Editing Spectral theorem (section)

== Finite-dimensional case ==
<!-- This section is linked from [[Singular value decomposition]] -->

=== Hermitian maps and Hermitian matrices ===
We begin by considering a [[Hermitian matrix]] on <math>\mathbb{C}^n</math> (but the following discussion will be adaptable to the more restrictive case of [[symmetric matrix|symmetric matrices]] on {{nowrap|<math>\mathbb{R}^n</math>).}} We consider a [[Hermitian operator|Hermitian map]] {{math|''A''}} on a finite-dimensional [[complex number|complex]] [[inner product space]] {{math|''V''}} endowed with a [[Definite bilinear form|positive definite]] [[sesquilinear form|sesquilinear]] [[inner product]] <math>\langle\cdot,\cdot\rangle</math>. The Hermitian condition on <math>A</math> means that for all {{math|''x'', ''y'' ∈ ''V''}},
<math display="block"> \langle A x, y \rangle = \langle x, A y \rangle.</math>

An equivalent condition is that {{math|1=''A''<sup>*</sup> = ''A''}}, where {{math|''A''<sup>*</sup>}} is the [[Hermitian conjugate]] of {{math|''A''}}. In the case that {{math|''A''}} is identified with a Hermitian matrix, the matrix of {{math|''A''<sup>*</sup>}} is equal to its [[conjugate transpose]]. (If {{math|''A''}} is a [[real matrix]], then this is equivalent to {{math|1=''A''<sup>T</sup> = ''A''}}, that is, {{math|''A''}} is a [[symmetric matrix]].)

This condition implies that all eigenvalues of a Hermitian map are real: To see this, it is enough to apply it to the case when {{math|1=''x'' = ''y''}} is an eigenvector. (Recall that an [[eigenvector]] of a linear map {{math|''A''}} is a non-zero vector {{math|''v''}} such that {{math|1=''Av'' = ''λv''}} for some scalar {{math|''λ''}}. The value {{math|''λ''}} is the corresponding [[eigenvalue]]. Moreover, the [[eigenvalues]] are roots of the [[characteristic polynomial]].)

{{math theorem | math_statement = If {{math|''A''}} is Hermitian on {{math|''V''}}, then there exists an [[orthonormal basis]] of {{math|''V''}} consisting of eigenvectors of {{math|''A''}}. Each eigenvalue of {{math|''A''}} is real.}}

We provide a sketch of a proof for the case where the underlying field of scalars is the [[complex number]]s.

By the [[fundamental theorem of algebra]], applied to the [[characteristic polynomial]] of {{math|''A''}}, there is at least one complex eigenvalue {{math|''λ''<sub>1</sub>}} and corresponding eigenvector {{math|''v''<sub>1</sub>}}, which must by definition be non-zero. Then since 
<math display="block">\lambda_1 \langle v_1, v_1 \rangle = \langle A (v_1), v_1 \rangle = \langle v_1, A(v_1) \rangle = \bar\lambda_1 \langle v_1, v_1 \rangle,</math> 
we find that {{math|''λ''<sub>1</sub>}} is real. Now consider the space <math>\mathcal{K}^{n-1} = \text{span}(v_1)^\perp</math>, the [[orthogonal complement]] of {{math|''v''<sub>1</sub>}}. By Hermiticity, <math>\mathcal{K}^{n-1}</math> is an [[invariant subspace]] of {{math|''A''}}. To see that, consider any <math>k \in \mathcal{K}^{n-1}</math> so that <math> \langle k, v_1 \rangle = 0 </math> by definition of <math>\mathcal{K}^{n-1}</math>.  To satisfy invariance, we need to check if <math>A(k) \in \mathcal{K}^{n-1}</math>. This is true because, <math>\langle A(k), v_1 \rangle = \langle k, A(v_1) \rangle = \langle k, \lambda_1 v_1 \rangle = 0</math>. Applying the same argument to <math>\mathcal{K}^{n-1}</math> shows that {{math|''A''}} has at least one real eigenvalue <math>\lambda_2</math> and corresponding eigenvector <math>v_2 \in \mathcal{K}^{n-1} \perp v_1</math>. This can be used to build another invariant subspace <math>\mathcal{K}^{n-2} = \text{span}(\{v_1, v_2\})^\perp</math>. Finite induction then finishes the proof.

The matrix representation of {{math|''A''}} in a basis of eigenvectors is diagonal, and by the construction the proof gives a basis of mutually orthogonal eigenvectors; by choosing them to be unit vectors one obtains  an orthonormal basis of eigenvectors. {{math|''A''}} can be written as a linear combination of pairwise orthogonal projections, called its '''spectral decomposition'''. Let
<math display="block">V_{\lambda} = \{v \in V: A v = \lambda v\}</math>
be the eigenspace corresponding to an eigenvalue <math>\lambda</math>. Note that the definition does not depend on any choice of specific eigenvectors. In general, {{math|''V''}} is the orthogonal direct sum of the spaces <math>V_{\lambda}</math> where the <math>\lambda</math> ranges over the [[Spectrum of a matrix|spectrum]] of <math>A</math>.

When the matrix being decomposed is Hermitian, the spectral decomposition is a special case of the [[Schur decomposition]] (see the proof in case of [[#Normal matrices|normal matrices]] below).

=== Spectral decomposition and the singular value decomposition ===

The spectral decomposition is a special case of the [[singular value decomposition]], which states that any matrix  <math>A \in \mathbb{C}^{m \times n}</math> can be expressed as 
<math>A = U\Sigma V^{*}</math>, where <math>U \in \mathbb{C}^{m \times m}</math> and <math>V \in \mathbb{C}^{n \times n}</math> are [[unitary matrices]] and <math>\Sigma \in \mathbb{R}^{m \times n}</math> is a diagonal matrix. The diagonal entries of <math>\Sigma</math> are uniquely determined by <math>A</math> and are known as the [[singular values]] of <math>A</math>. If <math>A</math> is Hermitian, then <math>A^* = A</math> and <math>V \Sigma U^* = U \Sigma V^*</math> which implies <math>U = V</math>.

=== Normal matrices ===
{{main|Normal matrix}}
The spectral theorem extends to a more general class of matrices. Let {{math|''A''}} be an operator on a finite-dimensional inner product space. {{math|''A''}} is said to be [[normal matrix|normal]]  if {{math|1=''A''<sup>*</sup>''A'' = ''AA''<sup>*</sup>}}. 

One can show that {{math|''A''}} is normal if and only if it is unitarily diagonalizable using the [[Schur decomposition]]. That is, any matrix can be written as {{math|1=''A'' = ''UTU''<sup>*</sup>}}, where {{math|''U''}} is unitary and {{math|''T''}} is [[upper triangular]].
If {{math|''A''}} is normal, then one sees that {{math|1=''TT''<sup>*</sup> = ''T''<sup>*</sup>''T''}}. Therefore, {{math|''T''}} must be diagonal since a normal upper triangular matrix is diagonal (see [[normal matrix#Consequences|normal matrix]]). The converse is obvious.

In other words, {{math|''A''}} is normal if and only if there exists a [[unitary matrix]] {{math|''U''}} such that
<math display="block">A = U D U^*,</math>
where {{math|''D''}} is a [[diagonal matrix]]. Then, the entries of the diagonal of {{math|''D''}} are the [[eigenvalue]]s of {{math|''A''}}. The column vectors of {{math|''U''}} are the eigenvectors of {{math|''A''}} and they are orthonormal. Unlike the Hermitian case, the entries of {{math|''D''}} need not be real.