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Eigenspaces

Let A be an n x n matrix and consider the set E = { xε R n : A x = λ x}. If x ε E, then so is t x for any scalar t, since




Furthermore, if x1 and x2 are in E, then




These calculations show that E is closed under scalar multiplication and vector addition, so E is a subspace of R n . Clearly, the zero vector belongs to E; but more notably, the nonzero elements in E are precisely the eigenvectors of A corresponding to the eigenvalue λ. When the zero vector is adjoined to the collection of eigenvectors corresponding to a particular eigenvalue, the resulting collection,




forms a vector space called the eigenspace of A correspondign to the eigenvalue λ. Since it depends on both A and the selection of one of its eigenvalues, the notation



will be used to denote this space. Since the equation A x = λ x is equivalent to ( A − λ I) x = 0, the eigenspace Eλ( A) can also be characterized as the nullspace of A − λ I:



This observation provides an immediate proof that Eλ( A) is a subspace of R n .

Consider the matrix




The determination of the eigenvectors of A shows that its eigenspaces are



and



E−1( A) is the line in R2 through the origin and the point (1, 1), and E−2( A) is the line through the origin and the point (2, 3). Both of these eigenspaces are 1-dimensional subspaces of R2.

Example 1: Determine the eigenspaces of the matrix




First, form the matrix




The determinant will be computed by performing a Laplace expansion along the second row:




The roots of the characteristic equation,




are clearly λ = −1 and 3, with 3 being a double root; these are the eigenvalues of B. The associated eigenvectors can now be found. Substituting λ = −1 into the matrix B − λ I in (*) gives



which is the coefficient matrix for the equation ( B − λ I) x = 0 with λ = −1, which determines the eigenvectors corresponding to the eigenvalue λ = −1. These eigenvectors are the nonzero solutions of



The identical first and third equations imply that x1 + x3 = 0— that is, x3 = − x1—and the second equation says x2 = 0. Therefore, the eigenvectors of B associated with the eigenvalue λ = −1 are all vectors of the form ( x1, 0, − x1)T = x1(1, 0, −1)T for x1 ≠ 0. Removing the restriction that the scalar multiple be nonzero includes the zero vector and gives the full eigenspace:




Now, since




the eigenvectors corresponding to the eigenvalue λ = 3 are the nonzero solutions of



These equations imply that x3 = x1, and since there is no restriction on x2, this component is arbitrary. Therefore, the eigenvectors of B associated with λ = 3 are all nonzero vectors of the form (x1,x2,x1)T = x1(1,0,1)T + x2(0,1,0)T The inclusion of the zero vector gives the eigenspace:




Note that dim E−1( B) = 1 and dim E3( B) = 2.

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