From the Eigenvalue Equation to a Homogeneous System
An eigenvector is a nonzero vector whose direction is unchanged by a linear transformation: the transformation scales it by some number. Once a candidate eigenvalue \(\lambda\) is given, finding every vector that is scaled by \(\lambda\) is a linear-system problem. The key step is to move all terms of the eigenvalue equation to one side. This turns the search for eigenvectors into the familiar task of solving a homogeneous system.
Throughout, let \(\mathbf{A}\in\mathbb{R}^{n\times n}\) and \(\lambda\in\mathbb{R}\). A scalar \(\lambda\) is an eigenvalue of \(\mathbf{A}\) if there is a vector \(\mathbf{v}\ne\mathbf{0}\) such that \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Such a vector \(\mathbf{v}\) is an eigenvector for \(\lambda\). The zero vector satisfies the same equation, but it is not an eigenvector: the definition requires a nonzero vector.
The eigenspace contains the zero vector as well as all eigenvectors for \(\lambda\). In particular, the eigenspace is not the list of eigenvectors: it is a subspace containing every scalar multiple and every sum of its vectors. This is useful because a basis describes the whole eigenspace compactly, while its dimension measures how many independent directions it contains.
Proof. Take any \(\mathbf{v}\in E_{\lambda}\). By the definition of the eigenspace, \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Since \(\lambda\mathbf{v}=(\lambda\mathbf{I}_n)\mathbf{v}\), subtracting the right side gives \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{v}=\mathbf{0}\). Therefore \(\mathbf{v}\in\operatorname{Nul}(\mathbf{A}-\lambda\mathbf{I}_n)\). Conversely, if \(\mathbf{v}\in\operatorname{Nul}(\mathbf{A}-\lambda\mathbf{I}_n)\), then \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{v}=\mathbf{0}\), so \(\mathbf{A}\mathbf{v}-\lambda\mathbf{v}=\mathbf{0}\), and hence \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Thus \(\mathbf{v}\in E_{\lambda}\). Both sets contain exactly the same vectors, proving equality. \(\square\)
This identity explains the procedure: subtract \(\lambda\) from the diagonal entries of \(\mathbf{A}\), row-reduce the resulting matrix, and solve the homogeneous system. The solution set is the whole eigenspace, not just a selection of eigenvectors. If \(\lambda\) is not an eigenvalue, the corresponding homogeneous system has only the zero solution; equivalently, it has no nonzero eigenvector. The method here takes a candidate \(\lambda\) as its input.
Finding a Basis and Dimension
To report an eigenspace fully, give a basis and its dimension. A basis is a linearly independent list that spans the space, meaning every vector in the space is a linear combination of the listed vectors. The dimension of a finite-dimensional space is the number of vectors in any basis. In practice, solving the row-reduced homogeneous system writes every solution in terms of free variables. Setting one free variable to \(1\) at a time, with the others set to \(0\), produces basis vectors for the solution space.
Proof. Each row operation replaces a system by an equivalent system: swapping equations, multiplying an equation by a nonzero scalar, or adding a multiple of one equation to another does not change which vectors satisfy all the equations. Each operation is reversible, so the original and row-reduced homogeneous systems have exactly the same solutions. In the reduced system, assign arbitrary values to the free variables. The equations then determine the pivot variables, so each solution is a linear combination of the solutions formed by setting one free variable to \(1\) and the others to \(0\). These vectors span the solution set because the arbitrary free-variable values are precisely their coefficients. They are independent because in a linear combination equal to zero, the coordinate corresponding to each free variable equals that vector's coefficient; all such coordinates being zero forces every coefficient to be zero. They therefore form a basis, and their number is the number of free variables. By the theorem above, the solution set is \(E_{\lambda}\). \(\square\)
Worked Eigenspace Calculations
Worked Example: A One-Dimensional Eigenspace
Let \(\mathbf{A}=\begin{bmatrix}4&2\\1&3\end{bmatrix}\), and compute \(E_5\). First form the shifted matrix:
Swap the two rows, then add the new first row to the second row:
The reduced system is \(x_1-2x_2=0\), so \(x_1=2x_2\). Set \(x_2=t\). Then \(\mathbf{x}=t\begin{bmatrix}2\\1\end{bmatrix}\), and a basis is \(\left(\begin{bmatrix}2\\1\end{bmatrix}\right)\). Therefore \(\dim E_5=1\).
Check: \(\mathbf{A}\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}4(2)+2(1)\\1(2)+3(1)\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}=5\begin{bmatrix}2\\1\end{bmatrix}\). The basis vector is nonzero and is indeed an eigenvector for \(5\).
