From an Eigenvalue to a Homogeneous System
Once an eigenvalue is known, finding its eigenvectors is a linear-system problem. The key is to solve for all vectors that satisfy the eigenvector equation, not just to find one nonzero example. Reduced row echelon form (RREF) makes that full set visible: pivot variables are determined by free variables, and each independent parameter contributes one vector to a basis for the eigenspace.
Let \(\mathbb{F}\) be \(\mathbb{R}\) or \(\mathbb{C}\), let \(\mathbf{A}\in\mathbb{F}^{n\times n}\), and let \(\lambda\in\mathbb{F}\). An eigenvalue \(\lambda\) of \(\mathbf{A}\) is a scalar for which there is a nonzero vector \(\mathbf{x}\) satisfying \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). The eigenspace \(E_{\lambda}\) consists of the zero vector together with all eigenvectors for \(\lambda\).
The equality follows by subtracting \(\lambda\mathbf{x}\) from both sides of \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). The zero vector belongs to \(E_{\lambda}\), but it is not an eigenvector; eigenvectors are, by definition, nonzero. Since the eigenvalue is assumed known, the homogeneous system has at least one nonzero solution. If \(\lambda\) is not an eigenvalue, its only solution is the zero vector, so this method produces no eigenvectors.
Proof. If \(\mathbf{x}\) is an eigenvector for \(\lambda\), then \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\) and \(\mathbf{x}\ne\mathbf{0}\). Subtracting \(\lambda\mathbf{x}\) gives \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\), so \(\mathbf{x}\) is a nonzero solution of the stated system. Conversely, if \(\mathbf{x}\ne\mathbf{0}\) and \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\), then \(\mathbf{A}\mathbf{x}-\lambda\mathbf{x}=\mathbf{0}\), hence \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). This is exactly the definition of an eigenvector for \(\lambda\). \(\square\)
How Free Variables Describe the Eigenspace
Row-reduce \(\mathbf{A}-\lambda\mathbf{I}_n\) and solve the resulting homogeneous system. A pivot variable is the variable in a pivot column; a free variable is the variable in a column without a pivot. In RREF, each pivot variable can be expressed in terms of the free variables. Assign a distinct parameter to each free variable and write the solution vector as a linear combination of fixed vectors.
For example, if the equations give \(x_1=2s-t\), \(x_2=s\), and \(x_3=t\), then \(\mathbf{x}=s(2,1,0)^{\top}+t(-1,0,1)^{\top}\). The two vectors are the parameter vectors: one is obtained by setting \(s=1,t=0\), and the other by setting \(s=0,t=1\). These vectors span every solution, not merely some of them.
Proof. Write \(\mathbf{R}=\operatorname{rref}(\mathbf{A}-\lambda\mathbf{I}_n)\). Each elementary row operation is reversible: swapping rows reverses itself, scaling a row by a nonzero scalar can be reversed by scaling by its reciprocal, and adding a multiple of one row to another can be reversed by subtracting that multiple. Therefore each operation preserves the solutions of the homogeneous system, and \(\mathbf{R}\mathbf{x}=\mathbf{0}\) has exactly the same solutions as \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\).
Let the free variables be \(t_1,\dots,t_k\). Because the system is homogeneous, each pivot variable is a linear combination of \(t_1,\dots,t_k\). Thus every solution vector can be written \(\mathbf{x}=t_1\mathbf{v}_1+\cdots+t_k\mathbf{v}_k\), where \(\mathbf{v}_j\) is the solution obtained by setting \(t_j=1\) and all other free variables to \(0\). This proves that these vectors span the solution set. To prove independence, suppose \(c_1\mathbf{v}_1+\cdots+c_k\mathbf{v}_k=\mathbf{0}\). In the free-variable coordinates, \(\mathbf{v}_j\) has value \(1\) in coordinate \(t_j\) and \(0\) in every other free coordinate. Therefore the left side has free coordinate \(t_j\) equal to \(c_j\). Since the result is \(\mathbf{0}\), every \(c_j=0\). The vectors are independent and span \(E_{\lambda}\), so they form a basis. Because \(\lambda\) is an eigenvalue, there is a nonzero solution; consequently at least one free variable occurs, and the construction does not give an empty list. \(\square\)
Worked Eigenspace Calculations
Worked Example: One Free Variable
Find a basis for the eigenspace of \(\mathbf{A}=\begin{bmatrix}4&2\\1&3\end{bmatrix}\) for the known eigenvalue \(\lambda=5\), and check the resulting eigenvector.
Subtract \(5\) from the diagonal entries to form the shifted matrix. Row-reduce it, first swapping the rows and then adding the new first row to the second:
The first variable is a pivot variable and the second is free. The equation is \(x_1-2x_2=0\), so let \(x_2=t\). Then \(x_1=2t\), and every vector in the eigenspace has the form \(\mathbf{x}=t(2,1)^{\top}\). A basis is therefore the one-vector list \(((2,1)^{\top})\).
