\1\end{bmatrix}\). The ordered basis is the standard basis here, so the coordinate matrix is \(\begin{bmatrix}2&1\\-1&2\end{bmatrix}\), as the theorem predicts.
Check: The first column of the claimed coordinate matrix gives \(2\mathbf{x}-\mathbf{y}=(2,-1)^{\top}=\mathbf{A}\mathbf{x}\), and its second gives \(\mathbf{x}+2\mathbf{y}=(1,2)^{\top}=\mathbf{A}\mathbf{y}\). Both basis-vector images match.
Verifying the Block in a Larger Space
The matrix \(\mathbf{A}\) need not be two-by-two. The block describes its action on the two-dimensional span \(W\), using \((\mathbf{x},\mathbf{y})\) as coordinates; it does not claim that the whole matrix has size two or that these vectors form a basis of all of \(\mathbb{R}^n\). The next example checks the real equations directly in \(\mathbb{R}^3\).
Worked Example: A Two-Dimensional Block Inside a Three-Dimensional Matrix
Let \(\mathbf{A}=\begin{bmatrix}1&1&0\\0&1&2\\-1&0&2\end{bmatrix}\), \(\mathbf{x}=\begin{bmatrix}1\\1\\0\end{bmatrix}\), and \(\mathbf{y}=\begin{bmatrix}0\\1\\1\end{bmatrix}\). Verify that \(\mathbf{z}=\mathbf{x}+i\mathbf{y}\) is an eigenvector for \(\lambda=2+i\), and find the real coordinate block.
Multiplying by the columns of \(\mathbf{A}\), or by its rows, gives \[ \mathbf{A}\mathbf{x} = \begin{bmatrix}1+1\\0+1\\-1+0\end{bmatrix} = \begin{bmatrix}2\\1\\-1\end{bmatrix}, \qquad \mathbf{A}\mathbf{y} = \begin{bmatrix}1+0\\1+2\\0+2\end{bmatrix} = \begin{bmatrix}1\\3\\2\end{bmatrix}. \] These are \(2\mathbf{x}-\mathbf{y}\) and \(\mathbf{x}+2\mathbf{y}\), respectively. For a direct complex check, \[ \mathbf{A}(\mathbf{x}+i\mathbf{y}) =\mathbf{A}\mathbf{x}+i\mathbf{A}\mathbf{y} =\begin{bmatrix}2+i\\1+3i\\-1+2i\end{bmatrix}. \] Also, \[ (2+i)(\mathbf{x}+i\mathbf{y}) =(2\mathbf{x}-\mathbf{y})+i(\mathbf{x}+2\mathbf{y}) =\begin{bmatrix}2+i\\1+3i\\-1+2i\end{bmatrix}. \] The vector \(\mathbf{z}\) is nonzero, so this verifies the eigenpair. Since \(\mathbf{x}\) and \(\mathbf{y}\) are independent, their ordered pair is a basis of their span: if \(c\mathbf{x}+d\mathbf{y}=\mathbf{0}\), the first coordinate gives \(c=0\), and then the second gives \(d=0\). Therefore the block in that basis is \(\begin{bmatrix}2&1\\-1&2\end{bmatrix}\).
Check: Applying the block’s first column to the basis means \(\mathbf{A}\mathbf{x}=2\mathbf{x}-\mathbf{y}=(2,1,-1)^{\top}\). Its second column means \(\mathbf{A}\mathbf{y}=\mathbf{x}+2\mathbf{y}=(1,3,2)^{\top}\). These agree with the direct products above.
Why the Basis Convention Changes the Signs
The displayed block is tied to the ordered basis \((\mathbf{x},\mathbf{y})\). If the second basis vector is changed to \(-\mathbf{y}\), the same action has different coordinate columns. Write \(\mathcal{C}=(\mathbf{x},-\mathbf{y})\). The real equations become \(\mathbf{A}\mathbf{x}=a\mathbf{x}+b(-\mathbf{y})\) and \(\mathbf{A}(-\mathbf{y})=-b\mathbf{x}+a(-\mathbf{y})\). Thus the columns in this new order are \((a,b)^{\top}\) and \((-b,a)^{\top}\), respectively.
Worked Example: Reversing the Sign of the Imaginary-Part Vector
Use the matrix and vectors from the first example, but take the ordered basis \(\mathcal{C}=(\mathbf{x},-\mathbf{y})\). Find and check the coordinate matrix.
