Eigenvectors and Eigenvalues

Definition Let $A$ be a real $n \times n$ matrix. A vector $\mathbf{v}$ in $\mathbb{R}^n$ is an eigenvector of $A$ if $\mathbf{v} \ne \mathbf{0}$ and there is a real number $\lambda$ with

The number $\lambda$ is the eigenvalue of $A$ belonging to $\mathbf{v}$, and a real number is an eigenvalue of $A$ when at least one eigenvector belongs to it. The matrix must be square, because $A\mathbf{v}$ and $\mathbf{v}$ have to lie in the same space. Each eigenvector has exactly one eigenvalue, since $\lambda\mathbf{v} = \mu\mathbf{v}$ with $\mathbf{v} \ne \mathbf{0}$ forces $\lambda = \mu$. Every nonzero multiple $c\mathbf{v}$ of an eigenvector is again an eigenvector with the same eigenvalue, because linearity gives $A(c\mathbf{v}) = cA\mathbf{v} = \lambda(c\mathbf{v})$. On this page matrices, vectors and eigenvalues are real.

Definition An eigenpair of a real $n \times n$ matrix $A$ is a pair $(\lambda, \mathbf{v})$ made of a real number $\lambda$ and a nonzero vector $\mathbf{v}$ in $\mathbb{R}^n$ with $A\mathbf{v} = \lambda\mathbf{v}$. To check a proposed eigenpair, confirm that $\mathbf{v}$ is not the zero vector, then compute $A\mathbf{v}$ and compare it with $\lambda\mathbf{v}$ in every entry. To decide whether a given nonzero vector $\mathbf{v}$ is an eigenvector at all, compute $A\mathbf{v}$ and ask whether one scalar $\lambda$ satisfies $A\mathbf{v} = \lambda\mathbf{v}$ in every entry at once. If $(\lambda, \mathbf{v})$ is an eigenpair, so is $(\lambda, c\mathbf{v})$ for every $c \ne 0$. When $\lambda$ is an eigenvalue, all solutions of $A\mathbf{v} = \lambda\mathbf{v}$, the zero vector included, form the null space $\operatorname{Null}(A - \lambda I)$, called the eigenspace of $\lambda$, and its nonzero members are the eigenvectors for $\lambda$.

Definition Let $A$ be a real $n \times n$ matrix and let $L = \operatorname{span}\{\mathbf{v}\}$ be a line through the origin, with $\mathbf{v} \ne \mathbf{0}$. The line $L$ is invariant under $A$ if $A\mathbf{x}$ lies in $L$ for every $\mathbf{x}$ in $L$. The line $L$ is invariant exactly when $\mathbf{v}$ is an eigenvector of $A$. If $L$ is invariant, then $A\mathbf{v}$ lies in $\operatorname{span}\{\mathbf{v}\}$, so $A\mathbf{v} = \lambda\mathbf{v}$ for some $\lambda$. Conversely, if $A\mathbf{v} = \lambda\mathbf{v}$, then $A(c\mathbf{v}) = (c\lambda)\mathbf{v}$ lies in $L$ for every $c$. Invariance asks only that $A$ map $L$ into $L$. When $\lambda \ne 0$ the image of $L$ is all of $L$, and when $\lambda = 0$ the whole line goes to the origin.

Definition A negative eigenvalue is an eigenvalue $\lambda < 0$. If $A\mathbf{v} = \lambda\mathbf{v}$ with $\mathbf{v} \ne \mathbf{0}$ and $\lambda < 0$, then $A\mathbf{v}$ points opposite to $\mathbf{v}$ and has length $\lVert A\mathbf{v} \rVert = \lvert \lambda \rvert \, \lVert \mathbf{v} \rVert$. The line $\operatorname{span}\{\mathbf{v}\}$ is still invariant. When $\lambda \ne 0$, its sign decides the orientation along the line, kept for $\lambda > 0$ and reversed for $\lambda < 0$. For every eigenvalue the length factor is $\lvert \lambda \rvert$. These statements concern vectors on that line only. They say nothing about how $A$ changes the lengths or angles of other vectors.

Definition Let $A$ be a real $n \times n$ matrix. The number $0$ is an eigenvalue of $A$ exactly when $A\mathbf{v} = \mathbf{0}$ for some nonzero $\mathbf{v}$, that is, when $\operatorname{Null} A \ne \{\mathbf{0}\}$. For a square matrix this happens exactly when $A$ is singular, which is the same as $\det A = 0$. The eigenvectors for $0$ are the nonzero vectors of $\operatorname{Null} A$. The zero vector is never an eigenvector, because $A\mathbf{0} = \lambda\mathbf{0}$ holds for every $\lambda$ and would single out no eigenvalue and no direction. Zero can be an eigenvalue, but the zero vector cannot be an eigenvector.

Definition Let $A$ be a real $n \times n$ matrix, and let $\mathbf{v}_1, \dots, \mathbf{v}_k$ be eigenvectors of $A$ with eigenvalues $\lambda_1, \dots, \lambda_k$ that are pairwise distinct, so $\lambda_i \ne \lambda_j$ whenever $i \ne j$. Then $\mathbf{v}_1, \dots, \mathbf{v}_k$ are linearly independent. In particular $A$ has at most $n$ distinct eigenvalues, because $\mathbb{R}^n$ holds at most $n$ independent vectors.

For two vectors, let $A\mathbf{v} = \lambda\mathbf{v}$ and $A\mathbf{w} = \mu\mathbf{w}$ with $\lambda \ne \mu$, and suppose $a\mathbf{v} + b\mathbf{w} = \mathbf{0}$. Applying $A$ gives $a\lambda\mathbf{v} + b\mu\mathbf{w} = \mathbf{0}$. Subtracting $\lambda$ times the original relation leaves

Since $\mathbf{w} \ne \mathbf{0}$ and $\mu - \lambda \ne 0$, this forces $b = 0$, and then $a\mathbf{v} = \mathbf{0}$ with $\mathbf{v} \ne \mathbf{0}$ forces $a = 0$. For $k$ vectors, argue by induction. A single nonzero vector is independent. Assume the statement for $k - 1$ vectors and let $c_1\mathbf{v}_1 + \cdots + c_k\mathbf{v}_k = \mathbf{0}$. Applying $A$ and subtracting $\lambda_k$ times the relation removes $\mathbf{v}_k$,

By the induction hypothesis every $c_i(\lambda_i - \lambda_k)$ is $0$, and since $\lambda_i \ne \lambda_k$, every $c_i$ with $i < k$ is $0$. Then $c_k\mathbf{v}_k = \mathbf{0}$ gives $c_k = 0$. The converse is false. Independent eigenvectors can share an eigenvalue, as in the example below.