Check: \(\mathbf{A}(2,1)^{\top}=(4\cdot2+2\cdot1,\;1\cdot2+3\cdot1)^{\top}=(10,5)^{\top}=5(2,1)^{\top}\). The basis vector is nonzero and satisfies the eigenvector equation.
Worked Example: Two Free Variables
Find a basis for the eigenspace of \(\mathbf{A}=\begin{bmatrix}3&1&0\\0&2&0\\0&0&2\end{bmatrix}\) for \(\lambda=2\).
The shifted matrix and its RREF are
The equation is \(x_1+x_2=0\), so \(x_1=-x_2\). Let \(x_2=s\) and \(x_3=t\). The full solution is \(\mathbf{x}=(-s,s,t)^{\top}=s(-1,1,0)^{\top}+t(0,0,1)^{\top}\). Thus a basis is \(((-1,1,0)^{\top},(0,0,1)^{\top})\).
Check: Multiplying each basis vector by \(\mathbf{A}\) gives \(\mathbf{A}(-1,1,0)^{\top}=(-2,2,0)^{\top}=2(-1,1,0)^{\top}\) and \(\mathbf{A}(0,0,1)^{\top}=(0,0,2)^{\top}=2(0,0,1)^{\top}\). Each is a nonzero eigenvector for \(\lambda=2\), and the free-variable theorem shows that their combinations give every vector in the eigenspace.
Worked Example: Reducing Repeated Rows
Find the full eigenspace for \(\lambda=2\) of \(\mathbf{A}=\begin{bmatrix}3&1&1\\1&3&1\\1&1&3\end{bmatrix}\).
Subtract \(2\) from each diagonal entry. Subtract row 1 from row 2 and row 3, then the first row is already the only nonzero row:
The equation is \(x_1+x_2+x_3=0\). Choose \(x_2=s\) and \(x_3=t\), giving \(x_1=-s-t\). Thus \(\mathbf{x}=s(-1,1,0)^{\top}+t(-1,0,1)^{\top}\), and a basis is \(((-1,1,0)^{\top},(-1,0,1)^{\top})\).
Check: \(\mathbf{A}(-1,1,0)^{\top}=(-2,2,0)^{\top}\), and \(\mathbf{A}(-1,0,1)^{\top}=(-2,0,2)^{\top}\). These are respectively \(2(-1,1,0)^{\top}\) and \(2(-1,0,1)^{\top}\). Both basis vectors are nonzero and satisfy the required equation.
A Common Mistake: Finding One Vector but Missing the Eigenspace
In the second example, the equation \(x_1+x_2=0\) has two free variables, \(x_2\) and \(x_3\). Setting both to \(1\) produces \((-1,1,1)^{\top}\), which is a valid eigenvector, but it is not by itself a basis for the eigenspace: one vector cannot describe independently varying \(s\) and \(t\). The correct parameter form is \((-s,s,t)^{\top}=s(-1,1,0)^{\top}+t(0,0,1)^{\top}\), so two basis vectors are required.
A separate row-operation error can also change the answer. In the first example, after swapping rows the matrix is \(\begin{bmatrix}1&-2\\-1&2\end{bmatrix}\). Subtracting row 1 from row 2 does not create a zero row: the second row becomes \((-1,2)-(1,-2)=(-2,4)\). The correct operation is \(R_2\to R_2+R_1\), which gives \((0,0)\). Check each row operation entry by entry before reading off free variables.
Why the Parametric Form Is Useful
The eigenspace is the complete collection of vectors associated with one eigenvalue, whereas an individual eigenvector is just one nonzero member of that collection. Its basis records all the independent directions in the eigenspace. When there are several free variables, giving a single eigenvector leaves out directions; when there is one free variable, every eigenvector for that eigenvalue is a scalar multiple of the single basis vector. In either case, the RREF gives a systematic description of every solution.
Use the method in this order: form \(\mathbf{A}-\lambda\mathbf{I}_n\), row-reduce it, identify free variables, express pivot variables in terms of parameters, and split the resulting vector into parameter vectors. Finally, multiply each proposed basis vector by the original \(\mathbf{A}\) to check that it satisfies \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). The row-reduction argument guarantees that the basis spans the full eigenspace, while this multiplication checks the calculation against the defining equation.
Check Your Understanding
For each problem, show the shifted matrix, determine the free variables, and check any proposed basis vectors in the eigenvector equation.
- For \(\mathbf{A}=\begin{bmatrix}2&1\\1&2\end{bmatrix}\) and the known eigenvalue \(\lambda=3\), find a basis for \(E_{\lambda}\).
- For \(\mathbf{A}=\begin{bmatrix}4&0&0\\0&2&1\\0&1&2\end{bmatrix}\) and \(\lambda=4\), find the eigenspace in parametric form.
- For \(\mathbf{A}=\begin{bmatrix}2&1&0\\1&2&0\\0&0&3\end{bmatrix}\) and \(\lambda=1\), find a basis for \(E_{\lambda}\).
- Explain why setting every free variable to \(1\) gives a solution but does not in general produce a basis.
- For the matrix in question 3, state what the RREF would tell you if the shifted matrix had a pivot in every column.
