Definition

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A nonzero vector which is scaled by a linear transformation is an eigenvector for that transformation.

If $A$ is an $m\times n$ matrix, $\bf v$ an $n\times 1$ non-zero vector, we say that $\bf v$ is an eigenvector of A with eigenvalue $\lambda$ if one has the equality $A*{\bf v} = \lambda {\bf v}$ In other words, if multiplying $\bf v$ on the left by the matrix $A$ has the same effect as multiplying it by the scalar $\lambda$ . We note first that, in order for this to be at all possible, $A*{\bf v}$ must also be an $n\times 1$ vector; in other words, $A$ must be a square matrix.

Given a square matrix, then, the eigenvalue problem is to find a complete description of the eigenvalues and associated eigenvectors for that matrix. In this section we will present a systematic way of doing this.

Before proceeding with examples, we note that

If $\bf v$ is an eigenvector of a matrix $A$ , the eigenvalue associated with it is unique.

Proof: Suppose $\lambda _1{\bf v} = A*{\bf v} = \lambda _2{\bf v}$ . Then $\lambda _1{\bf v} - \lambda _2{\bf v} = (\lambda _1 - \lambda _2){\bf v} = {\bf 0}$ . But since ${\bf v}\ne {\bf 0}$ , the only way this could happen is if the coefficient $(\lambda _1 - \lambda _2)$ is equal to zero, or equivalently, if $\lambda _1 = \lambda _2$ . $\blacksquare$

Let $A = I^{n\times n}$ be the $n\times n$ identity matrix. Then for any ${\bf v}\in \mathbb R^n$ , one has $A*{\bf v} = I*{\bf v} = {\bf v} = 1{\bf v}$ . So in this case, we see that every non-zero vector in $\mathbb R^n$ is an eigenvector of $A$ with corresponding eigenvalue $1$ .

Let $B = \begin {bmatrix}2 & 0\\0 & 3\end {bmatrix}$ . Then $B*{\bf e}_1 = 2{\bf e}_1$ , and $B*{\bf e}_2 = 3{\bf e}_2$ . Here the two standard basis vectors of $\mathbb R^2$ are eigenvectors of $B$ , but with different eigenvalues. Note that, unlike the matrix $A$ in the previous example, not every vector in $\mathbb R^2$ is an eigenvector of $B$ . In particular, if we let ${\bf v} = {\bf e}_1 + {\bf e}_2 = \begin {bmatrix} 1\\ 1\end {bmatrix}$ , then $B*{\bf v} = \begin {bmatrix}2\\ 3\end {bmatrix}$ , which cannot be written as a scalar multiple of $\bf v$ .

So how does one go about finding the eigenvalues of a matrix? In order to understand more clearly what it is we are looking for, we consider a reformulation of the defining equation above. First, we note that the scalar product $\lambda {\bf v}$ can be rewritten as a matrix product $\lambda {\bf v} = (\lambda I)*{\bf v}$ From this we have the following equivalent statements: $A*{\bf v} = \lambda {\bf v}\quad \Leftrightarrow \quad A*{\bf v} = (\lambda I)*{\bf v}\quad \Leftrightarrow \quad (A - \lambda I)*{\bf v} = {\bf 0}\quad \Leftrightarrow \quad {\bf v}\in N(A - \lambda I)$

Thus

A non-zero vector $\bf v$ is an eigenvector of $A$ with eigenvalue $\lambda$ if and only if it lies in the nullspace of the matrix $A-\lambda I$ . In particular, in order for such an vector to exist, the matrix $A-\lambda I$ must be singular.

This suggests that in order to solve the eigenvalue problem for a matrix $A$ , we should first determine the values $\lambda$ for which $A - \lambda I$ is singular; then, for each such $\lambda$ , determine a basis for the nullspace of $A - \lambda I$ . In other words, eigenvalues first, eigenvectors second. And to identify the values $\lambda$ for which $A - \lambda I$ is singular, we will use the determinant.

The above discussion for computing eigenvalues of square matrices can be easily extended to linear transformations.

If $L:V\to V$ is a linear transformation, we say $0\ne {\bf v}\in V$ is an eigenvector of $L$ with eigenvalue $\lambda$ iff $L({\bf v}) = \lambda {\bf v}$ .

Note that this definition is basis-free; it does not require the use of a particular basis. However, suppose we are given a basis $S = \{{\bf v}_1,\dots ,{\bf v}_n\}$ for $V$ . Let $A = {}_{S}L_{S}$ be the matrix representation of $L$ with respect to $S$ in both the domain and range, so that for all ${\bf v}\in V$ one has ${}_{S}L({\bf v}) = A*{}_{S}{\bf v}$ Then

$\bf v$ is an eigenvector for $L$ with eigenvalue $\lambda$ precisely when the coordinate vector ${}_{S}{\bf v}$ is an eigenvalue for $A$ with eigenvalue $\lambda$ .

In this way, fixing a basis for $V$ allows us to determine the eigenvalues and eigenvectors for $L$ by doing the same for the matrix representation $A$ of $L$ . We will investigate this further in the following sections.