The Defining Equation
A real square matrix acts on every vector in its domain, but it does not act in the same way on every vector. Depending on the matrix and the vector, the output may point in a different direction, or it may be a scalar multiple of the input. Eigenvectors are the nonzero vectors for which the second possibility occurs. The defining equation makes this special behavior precise and also explains why an arbitrary vector in the domain is not automatically an eigenvector.
Throughout this tutorial, \(\mathbf{A}\in\mathbb{R}^{n\times n}\), and vectors are columns in \(\mathbb{R}^n\). The matrix-vector product \(\mathbf{A}\mathbf{x}\) is the vector produced when \(\mathbf{A}\) acts on \(\mathbf{x}\). A scalar \(\lambda\in\mathbb{R}\) multiplies a vector by scaling its entries.
The equation says that \(\mathbf{A}\mathbf{x}\) and \(\mathbf{x}\) lie on the same line through the origin, unless \(\mathbf{A}\mathbf{x}=\mathbf{0}\). When \(\lambda>0\), the output points in the same direction as \(\mathbf{x}\); when \(\lambda<0\), it points in the opposite direction; when \(\lambda=0\), the output is the zero vector. The equation does not say that every vector is an eigenvector, or that every matrix has a real eigenvector.
The zero vector must be excluded. For every real \(\lambda\), the equation \(\mathbf{A}\mathbf{0}=\lambda\mathbf{0}\) holds, because both sides equal \(\mathbf{0}\). If the zero vector were allowed, the same zero vector would satisfy the equation for every scalar and would tell us nothing about the action of \(\mathbf{A}\). A valid eigenvector is therefore always nonzero, even when its image under \(\mathbf{A}\) is zero.
Finding the Eigenvectors for a Given Eigenvalue
For a fixed real number \(\lambda\), rearranging the defining equation gives \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\), where \(\mathbf{I}_n\) is the \(n\times n\) identity matrix. Thus the eigenvectors for \(\lambda\) are precisely the nonzero solutions of this homogeneous system. The zero vector is a solution too, but it is not an eigenvector.
Proof. Suppose \(\lambda\) is an eigenvalue. By definition, there is a nonzero \(\mathbf{x}\) such that \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). Subtracting \(\lambda\mathbf{x}\) from both sides gives \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\), so the homogeneous system has a nonzero solution. Conversely, suppose that system has a nonzero solution \(\mathbf{x}\). Expanding the left side gives \(\mathbf{A}\mathbf{x}-\lambda\mathbf{x}=\mathbf{0}\), hence \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). Since \(\mathbf{x}\ne\mathbf{0}\), the definition says that \(\lambda\) is an eigenvalue and \(\mathbf{x}\) is its eigenvector. Finally, the null space consists of every solution of the homogeneous equation, including \(\mathbf{0}\); the nonzero solutions are exactly the eigenvectors for \(\lambda\). \(\square\)
This theorem is an equivalence: a nonzero solution gives an eigenvector, and every eigenvector gives a nonzero solution. It also explains the role of row reduction when a candidate \(\lambda\) is already known: row-reduce \(\mathbf{A}-\lambda\mathbf{I}_n\) and solve its homogeneous system. If there is no free variable, the only solution is zero and \(\lambda\) is not an eigenvalue. If there is a free variable, a nonzero solution can be obtained by choosing a nonzero value for at least one free variable.
Worked Example: Two Eigenvalues and Their Eigenvectors
For \(\mathbf{A}=\begin{bmatrix}2&1\\0&3\end{bmatrix}\), find the eigenvectors associated with \(\lambda=2\) and \(\lambda=3\).
For \(\lambda=2\), subtract \(2\mathbf{I}_2\) from \(\mathbf{A}\):
Writing \(\mathbf{x}=\begin{bmatrix}x_1\\x_2\end{bmatrix}\), the equation \((\mathbf{A}-2\mathbf{I}_2)\mathbf{x}=\mathbf{0}\) gives \(x_2=0\) from each row. The variable \(x_1\) is free, so the solutions are \(\mathbf{x}=\begin{bmatrix}t\\0\end{bmatrix}\). Every choice \(t\ne0\) is an eigenvector for \(\lambda=2\). For example, \(\mathbf{x}=\begin{bmatrix}1\\0\end{bmatrix}\) satisfies
For \(\lambda=3\), the corresponding system is
The first row gives \(-x_1+x_2=0\), so \(x_2=x_1\). The solutions are \(\mathbf{x}=\begin{bmatrix}t\\t\end{bmatrix}\), and every \(t\ne0\) gives an eigenvector. Taking \(t=1\), check the equation directly:
Thus \(\lambda=2\) has eigenvectors of the form \(\begin{bmatrix}t\\0\end{bmatrix}\) for \(t\ne0\), and \(\lambda=3\) has eigenvectors of the form \(\begin{bmatrix}t\\t\end{bmatrix}\) for \(t\ne0\). In both cases, the zero solution belongs to the null space but is excluded from the list of eigenvectors.
