Editing Diagonalizable matrix (section)

== Examples ==

=== Diagonalizable matrices ===
* [[Involution (mathematics)|Involution]]s are diagonalizable over the reals (and indeed any field of characteristic not 2), with ±1 on the diagonal.
* Finite order [[endomorphism]]s are diagonalizable over <math>\mathbb{C}</math> (or any algebraically closed field where the characteristic of the field does not divide the order of the endomorphism) with [[roots of unity]] on the diagonal. This follows since the minimal polynomial is [[separable polynomial|separable]], because the roots of unity are distinct.
* [[Projection (linear algebra)|Projections]] are diagonalizable, with 0s and 1s on the diagonal.
* Real [[symmetric matrices]] are diagonalizable by [[orthogonal matrix|orthogonal matrices]]; i.e., given a real symmetric matrix {{nowrap|<math>A</math>,}} <math>Q^{\mathrm T}AQ</math> is diagonal for some orthogonal matrix {{nowrap|<math>Q</math>.}} More generally, matrices are diagonalizable by [[unitary matrix|unitary matrices]] if and only if they are [[normal matrix|normal]]. In the case of the real symmetric matrix, we see that {{nowrap|<math>A=A^{\mathrm T}</math>,}} so clearly <math>AA^{\mathrm T} = A^{\mathrm T}A</math> holds. Examples of normal matrices are real symmetric (or [[Skew-symmetric matrix|skew-symmetric]]) matrices (e.g. covariance matrices) and [[Hermitian matrix|Hermitian matrices]] (or skew-Hermitian matrices). See [[spectral theorem]]s for generalizations to infinite-dimensional vector spaces.

=== Matrices that are not diagonalizable ===
In general, a [[rotation matrix]] is not diagonalizable over the reals, but all [[rotation matrix#Independent planes|rotation matrices]] are diagonalizable over the complex field. Even if a matrix is not diagonalizable, it is always possible to "do the best one can", and find a matrix with the same properties consisting of eigenvalues on the leading diagonal, and either ones or zeroes on the superdiagonal – known as [[Jordan Normal Form|Jordan normal form]].

Some matrices are not diagonalizable over any field, most notably nonzero [[nilpotent matrix|nilpotent matrices]]. This happens more generally if the [[Eigenvalues and eigenvectors#Algebraic multiplicity|algebraic and geometric multiplicities]] of an eigenvalue do not coincide. For instance, consider

:<math> C = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}. </math>

This matrix is not diagonalizable: there is no matrix <math>U</math> such that <math>U^{-1}CU</math> is a diagonal matrix. Indeed, <math>C</math> has one eigenvalue (namely zero) and this eigenvalue has algebraic multiplicity 2 and geometric multiplicity 1.

Some real matrices are not diagonalizable over the reals. Consider for instance the matrix

:<math> B = \left[\begin{array}{rr} 0 & 1 \\ \!-1 & 0 \end{array}\right]. </math>

The matrix <math>B</math> does not have any real eigenvalues, so there is no '''real''' matrix <math>Q</math> such that <math>Q^{-1}BQ</math> is a diagonal matrix. However, we can diagonalize <math>B</math> if we allow complex numbers. Indeed, if we take

:<math> Q = \begin{bmatrix} 1 & i \\ i & 1 \end{bmatrix}, </math>

then <math>Q^{-1}BQ</math> is diagonal. It is easy to find that <math>B</math> is the rotation matrix which rotates counterclockwise by angle <math display="inline">\theta = -\frac{\pi}{2}</math>

Note that the above examples show that the sum of diagonalizable matrices need not be diagonalizable.

=== How to diagonalize a matrix ===
Diagonalizing a matrix is the same process as finding its [[eigenvalues and eigenvectors]], in the case that the eigenvectors form a basis. For example, consider the matrix

:<math>A=\left[\begin{array}{rrr}
0 & 1 & \!\!\!-2\\
0 & 1 & 0\\
1 & \!\!\!-1 & 3
\end{array}\right].</math>

The roots of the [[characteristic polynomial]] <math>p(\lambda)=\det(\lambda I-A)</math> are the eigenvalues {{nowrap|<math>\lambda_1 = 1,\lambda_2 = 1,\lambda_3 = 2</math>.}} Solving the linear system <math>\left(1I-A\right) \mathbf{v} = \mathbf{0}</math> gives the eigenvectors <math>\mathbf{v}_1 = (1,1,0)</math> and {{nowrap|<math>\mathbf{v}_2 = (0,2,1)</math>,}} while <math>\left(2I-A\right)\mathbf{v} = \mathbf{0}</math> gives {{nowrap|<math>\mathbf{v}_3 = (1,0,-1)</math>;}} that is, <math>A \mathbf{v}_i = \lambda_i \mathbf{v}_i</math> for {{nowrap|<math>i = 1,2,3</math>.}} These vectors form a basis of {{nowrap|<math>V = \mathbb{R}^3</math>,}} so we can assemble them as the column vectors of a [[Change of basis|change-of-basis]] matrix <math>P</math> to get:
<math display="block">P^{-1}AP =
\left[\begin{array}{rrr}
1 & 0 & 1\\
1 & 2 & 0\\
0 & 1 & \!\!\!\!-1
\end{array}\right]^{-1}

\left[\begin{array}{rrr}
0 & 1 & \!\!\!-2\\
0 & 1 & 0\\
1 & \!\!\!-1 & 3
\end{array}\right]

\left[\begin{array}{rrr}
1 & \,0 & 1\\
1 & 2 & 0\\
0 & 1 & \!\!\!\!-1
\end{array}\right]
=
\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix} = D .</math>
We may see this equation in terms of transformations: <math>P</math> takes the standard basis to the eigenbasis, {{nowrap|<math>P \mathbf{e}_i = \mathbf{v}_i</math>,}} so we have:
<math display="block">P^{-1} AP \mathbf{e}_i =
P^{-1} A \mathbf{v}_i =
P^{-1} (\lambda_i\mathbf{v}_i) =
\lambda_i\mathbf{e}_i,</math>
so that <math>P^{-1} AP</math> has the standard basis as its eigenvectors, which is the defining property of {{nowrap|<math>D</math>.}} 

Note that there is no preferred order of the eigenvectors in {{nowrap|<math>P</math>;}} changing the order of the [[eigenvectors]] in <math>P</math> just changes the order of the [[eigenvalues]] in the diagonalized form of {{nowrap|<math>A</math>.}}<ref>{{cite book| last1=Anton |first1=H. |last2= Rorres|first2= C. |title=Elementary Linear Algebra (Applications Version) | url=https://archive.org/details/studentsolutions00grob | url-access=registration |publisher=John Wiley & Sons|edition=8th|date=22 Feb 2000| ISBN= 978-0-471-17052-5}}</ref>