From an Eigenvalue to an Eigenspace Basis
Knowing an eigenvalue tells us that some nonzero vectors keep their direction under multiplication by a matrix, changing only by a scalar factor. But when an eigenvalue is repeated, there may be more than one independent direction with that property. To find all of them, and to know how many independent eigenvectors they provide, we solve a homogeneous system. The main task is to turn its free variables into a basis for the eigenspace.
Let \(\mathbf{A}\in\mathbb{F}^{n\times n}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\), and let \(\lambda\in\mathbb{F}\) be an eigenvalue of \(\mathbf{A}\). An eigenvector for \(\lambda\) is a nonzero vector \(\mathbf{x}\) satisfying \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). The eigenspace also includes the zero vector, so it is found by solving the equation for all \(\mathbf{x}\), not just the nonzero ones.
The equality in the definition follows by moving \(\lambda\mathbf{x}\) to the left: \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\) exactly when \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\). Thus the method is to form \(\mathbf{A}-\lambda\mathbf{I}_n\), row-reduce it, and describe every solution using its free variables. The pivot columns identify pivot variables; they are not, by themselves, the vectors in a basis for the null space.
Building a Basis from Free Variables
Proof. Row operations are reversible: swapping two 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 the original and reduced homogeneous systems have the same solution vectors. In the reduced system, each pivot variable is determined by the free variables. Let the free variables be \(t_1,\dots,t_q\), and let \(\mathbf{v}_j\) be the solution obtained by setting \(t_j=1\) and every other free variable to \(0\). Any solution with free-variable values \(t_1,\dots,t_q\) is \(\sum_{j=1}^q t_j\mathbf{v}_j\), because this combination has exactly those free-variable values and the pivot variables are uniquely determined by them. Hence the vectors span the null space. If \(\sum_{j=1}^q c_j\mathbf{v}_j=\mathbf{0}\), its \(j\)th free-variable coordinate is \(c_j\), so every \(c_j=0\). The vectors are independent. If \(q=0\), there are no free variables and every variable is a pivot variable determined to be \(0\), so the null space contains only \(\mathbf{0}\). \(\square\)
Apply this result with \(\mathbf{M}=\mathbf{A}-\lambda\mathbf{I}_n\). A free variable in the reduced system contributes one basis vector to the eigenspace, and the number of free variables is the dimension of that eigenspace. To construct a basis vector, record all variable values, including the pivot-variable values determined by the chosen free variable.
Worked Example: A Repeated Eigenvalue with a One-Dimensional Eigenspace
Find a basis for each eigenspace of \(\mathbf{A}=\begin{bmatrix}3&1&0\\0&3&0\\0&0&1\end{bmatrix}\).
The characteristic polynomial is \(\det(\mathbf{A}-\lambda\mathbf{I}_3)=(3-\lambda)^2(1-\lambda)\), so the eigenvalues are \(3\) and \(1\). For \(\lambda=3\),
Here the row swap moves the nonzero third row into the second position, and scaling that row gives a leading \(1\). The equations are \(x_2=0\) and \(x_3=0\), while \(x_1\) is free. Setting \(x_1=1\) gives \(\mathbf{v}_1=(1,0,0)^{\top}\), so \(E_3=\operatorname{span}\{(1,0,0)^{\top}\}\).
For \(\lambda=1\),
The equations give \(x_1=x_2=0\), with \(x_3\) free. Thus \(E_1=\operatorname{span}\{(0,0,1)^{\top}\}\). As a check, \(\mathbf{A}(1,0,0)^{\top}=(3,0,0)^{\top}=3(1,0,0)^{\top}\), and \(\mathbf{A}(0,0,1)^{\top}=(0,0,1)^{\top}=1(0,0,1)^{\top}\).
Repeated Eigenvalues Can Have Several Independent Directions
An eigenvalue's repetition in the characteristic polynomial does not, by itself, tell us how many independent eigenvectors belong to it. Row reduction gives the relevant count directly: it is the number of free variables in \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\).
Worked Example: Two Basis Vectors for One Eigenvalue
Find a basis for the eigenspace of \(\lambda=2\) for \(\mathbf{A}=\begin{bmatrix}2&1&0\\0&2&0\\0&0&2\end{bmatrix}\).
Subtract \(2\mathbf{I}_3\) and reduce:
The equation is \(x_2=0\). Both \(x_1\) and \(x_3\) are free. Taking \((x_1,x_3)=(1,0)\) and \((0,1)\) gives the basis vectors \((1,0,0)^{\top}\) and \((0,0,1)^{\top}\). Therefore \(E_2=\operatorname{span}\{(1,0,0)^{\top},(0,0,1)^{\top}\}\).
Check: \(\mathbf{A}(1,0,0)^{\top}=(2,0,0)^{\top}=2(1,0,0)^{\top}\), and \(\mathbf{A}(0,0,1)^{\top}=(0,0,2)^{\top}=2(0,0,1)^{\top}\). The two vectors are independent because a linear combination \(c_1(1,0,0)^{\top}+c_2(0,0,1)^{\top}=\mathbf{0}\) has first coordinate \(c_1=0\) and third coordinate \(c_2=0\).