What Makes a Vector an Eigenvector?
A proposed vector must be tested against the equation, not against its appearance. Compute \(\mathbf{A}\mathbf{x}\) and compare it with \(\mathbf{x}\): there must be one scalar \(\lambda\) that multiplies every component of \(\mathbf{x}\) to give the corresponding component of \(\mathbf{A}\mathbf{x}\). If \(\mathbf{x}\ne\mathbf{0}\) and such a scalar exists, then \(\mathbf{x}\) is an eigenvector. If no single scalar works, it is not.
Worked Example: A Vector in the Domain That Is Not an Eigenvector
Use the same matrix \(\mathbf{A}=\begin{bmatrix}2&1\\0&3\end{bmatrix}\) and test \(\mathbf{x}=\begin{bmatrix}1\\2\end{bmatrix}\).
First calculate the matrix-vector product:
If \(\mathbf{x}\) were an eigenvector, then \(\begin{bmatrix}4\\6\end{bmatrix}=\lambda\begin{bmatrix}1\\2\end{bmatrix}\), which requires \(4=\lambda\) from the first component and \(6=2\lambda\), or \(\lambda=3\), from the second. No scalar can equal both \(4\) and \(3\). Therefore \(\mathbf{x}\) is not an eigenvector of \(\mathbf{A}\).
Check: using \(\lambda=4\) gives \(\lambda\mathbf{x}=\begin{bmatrix}4\\8\end{bmatrix}\ne\begin{bmatrix}4\\6\end{bmatrix}\); using \(\lambda=3\) gives \(\lambda\mathbf{x}=\begin{bmatrix}3\\6\end{bmatrix}\ne\begin{bmatrix}4\\6\end{bmatrix}\). The proposed vector is in the domain \(\mathbb{R}^2\), but it does not satisfy the defining equation for any real \(\lambda\).
For a nonzero vector, there can be at most one eigenvalue associated with it. This is useful when checking computations: a single nonzero vector cannot have two different eigenvalues for the same matrix.
Proof. The two equalities imply \(\lambda\mathbf{x}=\mu\mathbf{x}\), so \((\lambda-\mu)\mathbf{x}=\mathbf{0}\). Since \(\mathbf{x}\ne\mathbf{0}\), at least one component \(x_i\) is nonzero. The \(i\)-th component of this vector equation is \((\lambda-\mu)x_i=0\). Dividing by \(x_i\ne0\) gives \(\lambda-\mu=0\), so \(\lambda=\mu\). \(\square\)
Eigenspaces and Scaling
For a fixed eigenvalue \(\lambda\), its eigenspace \(E_{\lambda}\) includes the zero vector as well as the eigenvectors associated with \(\lambda\). This naming is consistent with the null-space description: an eigenspace is the solution set of a homogeneous system. In particular, if \(\mathbf{x}\) is an eigenvector for \(\lambda\), every nonzero scalar multiple \(c\mathbf{x}\) is another eigenvector for the same eigenvalue, because the defining equation is preserved under scaling.
Proof. The null space of a matrix is the set of vectors sent to \(\mathbf{0}\) by that matrix. It contains \(\mathbf{0}\), since \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{0}=\mathbf{0}\). If \(\mathbf{u},\mathbf{v}\in E_{\lambda}\), then \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{u}=\mathbf{0}\) and \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{v}=\mathbf{0}\). By distributivity of matrix multiplication, \((\mathbf{A}-\lambda\mathbf{I}_n)(\mathbf{u}+\mathbf{v})=\mathbf{0}+\mathbf{0}=\mathbf{0}\), so \(\mathbf{u}+\mathbf{v}\in E_{\lambda}\). For any real scalar \(c\), \((\mathbf{A}-\lambda\mathbf{I}_n)(c\mathbf{u})=c(\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{u}=c\mathbf{0}=\mathbf{0}\), so \(c\mathbf{u}\in E_{\lambda}\). Thus \(E_{\lambda}\) contains zero and is closed under addition and scalar multiplication, which proves it is a subspace. By the defining equation, its nonzero vectors are exactly the eigenvectors for \(\lambda\). \(\square\)
When \(\mathbf{x}\) is an eigenvector and \(c\ne0\), the scaling claim can also be checked directly: \(\mathbf{A}(c\mathbf{x})=c\mathbf{A}\mathbf{x}=c\lambda\mathbf{x}=\lambda(c\mathbf{x})\), and \(c\mathbf{x}\ne\mathbf{0}\). The restriction \(c\ne0\) matters: choosing \(c=0\) produces the zero vector, which lies in the eigenspace but is not an eigenvector.
