Eigenpairs As Scalar Actions

When a Matrix Acts by Scaling

A matrix usually changes a vector’s direction as well as its length. Some nonzero vectors, however, are sent to scalar multiples of themselves. On such a vector, the matrix acts in a particularly simple way: it scales the vector, possibly reversing its direction or sending it to the zero vector. Eigenvalues and eigenvectors name the scalar and the nonzero vector in this relationship.

Let \(\mathbb{F}\) be either \(\mathbb{R}\) or \(\mathbb{C}\), and let \(\mathbf{A}\in\mathbb{F}^{n\times n}\). A vector in \(\mathbb{F}^n\) is a column. The requirement that \(\mathbf{A}\) be square matters because \(\mathbf{A}\mathbf{v}\) and \(\mathbf{v}\) must belong to the same vector space in order to compare them by scalar multiplication.

The equation says that the output vector equals the input vector multiplied by one scalar. It does not require \(\lambda\) to be nonzero: if \(\lambda=0\), the matrix sends the eigenvector to \(\mathbf{0}\). The vector itself must be nonzero. Without that condition, the equation would hold for \(\mathbf{v}=\mathbf{0}\) for every scalar, making it useless for identifying eigenvalues.

For a proposed pair, the direct test is to calculate \(\mathbf{A}\mathbf{v}\) and compare every component with the same scalar multiple \(\lambda\mathbf{v}\). The scalar must work in all components at once. A matrix-vector product that merely has some components in common with a multiple of \(\mathbf{v}\) is not enough.

Checking Eigenpairs in Two Dimensions

A proposed vector need not be an eigenvector just because the matrix product is easy to calculate. For instance, for the same matrix, \(\mathbf{u}=\begin{bmatrix}1\\1\end{bmatrix}\) gives \(\mathbf{A}\mathbf{u}=\begin{bmatrix}2\\-1\end{bmatrix}\). If this were \(\lambda\mathbf{u}=\begin{bmatrix}\lambda\\\lambda\end{bmatrix}\), the first component would force \(\lambda=2\), while the second would force \(\lambda=-1\). No one scalar satisfies both conditions, so \(\mathbf{u}\) is not an eigenvector.

What Is Fixed by an Eigenpair?

The vector in an eigenpair determines its eigenvalue uniquely. This does not mean that an eigenvalue has only one eigenvector: multiplying an eigenvector by a nonzero scalar gives another eigenvector for the same eigenvalue. The distinction is between the scalar being determined by a fixed nonzero vector and the vector being determined by the scalar.

Proof. The two assumed equations have the same left-hand side, so \(\lambda\mathbf{v}=\mu\mathbf{v}\), and hence \((\lambda-\mu)\mathbf{v}=\mathbf{0}\). Since \(\mathbf{v}\ne\mathbf{0}\), at least one component \(v_i\) is nonzero. The \(i\)th component of the last equation is \((\lambda-\mu)v_i=0\). A field has no zero divisors, so \(v_i\ne0\) implies \(\lambda-\mu=0\), or \(\lambda=\mu\). \(\square\)

There is also a direct scaling property. If \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) and \(c\in\mathbb{F}\) is nonzero, then \(\mathbf{A}(c\mathbf{v})=c\mathbf{A}\mathbf{v}=c\lambda\mathbf{v}=\lambda(c\mathbf{v})\), and \(c\mathbf{v}\ne\mathbf{0}\). So \(c\mathbf{v}\) is another eigenvector for \(\lambda\). The nonzero requirement on \(c\) ensures that the new vector remains an eigenvector rather than becoming \(\mathbf{0}\).

Eigenpairs and a Change of Basis

A matrix for a linear operator depends on the basis used for coordinates, while the operator’s action does not. Earlier material on similarity and change of basis gives the relation between two matrices of the same operator: if \(\mathbf{C}=\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\) with \(\mathbf{P}\) invertible, the coordinates of a vector in the two bases are related by \(\mathbf{P}^{-1}\). The next result spells out how that relation carries an eigenpair across the change of coordinates.

Proof. Fix \(\mathbf{v}\ne\mathbf{0}\). Since \(\mathbf{P}\) is invertible, \(\mathbf{P}^{-1}\mathbf{v}\ne\mathbf{0}\): if \(\mathbf{P}^{-1}\mathbf{v}=\mathbf{0}\), multiplying by \(\mathbf{P}\) would give \(\mathbf{v}=\mathbf{0}\), a contradiction. By substitution of \(\mathbf{C}=\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\),

If \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\), the right-hand side is \(\mathbf{P}^{-1}(\lambda\mathbf{v})=\lambda\mathbf{P}^{-1}\mathbf{v}\), giving the equation for \(\mathbf{C}\). Conversely, if \(\mathbf{C}(\mathbf{P}^{-1}\mathbf{v})=\lambda(\mathbf{P}^{-1}\mathbf{v})\), the displayed identity gives \(\mathbf{P}^{-1}\mathbf{A}\mathbf{v}=\lambda\mathbf{P}^{-1}\mathbf{v}\). Multiplying both sides by \(\mathbf{P}\) yields \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). The transformed vector is nonzero in both directions, as established at the start. \(\square\)

A Three-Dimensional Check and a Common Pitfall

Common mistake: accepting the zero vector. The calculation \(\mathbf{A}\mathbf{0}=\lambda\mathbf{0}\) is true for every matrix and every scalar. A wrong conclusion would be that \(\mathbf{0}\) is an eigenvector for every \(\lambda\). The definition rules this out explicitly: an eigenvector must satisfy \(\mathbf{v}\ne\mathbf{0}\). The equation alone is not the whole eigenpair condition.

Another useful check is to compare all components using the same proposed scalar. For example, \(\begin{bmatrix}2\\-1\end{bmatrix}\) is not a scalar multiple of \(\begin{bmatrix}1\\1\end{bmatrix}\), because its components would require the scalar to be both \(2\) and \(-1\). Matching just one component does not establish an eigenpair.

These examples focus on verifying a proposed scalar action. They do not give a method for finding every possible eigenvalue of an arbitrary matrix. The immediate habit to retain is precise: specify a nonzero column vector, multiply, and check whether one scalar reproduces the entire output. When the matrix is expressed in different bases, transform the vector along with the matrix; an invertible change of coordinates preserves the eigenvalue and does not turn a nonzero vector into zero.