The dimension of \(E_\lambda\), also called the geometric multiplicity of \(\lambda\), is the number of vectors in any basis of that eigenspace. In these examples, the eigenspace for \(3\) has dimension \(1\), while the eigenspace for \(2\) has dimension \(2\). When counting eigenvectors available for a diagonalization, count independent eigenvectors—not every nonzero scalar multiple of a basis vector, since those multiples do not add new independent directions.
Counting Independent Eigenvectors Across Eigenspaces
A basis of each eigenspace provides a list of independent eigenvectors for that eigenvalue. To combine these lists for distinct eigenvalues, use the result that eigenspaces for distinct eigenvalues form a direct sum. In practical terms, this means that the union of bases for distinct eigenspaces is linearly independent. Thus, if the distinct eigenvalues are \(\lambda_1,\dots,\lambda_s\), the number of independent eigenvectors in this combined list is \(\dim E_{\lambda_1}+\cdots+\dim E_{\lambda_s}\).
This count measures how many independent eigenvectors have been found from the eigenspaces. In an \(n\)-dimensional space, a basis made of eigenvectors must contain \(n\) independent vectors. A total below \(n\) cannot supply such a basis; a total of \(n\) means the combined eigenspace bases already provide \(n\) independent vectors. This is why it is useful to row-reduce for every distinct eigenvalue, rather than counting eigenvalues alone.
Worked Example: Adding the Contributions from Distinct Eigenspaces
For \(\mathbf{A}=\begin{bmatrix}1&0&1\\0&1&1\\0&0&3\end{bmatrix}\), find bases for its eigenspaces and count the independent eigenvectors supplied by them.
The characteristic polynomial is \((1-\lambda)^2(3-\lambda)\), so the eigenvalues are \(1\) and \(3\). For \(\lambda=1\),
For example, subtract the first row from the second and subtract twice the first row from the third. The only equation is \(x_3=0\), so \(x_1,x_2\) are free. Setting them in turn to \((1,0)\) and \((0,1)\) gives a basis \(\{(1,0,0)^{\top},(0,1,0)^{\top}\}\) for \(E_1\).
For \(\lambda=3\),
Scale the first two rows by \(-\tfrac12\). The equations are \(x_1=\tfrac12x_3\) and \(x_2=\tfrac12x_3\). Set the free variable \(x_3=2\) to obtain the integer basis vector \((1,1,2)^{\top}\). Hence \(E_3=\operatorname{span}\{(1,1,2)^{\top}\}\).
Check: \(\mathbf{A}(1,0,0)^{\top}=(1,0,0)^{\top}\), \(\mathbf{A}(0,1,0)^{\top}=(0,1,0)^{\top}\), and \(\mathbf{A}(1,1,2)^{\top}=(3,3,6)^{\top}=3(1,1,2)^{\top}\). The two bases together give \(2+1=3\) independent eigenvectors: independence within each basis follows from its construction, and the eigenspaces belong to distinct eigenvalues, so their combined list is independent.
Common Pitfall: Pivot Columns Are Not Null-Space Basis Vectors
A pivot column identifies a variable that is determined by the equations; it does not tell you to choose the corresponding standard basis vector. For the \(\lambda=1\) calculation in the last example, the RREF of \(\mathbf{A}-\mathbf{I}_3\) has a pivot in column \(3\). A wrong conclusion would be that \((0,0,1)^{\top}\) is a basis vector for \(E_1\). But \(\mathbf{A}(0,0,1)^{\top}=(1,1,3)^{\top}\), which is not \((0,0,1)^{\top}\). The correct equations give \(x_3=0\), with \(x_1\) and \(x_2\) free, so a basis is \(\{(1,0,0)^{\top},(0,1,0)^{\top}\}\).
To avoid this error, read the RREF as a system of equations, assign values to its free variables, and then write the complete solution vectors. Finally, verify each proposed basis vector by substituting it into \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). A row-reduction answer should satisfy both checks: the vectors span all solutions, and each vector is genuinely an eigenvector for the stated eigenvalue.
Check Your Understanding
For each question, use the equations from the reduced matrix and check any proposed eigenvectors in the defining equation.
- For \(\mathbf{A}=\begin{bmatrix}4&0\\0&1\end{bmatrix}\), find a basis for each eigenspace.
- If the RREF of \(\mathbf{A}-\lambda\mathbf{I}_4\) has two free variables, how many vectors are in the basis produced by the free-variable method?
- For \(\mathbf{A}=\begin{bmatrix}2&1\\0&2\end{bmatrix}\), find a basis for \(E_2\) and state how many independent eigenvectors it provides.
- Why can bases for eigenspaces of distinct eigenvalues be combined without introducing dependence?
- A student takes the pivot columns of \(\operatorname{rref}(\mathbf{A}-\lambda\mathbf{I})\) as an eigenspace basis. What should the student do instead?
