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#REDIRECT ] |
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{{merging|eigenvalue, eigenvector, and eigenspace}} |
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{{MathCOTW}} |
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{{Redirect category shell| |
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In ], the '''eigenvectors''' (from the ] ''eigen'' meaning "own") of a ] are non-zero ] which, when operated on by the operator, result in a ] multiple of themselves. The scalar is then called the ] associated with the eigenvector. |
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{{R to plural}} |
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{{R from subtopic}} |
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In ] and ] the eigenvectors of a ] or a ] often have important physical significance. In ] the eigenvectors of the governing equations typically correspond to natural modes of vibration in a body, and the eigenvalues to their frequencies. In ], operators correspond to ] variables, eigenvectors are also called '''eigenstates''', and the eigenvalues of an operator represent those values of the corresponding variable that have non-zero ] of being measured. |
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{{R with history}} |
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{{R unprintworthy}} |
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== Definition == |
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}} |
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Formally, we define eigenvectors and eigenvalues as follows: |
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If ''A'' : ''V'' <tt>-></tt> ''V'' is a linear operator on some ] ''V'', ''v'' is a non-zero ] in ''V'' and ''λ'' is a scalar (possibly zero) such that |
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: <math>Av = \lambda v\;</math> |
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then we say that ''v'' is an eigenvector of the operator ''A'', and its associated eigenvalue is ''λ''. Note that if ''v'' is an eigenvector with ] ''λ'', then any non-zero multiple of ''v'' is also an eigenvector with eigenvalue ''λ''. In fact, all the eigenvectors with associated eigenvalue ''λ'', together with ''0'', form a subspace of ''V'', the '''eigenspace''' for the eigenvalue ''λ''. |
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== Examples == |
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For ]s of two-]al space '''R'''<sup>2</sup> we can discern the following special cases: |
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* ]s: no eigenvectors. Example: Av=v+a |
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* ]s: no real eigenvectors, (Complex eigenvalue, eigenvector pairs exist). Example: A=(01)(10). |
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* ]: eigenvectors are perpendicular and parallel to the line of ], the eigenvalues are -1 and 1, respectively. Example: A=(10)(0 -1) |
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* uniform scaling: all vectors are eigenvectors, and the eigenvalue is the ].Example: Ax=cx |
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* projection onto a line: vectors on the line are eigenvectors with eigenvalue 1 and vectors perpendicular to the line are eigenvectors with eigenvalue 0. Example: A=(00)(01). |
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== Identifying eigenvectors == |
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For example, consider the ] |
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: <math>A = |
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\begin{bmatrix} |
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\; 0 & 1 & -1 \\ |
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\; 1 & 1 & \; 0 \\ |
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-1 & 0 & \; 1 |
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\end{bmatrix} |
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</math> |
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which represents a linear operator '''R'''<sup>3</sup> <tt>-></tt> '''R'''<sup>3</sup>. One can check that |
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: <math>A \begin{bmatrix} \; 1 \\ \; 1 \\ -1 \end{bmatrix} |
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= \begin{bmatrix} \; 2 \\ \; 2 \\ -2 \end{bmatrix} |
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= 2 \begin{bmatrix} \; 1 \\ \; 1 \\ -1 \end{bmatrix} |
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</math> |
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and therefore 2 is an eigenvalue of ''A'' and we have found a corresponding eigenvector. |
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== The characteristic polynomial == |
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An important tool for describing eigenvalues of square matrices is the ]: saying that ''λ'' is an eigenvalue of ''A'' is equivalent to stating that the ] (''A'' - ''λ''id<sub>''V''</sub>) ''v'' = 0 (where id<sub>''V''</sub> is the ]) has a non-zero solution ''v'' (namely an eigenvector), and so it is equivalent to the ] det(''A'' - ''λ'' id<sub>''V''</sub>) being zero. The function ''p''(''λ'') = det(''A'' - ''λ''id<sub>''V''</sub>) is a ] in ''λ'' since determinants are defined as sums of products. |
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This is the ''characteristic polynomial'' of ''A''; its zeros are precisely the eigenvalues of ''A''. |
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If ''A'' is an ''n''-by-''n'' matrix, then its characteristic polynomial has degree ''n'' and ''A'' can therefore have at most ''n'' eigenvalues. |
