Editing Diagonalizable matrix (section)

== Diagonalization ==
{{See also||Eigendecomposition of a matrix}}Consider the two following arbitrary bases <math>E = \{{ \boldsymbol{e}_i | \forall i \in [n] } \}  </math> and <math>F = \{ {\boldsymbol{\alpha}_i | \forall i \in [n] } \} </math>. Suppose that there exists a linear transformation represented by a matrix <math>A_E </math> which is written with respect to basis E. Suppose also that there exists the following eigen-equation:

<math>A_E \boldsymbol{\alpha}_{E,i} = \lambda_i \boldsymbol{\alpha}_{E,i} </math>

The alpha eigenvectors are written also with respect to the E basis. Since the set F is both a set of eigenvectors for matrix A and it spans some arbitrary vector space, then we say that there exists a matrix <math>D_F </math> which is a diagonal matrix that is similar to <math>A_E </math>. In other words, <math>A_E </math> is a diagonalizable matrix if the matrix is written in the basis F. We perform the change of basis calculation using the transition matrix <math>S </math>, which changes basis from E to F as follows:

<math>D_F = S_{E}^F \ A_E  \ S_{E}^{-1F} </math>,

where <math>S_{E}^F  </math> is the transition matrix from E-basis to F-basis. The inverse can then be equated to a new transition matrix <math>P  </math> which changes basis from F to E instead and so we have the following relationship : 

<math>S_{E}^{-1 F} = P_{F}^{E}  </math>

Both <math>S   </math> and <math>P  </math> transition matrices are invertible. Thus we can manipulate the matrices in the following fashion:<math display="block">\begin{align}
    D = S \ A_{E} \ S^{-1} \\
    D = P^{-1} \ A_{E} \ P 
\end{align}</math>The matrix <math>A_{E}   </math> will be denoted as <math>A   </math>, which is still in the E-basis. Similarly, the diagonal matrix is in the F-basis.

[[File:Diagonalization as rotation.gif|400px|thumb|right|The diagonalization of a symmetric matrix can be interpreted as a rotation of the axes to align them with the eigenvectors.]]
If a matrix <math>A</math> can be diagonalized, that is,

: <math>P^{-1}AP = \begin{bmatrix}
  \lambda_1 &         0 &  \cdots &         0 \\
          0 & \lambda_2 &  \cdots &         0 \\
     \vdots &    \vdots & \ddots &    \vdots \\
          0 &         0 &  \cdots & \lambda_n
\end{bmatrix} = D,</math>

then:

: <math>AP = P\begin{bmatrix}
  \lambda_1 &         0 &  \cdots &         0 \\
          0 & \lambda_2 &  \cdots &         0 \\
     \vdots &    \vdots & \ddots &    \vdots \\
          0 &         0 &  \cdots & \lambda_n
\end{bmatrix}.</math>

The transition matrix S has the E-basis vectors as columns written in the basis F. Inversely, the inverse transition matrix P has F-basis vectors <math>\boldsymbol{\alpha}_i  </math> written in the basis of E so that we can represent P in block matrix form in the following manner:

:<math>P = \begin{bmatrix} \boldsymbol{\alpha}_{E,1} & \boldsymbol{\alpha}_{E,2} & \cdots & \boldsymbol{\alpha}_{E,n} \end{bmatrix},</math>

as a result we can write:<math display="block">\begin{align}
    A \begin{bmatrix} \boldsymbol{\alpha}_{E,1} & \boldsymbol{\alpha}_{E,2} & \cdots & \boldsymbol{\alpha}_{E,n} \end{bmatrix} =  \begin{bmatrix} \boldsymbol{\alpha}_{E,1} & \boldsymbol{\alpha}_{E,2} & \cdots & \boldsymbol{\alpha}_{E,n} \end{bmatrix}D.
\end{align}</math>

In block matrix form, we can consider the A-matrix to be a matrix of 1x1 dimensions whilst P is a 1xn dimensional matrix. The D-matrix can be written in full form with all the diagonal elements as an nxn dimensional matrix:

<math> A \begin{bmatrix} \boldsymbol{\alpha}_{E,1} & \boldsymbol{\alpha}_{E,2} & \cdots & \boldsymbol{\alpha}_{E,n} \end{bmatrix}=  \begin{bmatrix} \boldsymbol{\alpha}_{E,1} & \boldsymbol{\alpha}_{E,2} & \cdots & \boldsymbol{\alpha}_{E,n} \end{bmatrix}
\begin{bmatrix}
  \lambda_1 &         0 &  \cdots &         0 \\
          0 & \lambda_2 &  \cdots &         0 \\
     \vdots &    \vdots & \ddots &    \vdots \\
          0 &         0 &  \cdots & \lambda_n
\end{bmatrix}.   </math>

Performing the above matrix multiplication we end up with the following result:<math display="block">\begin{align}
        A \begin{bmatrix} \boldsymbol{\alpha}_1 & \boldsymbol{\alpha}_2 & \cdots & \boldsymbol{\alpha}_n \end{bmatrix} =  \begin{bmatrix} \lambda_1 \boldsymbol{\alpha}_1 & \lambda_2\boldsymbol{\alpha}_2 & \cdots & \lambda_n \boldsymbol{\alpha}_n  \end{bmatrix}


\end{align}</math>Taking each component of the block matrix individually on both sides, we end up with the following:
:<math>A\boldsymbol{\alpha}_i = \lambda_i \boldsymbol{\alpha}_i \qquad (i=1,2,\dots,n).</math>

So the column vectors of <math>P</math> are [[right eigenvector]]s of {{nowrap|<math>A</math>,}} and the corresponding diagonal entry is the corresponding [[eigenvalue]]. The invertibility of <math>P</math> also suggests that the eigenvectors are [[linearly independent]] and form a basis of {{nowrap|<math>F^{n}</math>.}} This is the necessary and sufficient condition for diagonalizability and the canonical approach of diagonalization. The [[row vector]]s of <math>P^{-1}</math> are the [[left eigenvector]]s of {{nowrap|<math>A</math>.}}

When a complex matrix <math>A\in\mathbb{C}^{n\times n}</math> is a [[Hermitian matrix]] (or more generally a [[normal matrix]]), eigenvectors of <math>A</math> can be chosen to form an [[orthonormal basis]] of {{nowrap|<math>\mathbb{C}^n</math>,}} and <math>P</math> can be chosen to be a [[unitary matrix]]. If in addition, <math>A\in\mathbb{R}^{n\times n}</math> is a real [[symmetric matrix]], then its eigenvectors can be chosen to be an orthonormal basis of <math>\mathbb{R}^n</math> and <math>P</math> can be chosen to be an [[orthogonal matrix]].

For most practical work matrices are diagonalized numerically using computer software. [[eigenvalue algorithm|Many algorithms]] exist to accomplish this.