Spectral Theorem

Examples of operators to which the spectral theorem applies are Self-adjoint Operator s or more generally Normal Operator s on Hilbert Space s. The spectral theorem also provides a Canonical decomposition, called the spectral decomposition of the underlying vector space on which it acts. In this article we consider mainly the simplest kind of spectral theorem, that for a Self-adjoint operator on a Hilbert space. However, as noted above, the spectral theorem also holds for normal operators on a Hilbert space. FINITE-DIMENSIONAL CASE We begin by considering a symmetric operator ''A'' on a finite-dimensional Real or Complex Inner Product Space ''V'' with the standard Hermitian Inner Product ; in Dirac's Bra-ket Notation , the symmetry condition means : $\langle A x \mid y angle = \langle x \mid A y angle$ for all ''x'', ''y'' elements of ''V''. Recall that an Eigenvector of a linear operator ''A'' is a (non-zero) vector ''x'' such that ''Ax'' = ''rx'' for some scalar ''r''. The value ''r'' is the corresponding Eigenvalue . Theorem. There is an Orthonormal Basis of ''V'' consisting of eigenvectors of ''A''. Each eigenvalue is real. This result is of such importance in many parts of mathematics, that we provide a sketch of a proof for the case wherein the underlying field of scalars is the complex numbers. First we show that all the eigenvalues are real. Suppose that λ is an eigenvalue of ''A'' with corresponding eigenvector ''x''. Thus : $\overline{\lambda} \langle x \mid x angle= \langle A x \mid x angle = \langle x \mid A x angle = \lambda \langle x \mid x angle .$ Since ''x'' is non-zero, it follows that λ equals its own conjugate and is therefore real. To prove the existence of an eigenvector basis, we use induction on the dimension of ''V''. In fact it suffices to show ''A'' has at least one non-zero eigenvector ''e''. For then we can consider the space ''K'' of vectors ''v'' orthogonal to ''e''. This is finite-dimensional, and ''A'' has the property that it maps every vector ''w'' in ''K'' into ''K'': : $\langle A w \mid e angle = \langle w \mid A e angle = \lambda \langle w \mid e angle = 0.$ Moreover, ''A'' considered as a linear operator on ''K'' is also symmetric, so by the induction hypothesis there is a basis for ''V'' consisting of eigenvectors of ''A''. It remains, however, to show that ''A'' has at least one eigenvector. Since the ground field is Algebraically Closed , the polynomial function (called the Characteristic Polynomial of ''A'') : $p(\lambda) = \det(\lambda I - A)$ has a ''complex'' root ''r''. This implies the linear operator ''A'' − ''rI'' is not invertible and hence maps a non-zero vector ''e'' to 0. This vector ''e'' is a non-zero eigenvector of ''A''. This implies that ''r'' is an eigenvalue, so is actually a real number. This completes the proof. The spectral theorem is also true for symmetric operators on finite-dimensional real inner product spaces. The spectral decomposition of an operator ''A'' which has an orthonormal basis of eigenvectors is obtained by grouping together all vectors corresponding to the same eigenvalue. Thus : $V_\lambda = \{\,v \in V: A v = \lambda v\,\}.$ Note that these spaces are invariantly defined, in that the definition does not depend on any choice of specific eigenvectors. As an immediate consequence of the spectral theorem for symmetric operators we get the spectral decomposition theorem: ''V'' is the orthogonal direct sum of the spaces ''V''_λ where the index ranges over eigenvalues. Another equivalent formulation, letting ''P''_λ be the Orthogonal Projection onto ''V''_λ ( $P_\lambda P_\mu=0 \quad \mbox{if } \lambda eq \mu$ ) and λ₁,..., λ_''m'' the eigenvalues of ''A'', is : $A =\lambda_1 P_{\lambda_1} +\cdots+\lambda_m P_{\lambda_m}.$ If ''A'' is a normal operator on a finite-dimensional inner product space, ''A'' also has a spectral decomposition and the decomposition theorem holds for ''A''. The eigenvalues will be Complex Numbers in general. The proof is somewhat more complicated and is discussed in the ''Axler'' reference below. These results translate immediately into results about matrices: For any Normal Matrix ''A'', there exists a Unitary Matrix ''U'' such that \; where Λ is the Diagonal Matrix the entries of which are the Eigenvalue s of ''A''. Furthermore, any matrix which can be diagonalized in this way must be normal. The column vectors of ''U'' are the eigenvectors of ''A'' and they are Orthogonal . The spectral decomposition is a special case of the Schur Decomposition . It is also a special case of the Singular Value Decomposition . If ''A'' is a real symmetric matrix, it follows by the real version of the spectral theorem for symmetric operators that there is an Orthogonal Matrix ''U'' such that ''UAU^T'' is diagonal and all the eigenvalues of ''A'' are Real . THE SPECTRAL THEOREM FOR COMPACT SELF-ADJOINT OPERATORS In Hilbert spaces in general, the statement of the spectral theorem for Compact self-adjoint operators is virtually the same as in the finite-dimensional case. Theorem. Suppose ''A'' is a compact self-adjoint operator on a Hilbert space ''V''. There is an Orthonormal Basis of ''V'' consisting of eigenvectors of ''A''. Each eigenvalue is real. Again the key point is to prove the existence of at least one nonzero eigenvector. To prove this, we cannot rely on determinants to show existence of eigenvalues, but instead we use a maximization argument analogous to proving the Min-max Theorem for eigenvalues. Note that the above spectral theorem holds for real or complex Hilbert spaces. GENERALIZATION TO NON-SYMMETRIC MATRICES For a non-symmetric but square ( $N imes N$ dimensional) matrix $\mathbf{A}$ , the ''right eigenvectors'' $\mathbf{r}_{k}$ are defined by : $\mathbf{A} \cdot \mathbf{r}_{k} = \lambda_{k} \mathbf{r}_{k}$ whereas the ''left eigenvectors'' $\mathbf{l}_{k}$ are defined by : $\mathbf{l}_{k} \cdot \mathbf{A} = \lambda_{k} \mathbf{l}_{k}$ or, equivalently, : $\mathbf{A}^{T} \cdot \mathbf{l}_{k} = \lambda_{k} \mathbf{l}_{k}$ where $\mathbf{A}^{T}$ represents the Transpose of $\mathbf{A}$ . In these equations, the Eigenvalues $\lambda_{k}$ are the same, being the roots of the same Characteristic Polynomial :
	\det \left \mathbf{A}^{T} - \lambda I Ight	0

	CATEGORIES ABOUT SPECTRAL THEOREM

	spectral theory

	mathematical theorems

	linear algebra

	matrix theory

	singular value decomposition

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Spectral Theorem