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Returning to the example above, if we wanted to compute all of ''A'''s eigenvalues, we could determine the characteristic polynomial first: |
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:<math>p(x) = \det( A - xI) = |
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\begin{vmatrix} |
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-x & 1 & -1 \\ |
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1 & 1-x & 0 \\ |
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-1 & 0 & 1-x \end{vmatrix} = -x^3 + 2x^2 + x - 2 = -(x - 2) (x - 1) (x + 1)</math> |
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and we see that the eigenvalues of ''A'' are 2, 1 and -1. |
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The ] states that every square matrix satisfies its own characteristic polynomial, that is ''p''(''A'')=0. |
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(In practice, eigenvalues of large matrices are not computed using the characteristic polynomial. Faster and more numerically stable methods are available, for instance the ].) |
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== Complex eigenvectors == |
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Note that if ''A'' is a ] matrix, the characteristic polynomial will have real coefficients, but not all its roots will necessarily be real. The ] eigenvalues will all be associated to complex eigenvectors. |
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In general, if ''v''<sub>1</sub>, ..., ''v''<sub>''m''</sub> are eigenvectors with different eigenvalues λ<sub>1</sub>, ..., λ<sub>''m''</sub>, then the vectors ''v''<sub>1</sub>, ..., ''v''<sub>''m''</sub> are necessarily ]. |
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The ] for symmetric matrices states that, if ''A'' is a real symmetric ''n''-by-''n'' matrix, then all its eigenvalues are real, and there exist ''n'' linearly independent eigenvectors for ''A'' which are mutually ]. |
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Our example matrix from above is symmetric, and three mutually orthogonal eigenvectors of ''A'' are |
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:<math>v_1 = \begin{bmatrix}\; 1 \\ \;1 \\ -1 \end{bmatrix},\quad v_2 = \begin{bmatrix}\; 0\;\\ 1 \\ 1 \end{bmatrix},\quad v_3 = \begin{bmatrix}\; 2 \\ -1 \\ \; 1 \end{bmatrix}.</math> |
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These three vectors form a ] of '''R'''<sup>3</sup>. With respect to this basis, the ] represented by ''A'' takes a particularly simple form: every vector ''x'' in '''R'''<sup>3</sup> can be written uniquely as |
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:<math>x = x_1 v_1 + x_2 v_2 + x_3 v_3</math> |
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and then we have |
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:<math>A x = 2x_1 v_1 + x_2 v_2 - x_3 v_3.</math> |
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==Decomposition theorem== |
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An ''n'' by ''n'' matrix has ''n'' linearly independent real eigenvectors if and only if it can be decomposed into the form |
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:<math>A=U \Lambda U^{-1}\;</math> |
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and Λ is a ] with all of the eigenvalues on the diagonal. If ''A'' is ], then ''U'' is ], and if ''A'' is ], then ''U'' is ]. Such a matrix ''U'' does not always exist; for example |
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:<math>A=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right)</math> |
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has only one 1-dimensional eigenspace. In such a case, the ] must be used. |
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== Infinite-dimensional spaces == |
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The concept of eigenvectors can be extended to ]s |
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acting on infinite-dimensional ]s or |
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]s. |
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There are operators on Banach spaces which have no eigenvectors at all. For example, take the ] on the Hilbert space <math>\ell^2(\mathbf{Z})</math>; it is easy to see that any potential eigenvector can't be square-summable, so none exist. However, any bounded linear operator on a Banach space ''V'' does have non-empty '''spectrum'''. The spectrum σ(''A'') of the operator ''A'' : ''V'' → ''V'' is defined as |
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:<math>\sigma(A) = \{ \lambda \in \mathbf{C} | A - \lambda \mathrm{id}_V \mbox{ is not invertible }\}.</math> |
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Then σ(''A'') is a ] of complex numbers, and it is non-empty. When ''A'' is a ] (and in particular when ''A'' is an operator between ]-dimensional spaces as above), the spectrum of ''A'' is the same as the set of its eigenvalues. |
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The spectrum of an operator is an important property in ]. |
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==See also== |
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*] |
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*] |
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*] |
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==External links== |
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] ] |
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'''Bold text''' |
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