Here \(a=2\) and \(b=1\). We already have \(\mathbf{A}\mathbf{x}=2\mathbf{x}-\mathbf{y}\), which in the new basis is \(2\mathbf{x}+1(-\mathbf{y})\), so its coordinate column is \((2,1)^{\top}\). Also, \(\mathbf{A}(-\mathbf{y})=-\mathbf{A}\mathbf{y}=-\mathbf{x}-2\mathbf{y}=-\mathbf{x}+2(-\mathbf{y})\), so the second coordinate column is \((-1,2)^{\top}\). The resulting matrix is \(\begin{bmatrix}2&-1\\1&2\end{bmatrix}\).
Check: The first column represents \(2\mathbf{x}+(-\mathbf{y})=(2,-1)^{\top}=\mathbf{A}\mathbf{x}\). The second represents \(-\mathbf{x}+2(-\mathbf{y})=(-1,-2)^{\top}=\mathbf{A}(-\mathbf{y})\). Both images are correct in the stated basis.
A Converse Check
The real equations also recover the complex eigenpair. This gives a useful way to check a proposed block: its columns must encode the two real equations with signs consistent with the chosen order.
Proof. Because \(\mathbf{A}\) is real, it acts on the complex vector by \(\mathbf{A}(\mathbf{x}+i\mathbf{y})=\mathbf{A}\mathbf{x}+i\mathbf{A}\mathbf{y}\). Substitute the two assumed equations to obtain \[ \mathbf{A}(\mathbf{x}+i\mathbf{y}) =(a\mathbf{x}-b\mathbf{y})+i(b\mathbf{x}+a\mathbf{y}) =(a+ib)(\mathbf{x}+i\mathbf{y}). \] If \(\mathbf{x}+i\mathbf{y}\ne\mathbf{0}\), this is exactly the definition of an eigenvector equation with nonzero eigenvector, so the stated eigenpair follows. \(\square\)
Common Sign and Coordinate Pitfalls
Incorrect sign from complex multiplication. A common error is to write the real part of \((a+ib)(\mathbf{x}+i\mathbf{y})\) as \(a\mathbf{x}+b\mathbf{y}\). The factor \(i^2=-1\) makes that sign negative: the real part is \(a\mathbf{x}-b\mathbf{y}\). For the first example, the incorrect equation would predict \(\mathbf{A}\mathbf{x}=2\mathbf{x}+\mathbf{y}=(2,1)^{\top}\), but direct multiplication gives \(\mathbf{A}\mathbf{x}=(2,-1)^{\top}\). The corrected equation \(2\mathbf{x}-\mathbf{y}=(2,-1)^{\top}\) agrees.
Swapping rows and columns. A matrix’s columns are the coordinate vectors of the images of the ordered basis vectors. In the basis \((\mathbf{x},\mathbf{y})\), \(\mathbf{A}\mathbf{x}=a\mathbf{x}-b\mathbf{y}\) supplies the first column, not the first row. Reversing the column order or reading the coefficients as rows changes the represented action.
Assuming the real and imaginary parts always form a basis. Their independence was proved using \(b\ne0\). If \(b=0\), that argument does not apply, and a real eigenvector can have zero imaginary part. In the nonreal case considered here, the nonzero imaginary part of the eigenvalue is exactly what ensures independence and gives a genuine two-dimensional real coordinate block.
Check Your Understanding
Use the ordered basis explicitly in each answer, and check coordinate columns by applying the matrix to the basis vectors.
- If \(\lambda=3+2i\) and \(\mathbf{z}=\mathbf{x}+i\mathbf{y}\) is an eigenvector of a real matrix, write the two real equations for \(\mathbf{A}\mathbf{x}\) and \(\mathbf{A}\mathbf{y}\).
- For \(\mathbf{A}=\begin{bmatrix}1&2\\-2&1\end{bmatrix}\), verify that \(\begin{bmatrix}1\\i\end{bmatrix}\) is an eigenvector and state the coordinate block in the ordered basis of its real and imaginary parts.
- Explain why \(\mathbf{x},\mathbf{y}\) are independent when \(b\ne0\), even if neither vector is assumed nonzero in advance.
- For the basis \((\mathbf{x},-\mathbf{y})\), derive the coordinate columns from the two real equations rather than memorizing the block.
- Identify the sign error in the claim that the real part of \((a+ib)(\mathbf{x}+i\mathbf{y})\) is \(a\mathbf{x}+b\mathbf{y}\), and correct it.