Worked Example: A Two-Dimensional Eigenspace
Let \(\mathbf{A}=\begin{bmatrix}2&0&1\\0&2&1\\0&0&3\end{bmatrix}\), and compute \(E_2\). Subtract \(2\mathbf{I}_3\):
Subtract row 1 from rows 2 and 3. The resulting row-equivalent system is
The equation is \(x_3=0\); \(x_1\) and \(x_2\) are free. Thus every vector in the eigenspace has the form \(\begin{bmatrix}x_1\\x_2\\0\end{bmatrix}=x_1\begin{bmatrix}1\\0\\0\end{bmatrix}+x_2\begin{bmatrix}0\\1\\0\end{bmatrix}\). These two vectors are independent, so they form a basis and \(\dim E_2=2\).
Check: \(\mathbf{A}\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}2\\0\\0\end{bmatrix}=2\begin{bmatrix}1\\0\\0\end{bmatrix}\), and \(\mathbf{A}\begin{bmatrix}0\\1\\0\end{bmatrix}=\begin{bmatrix}0\\2\\0\end{bmatrix}=2\begin{bmatrix}0\\1\\0\end{bmatrix}\). By linearity, every linear combination of these basis vectors is also scaled by \(2\).
Worked Example: A Basis Not Aligned with the Coordinate Axes
Let \(\mathbf{A}=\begin{bmatrix}3&1\\0&2\end{bmatrix}\), and find a basis for \(E_2\). The shifted matrix and its reduced form are
Therefore \(E_2=\operatorname{span}\left\{\begin{bmatrix}-1\\1\end{bmatrix}\right\}\), and \(\dim E_2=1\).
Check: \(\mathbf{A}\begin{bmatrix}-1\\1\end{bmatrix}=\begin{bmatrix}3(-1)+1(1)\\0(-1)+2(1)\end{bmatrix}=\begin{bmatrix}-2\\2\end{bmatrix}=2\begin{bmatrix}-1\\1\end{bmatrix}\). This confirms the nonzero basis vector is an eigenvector for \(2\).
Common Mistake: Reducing the Wrong Matrix
The matrix to row-reduce is \(\mathbf{A}-\lambda\mathbf{I}_n\), not \(\mathbf{A}\) itself. For example, in the first worked example, row-reducing \(\mathbf{A}=\begin{bmatrix}4&2\\1&3\end{bmatrix}\) gives
A wrong calculation would conclude that the null space of \(\mathbf{A}\) is \(\{\mathbf{0}\}\), and therefore that there is no eigenvector. But this tests whether \(\mathbf{A}\mathbf{x}=\mathbf{0}\), not whether \(\mathbf{A}\mathbf{x}=5\mathbf{x}\). The correction is to subtract \(5\mathbf{I}_2\) first and solve \((\mathbf{A}-5\mathbf{I}_2)\mathbf{x}=\mathbf{0}\); as shown above, that gives the nonzero vector \(\begin{bmatrix}2\\1\end{bmatrix}\). Always keep the candidate eigenvalue in the shifted matrix.
A second point to keep straight is the role of the zero vector. It belongs to every null space and every eigenspace, but it cannot by itself establish that \(\lambda\) is an eigenvalue. To establish that, the null-space calculation must produce at least one nonzero vector. When reporting a basis, its vectors are necessarily nonzero; the zero vector is instead represented by taking every basis coefficient to be zero.
Why the Null-Space View Is Useful
The equation \(E_{\lambda}=\operatorname{Nul}(\mathbf{A}-\lambda\mathbf{I}_n)\) makes the full set of eigenvectors accessible through a systematic calculation. It gives more information than checking one proposed vector: row reduction identifies every solution, a basis records the independent directions, and the number of basis vectors gives the eigenspace dimension. The same steps work whether the solution is one-dimensional, as in the first and third examples, or has several independent directions, as in the second.
When using this method, keep the logical roles separate. The candidate \(\lambda\) determines the matrix to row-reduce; the homogeneous solutions determine the eigenspace; and nonzero solutions determine whether the candidate is an eigenvalue at all. A basis describes the eigenspace, while its dimension counts the basis vectors. These checks help ensure that a correct-looking row reduction answers the intended question.
Check Your Understanding
For each calculation, state the solution set and, when applicable, give a basis and dimension. Verify any proposed basis vector by multiplying by the original matrix.
- For \(\mathbf{A}=\begin{bmatrix}3&0\\0&1\end{bmatrix}\), find a basis and dimension for \(E_3\).
- For \(\mathbf{A}=\begin{bmatrix}1&1&0\\0&1&0\\0&0&1\end{bmatrix}\), find \(E_1\) by row-reducing \(\mathbf{A}-\mathbf{I}_3\).
- Explain why solving \(\mathbf{A}\mathbf{x}=\mathbf{0}\) does not generally compute the eigenspace for a specified \(\lambda\).
- If the reduced system for \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\) has two free variables, how many vectors are in the resulting basis, and what is the dimension of \(E_{\lambda}\)?