Worked Example: Scalar and Zero Matrices
Compare \(\mathbf{A}=4\mathbf{I}_2\) and \(\mathbf{B}=\mathbf{O}\), where \(\mathbf{O}\) is the \(2\times2\) zero matrix. Let \(\mathbf{x}=\begin{bmatrix}a\\b\end{bmatrix}\) be any nonzero vector in \(\mathbb{R}^2\).
For the scalar matrix, multiplication gives
Because \(\mathbf{x}\ne\mathbf{0}\), this verifies that every nonzero vector is an eigenvector of \(\mathbf{A}\) with eigenvalue \(4\). For the zero matrix,
Thus every nonzero vector is an eigenvector of \(\mathbf{B}\) with eigenvalue \(0\). In each case, the zero vector also satisfies the displayed equation, but it is not an eigenvector by definition. These examples show why one should not assume all matrices change the direction of their vectors: the exact action depends on the matrix.
Not Every Real Matrix Has a Real Eigenvector
The defining equation also makes it possible to show that a particular real matrix has no real eigenvectors. Such a conclusion concerns real vectors and real scalars. It does not claim that the corresponding matrix has no eigenvectors if one instead allows complex scalars and vectors.
Worked Example: A Real Matrix With No Real Eigenvectors
Show that \(\mathbf{A}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\) has no real eigenvector.
Suppose, seeking a contradiction, that a nonzero real vector \(\mathbf{x}=\begin{bmatrix}x_1\\x_2\end{bmatrix}\) and a real number \(\lambda\) satisfy the defining equation. Multiplying gives
Equating components yields \(-x_2=\lambda x_1\) and \(x_1=\lambda x_2\). If \(x_2=0\), the second equation gives \(x_1=0\), contradicting \(\mathbf{x}\ne\mathbf{0}\). Hence \(x_2\ne0\). Substitute \(x_1=\lambda x_2\) into the first equation: \(-x_2=\lambda^2x_2\). Dividing by \(x_2\ne0\) gives \(\lambda^2=-1\), which no real number satisfies. The assumption of a real eigenvector is impossible, so this matrix has no real eigenvectors.
Common Mistake: Testing Only One Component
A frequent error is to divide one component of \(\mathbf{A}\mathbf{x}\) by the corresponding component of \(\mathbf{x}\), call the result \(\lambda\), and stop. That procedure is valid only if the same scalar also works in every other component. For the matrix and vector in the earlier test, the first component suggests \(\lambda=4\), but the second suggests \(\lambda=3\). Therefore the proposed vector is not an eigenvector; it is incorrect to report either number as its eigenvalue.
There is a related mistake in which a student checks only that the equation holds for \(\mathbf{x}=\mathbf{0}\). That check cannot establish an eigenvalue, because \(\mathbf{A}\mathbf{0}=\lambda\mathbf{0}\) holds for every scalar \(\lambda\) and every matrix. To verify an eigenpair, check both parts: the vector must be nonzero, and direct substitution must give \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\).
The definition keeps three objects distinct: the matrix \(\mathbf{A}\), which acts on the whole domain; an eigenvalue \(\lambda\), which is a scalar; and an eigenvector \(\mathbf{x}\ne\mathbf{0}\), which is a particular input satisfying the equation. A matrix may have many eigenvectors for one eigenvalue, may have eigenvectors for several eigenvalues, or may have no real eigenvectors. The defining equation, together with the nonzero requirement, determines which case applies.
Check Your Understanding
For each question, use the defining equation and keep the nonzero condition in view.
- For \(\mathbf{A}=\begin{bmatrix}1&0\\0&5\end{bmatrix}\), test whether \(\begin{bmatrix}0\\2\end{bmatrix}\) is an eigenvector and, if so, state its eigenvalue.
- Explain why \(\mathbf{0}\) solves \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\) for every \(\lambda\), but is not an eigenvector.
- Find all vectors in \(E_2\) for \(\mathbf{A}=\begin{bmatrix}2&1\\0&3\end{bmatrix}\), and identify which of them are eigenvectors.
- Can a fixed nonzero vector be an eigenvector for two different eigenvalues of the same matrix? Justify your answer.
- For the zero matrix \(\mathbf{O}\in\mathbb{R}^{2\times2}\), state which nonzero vectors are eigenvectors and give their eigenvalue.
