From the Eigenvector Equation to a Homogeneous System
An eigenvector of a matrix is defined by an equation involving the matrix and an unknown vector. When the matrix or eigenvalue has complex entries, the same equation can be solved systematically by row reduction, provided the arithmetic and the choice of scalars are handled over \(\mathbb{C}\). The goal is not just to find one nonzero eigenvector: row reduction describes every vector in the eigenspace.
Let \(\mathbf{A}\in\mathbb{C}^{n\times n}\). A scalar \(\lambda\in\mathbb{C}\) is an eigenvalue of \(\mathbf{A}\) if there is a nonzero column vector \(\mathbf{v}\in\mathbb{C}^n\) such that \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Such a vector \(\mathbf{v}\) is an eigenvector for \(\lambda\). The set consisting of the zero vector and all eigenvectors for \(\lambda\) is the eigenspace \(E_\lambda\).
The conversion is just rearrangement: \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\) holds exactly when \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\). The zero vector always solves this system, but it is not an eigenvector. Thus a proposed \(\lambda\) is an eigenvalue precisely when the system has at least one nonzero solution. If the system has free variables, assign them arbitrary complex values and express the pivot variables in terms of them. This gives the whole eigenspace, including zero; the eigenvectors are the resulting vectors other than zero.
Why Row Reduction Gives the Eigenspace
Proof. First, each elementary row operation preserves the solution set of a homogeneous system over \(\mathbb{C}\). Swapping two equations changes only their order. Multiplying an equation by a nonzero complex scalar preserves its solutions because the operation can be undone by multiplying by the reciprocal scalar. Replacing one equation by itself plus a multiple of another also preserves solutions: the operation can be undone by subtracting that multiple of the unchanged equation. Therefore the original system and its RREF system have the same solutions. By rearranging the eigenvector equation, the original system has solution set \(E_\lambda\). In RREF, each pivot variable is determined by the free variables, and every choice of free variables produces a solution; conversely, every solution has some values for its free variables and must then have the corresponding pivot values. If there are no free variables, every variable is a pivot variable and the homogeneous system has only the zero solution. If there is a free variable, set it to \(1\), set the other free variables to \(0\), and solve for the pivot variables. This gives a solution with a nonzero free coordinate, so it is nonzero. Hence a nonzero solution exists exactly when there is a free variable, proving the eigenvalue criterion as well as the parameterisation claim. \(\square\)
This result explains why the scalar field matters. A free variable here can take any value in \(\mathbb{C}\), not only in \(\mathbb{R}\). A parameterisation that restricts free variables to real numbers may describe only part of a complex eigenspace. Also, the RREF must be computed using valid complex arithmetic: in particular, division by a nonzero complex number is allowed, and signs involving \(i\) must be tracked carefully.
A Two-Parameter Eigenspace
Worked Example: A Complex Eigenspace with Two Free Parameters
For \(\mathbf{A}=\begin{bmatrix}2+i&i&0\\2&1+3i&0\\0&0&1+i\end{bmatrix}\), find the eigenspace for \(\lambda=1+i\).
Subtract \((1+i)\mathbf{I}_3\) from \(\mathbf{A}\), then eliminate the second row by subtracting twice the first row:
The equation in RREF is \(x_1+ix_2=0\). Let \(x_2=s\) and \(x_3=t\), where \(s,t\in\mathbb{C}\). Then \(x_1=-is\), so
To check the result, let \(\mathbf{v}_1=(-i,1,0)^{\top}\) and \(\mathbf{v}_2=(0,0,1)^{\top}\). Direct multiplication gives \(\mathbf{A}\mathbf{v}_1=(1-i,1+i,0)^{\top}=(1+i)\mathbf{v}_1\), because \((1+i)(-i)=1-i\). It also gives \(\mathbf{A}\mathbf{v}_2=(0,0,1+i)^{\top}=(1+i)\mathbf{v}_2\). Every displayed linear combination is therefore in the eigenspace, and the row-reduction theorem shows there are no other solutions.
The parameters \(s\) and \(t\) are independent choices from \(\mathbb{C}\). Setting both to zero gives the zero vector, which belongs to the eigenspace but is not an eigenvector. Every other choice gives an eigenvector. Writing the result as a span makes its structure visible: the two displayed vectors generate all solutions by complex linear combinations.
A One-Parameter Eigenspace
Worked Example: A Complex Eigenvector with Three Coordinates
For \(\mathbf{A}=\begin{bmatrix}3-i&i&0\\0&3-i&2\\1&1+i&4-i\end{bmatrix}\), find the eigenspace for \(\lambda=2-i\).
