Revision as of 15:17, 5 February 2019 edit Marc van Leeuwen (talk \| contribs) Extended confirmed users, Pending changes reviewers 2,666 edits m →Hermitian maps and Hermitian matrices: solutions to ->roots of (relating to the characteristic polynomial) ← Previous edit		Revision as of 15:26, 5 February 2019 edit undo Marc van Leeuwen (talk \| contribs) Extended confirmed users, Pending changes reviewers 2,666 edits →Hermitian maps and Hermitian matrices: Rephrased; getting a diagonal matrix is not dependent on having the eigenvector basis be orthonormal Next edit →
Line 30: The spectral theorem holds also for symmetric maps on finite-dimensional real inner product spaces, but the existence of an eigenvector does not follow immediately from the [[fundamental theorem of algebra]]. To prove this, consider {{math\|''A''}} as a Hermitian matrix and use the fact that all eigenvalues of a Hermitian matrix are real. IfThe ~~one~~matrix ~~chooses the eigenvectors~~representation of {{math\|''A''}} asin ~~an orthonormal~~a basis of eigenvectors is diagonal, and by the ~~matrix~~construction ~~representation~~the proof gives a basis of ~~{{math\|''A''}}~~mutually inorthogonal ~~this~~eigenvectors; by choosing them to be unit vectors one obtains an orthonormal basis isof ~~diagonal~~eigenvectors. ~~Equivalently,~~ {{math\|''A''}} can be written as a linear combination of pairwise orthogonal projections, called its '''spectral decomposition'''. Let :<math> V_\lambda = \{\,v \in V: A v = \lambda v\,\}</math>

Spectral theorem: Difference between revisions