Eigenspaces

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The span of the eigenvectors associated with a fixed eigenvalue define the eigenspace corresponding to that eigenvalue.

Let $A$ be a real $n\times n$ matrix. As we saw above, $\lambda$ is an eigenvalue of $A$ iff $N(A-\lambda I)\ne 0$ , with the non-zero vectors in this nullspace comprising the set of eigenvectors of $A$ with eigenvalue $\lambda$ .

The eigenspace of $A$ corresponding to an eigenvalue $\lambda$ is $E_\lambda (A) := N(A-\lambda I)\subset \mathbb R^n$ .

Suppose $A = \begin {bmatrix} 2 & 0\\0 & 3\end {bmatrix}$ . Show that the two possible eigenvalues of $A$ are $\lambda = 2$ and $\lambda = 3$ . Then show the two corresponding eigenspaces for these eigenvalues are $E_2(A) = Span\{{\bf e}_1\}$ , $E_3(A) = Span\{{\bf e}_2\}$ .

$A = \begin {bmatrix} 2 & 5\\0 & 2\end {bmatrix}$ . Determine the eigenvalues of $A$ , and a minimal spanning set (basis) for each eigenspace.

Note that the dimension of the eigenspace corresponding to a given eigenvalue must be at least 1, since eigenspaces must contain non-zero vectors by definition.

More generally, if $L:V\to V$ is a linear transformation, and $\lambda$ is an eigenvalue of $L$ , then the eigenspace of $L$ corresponding to $\lambda$ is $E_{\lambda }(L) := ker(L - \lambda Id_V)\subset V$

For each eigenvalue $\lambda$ of $L$ , $E_{\lambda }(L)$ is a subspace of $V$ . Moreover, if $S$ is a basis for $V$ and $A = {}_{S}L_{S}$ is the representation of $L$ with respect to $S$ (in both domain and range), then ${\bf v}\in E_{\lambda }(L)$ iff ${}_{S}{\bf v}\in E_{\lambda }(A)$ .

In this way, fixing a basis for $V$ identifies the eigenspaces of $L$ (in $V$ ) with the eigenspaces of the matrix representation $A$ of $L$ (in $\mathbb R^n$ ).