Subtracting \((2-i)\mathbf{I}_3\) gives \(\begin{bmatrix}1&i&0\\0&1&2\\1&1+i&2\end{bmatrix}\). First subtract row 1 from row 3, then subtract row 2 from the new row 3. Finally, clear the \(i\) above the second pivot:
The operations are \(R_3\to R_3-R_1\), \(R_3\to R_3-R_2\), and \(R_1\to R_1-iR_2\), in that order. The equations are \(x_1-2ix_3=0\) and \(x_2+2x_3=0\). Set the free variable \(x_3=t\in\mathbb{C}\). Then \(x_1=2it\), \(x_2=-2t\), and
For a direct check, let \(\mathbf{v}=(2i,-2,1)^{\top}\). Then \(\mathbf{A}\mathbf{v}=(2+4i,-4+2i,2-i)^{\top}\). On the other hand, \((2-i)\mathbf{v}=(4i-2i^2,-4+2i,2-i)^{\top}=(2+4i,-4+2i,2-i)^{\top}\). Thus \(\mathbf{A}\mathbf{v}=(2-i)\mathbf{v}\), and every nonzero complex multiple of \(\mathbf{v}\) is also an eigenvector for \(2-i\).
In this example the free variable is complex even though the row-reduced equations have simple integer and Gaussian-integer coefficients. The full parameterisation is not merely the one vector obtained by setting \(t=1\): it includes every complex multiple, as well as zero. Row reduction identifies the number of free choices and the exact dependence of each pivot variable on them.
A Real Matrix with a Nonreal Eigenvalue
Worked Example: Solving Over \(\mathbb{C}\) for a Real Matrix
Let \(\mathbf{A}=\begin{bmatrix}0&0&2\\1&0&-1\\0&1&2\end{bmatrix}\). Find the eigenspace for the proposed eigenvalue \(\lambda=i\).
Although \(\mathbf{A}\) has real entries, solve the system over \(\mathbb{C}\). The matrix \(\mathbf{A}-i\mathbf{I}_3\) is row-reduced as follows. First swap rows 1 and 2, then use \(R_2\to R_2+iR_1\), followed by \(R_3\to R_3-R_2\), and finally \(R_1\to R_1+iR_2\):
Therefore \(x_1+2ix_3=0\) and \(x_2+(2-i)x_3=0\). With \(x_3=t\in\mathbb{C}\), this gives
Check \(\mathbf{v}=(-2i,-2+i,1)^{\top}\) by multiplying: \(\mathbf{A}\mathbf{v}=(2,-1-2i,i)^{\top}\). Also \(i\mathbf{v}=(2,-1-2i,i)^{\top}\), since \(i(-2i)=2\) and \(i(-2+i)=-1-2i\). Thus \(\mathbf{v}\ne\mathbf{0}\) is indeed an eigenvector for \(i\).
This calculation illustrates that real entries do not require eigenvectors to have real coordinates. The equation is being solved over \(\mathbb{C}\), so the entries of the eigenvector and the free parameters may be complex. The row-reduced system, rather than the field of the entries of \(\mathbf{A}\), determines the scalar choices allowed in the eigenspace.
Common Pitfall: Losing a Sign in Complex Division
In the last example, the first equation before row reduction is \(-ix_1+2x_3=0\). A tempting but incorrect rearrangement is \(x_1=2ix_3\). Substituting that claim into the equation gives \((-i)(2ix_3)+2x_3=2x_3+2x_3=4x_3\), which is not zero for general \(x_3\). The correct rearrangement is \(-ix_1=-2x_3\), so \(x_1=(-2x_3)/(-i)=-2ix_3\), because \(1/(-i)=i\). Substitution confirms \((-i)(-2ix_3)+2x_3=-2x_3+2x_3=0\).
A second error is to set free variables to real values without justification. In a complex eigenspace the free parameters range over \(\mathbb{C}\); choosing only real values can omit valid eigenvectors. A reliable check is to substitute the proposed parameterisation into \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\), then verify a nonzero sample vector directly in \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\).
Check Your Understanding
For each problem, work over \(\mathbb{C}\) and use the row-reduced equations to describe the full solution set.
- For \(\mathbf{A}=\begin{bmatrix}1&i\\0&2\end{bmatrix}\) and \(\lambda=1\), find \(E_1\) and verify a nonzero eigenvector.
- If the RREF of \(\mathbf{A}-\lambda\mathbf{I}\) has a pivot in every column, what is its homogeneous solution set? Is \(\lambda\) an eigenvalue?
- For \(\begin{bmatrix}1&i&0\\0&1&-i\\0&0&0\end{bmatrix}\mathbf{x}=\mathbf{0}\), use \(x_3=t\in\mathbb{C}\) to express \(x_1\) and \(x_2\) in terms of \(t\).
- Explain why restricting a free variable to \(\mathbb{R}\) may fail to describe an eigenspace over \(\mathbb{C}\).
- For the vector \((-2i,-2+i,1)^{\top}\) in the last worked example, check each coordinate of \(\mathbf{A}\mathbf{v}=i\mathbf{v}\).
