Editing Diagonalizable matrix (section)

== Application to matrix functions ==
Diagonalization can be used to efficiently compute the powers of a matrix {{nowrap|<math>A = PDP^{-1}</math>:}}

: <math>\begin{align} 
  A^k &= \left(PDP^{-1}\right)^k = \left(PDP^{-1}\right) \left(PDP^{-1}\right) \cdots \left(PDP^{-1}\right) \\
      &= PD\left(P^{-1}P\right) D \left(P^{-1}P\right) \cdots \left(P^{-1}P\right) D P^{-1} = PD^kP^{-1},
\end{align}</math>

and the latter is easy to calculate since it only involves the powers of a diagonal matrix. For example, for the matrix <math>A</math> with eigenvalues <math>\lambda = 1,1,2</math> in the example above we compute:

: <math>\begin{align}
  A^k = PD^kP^{-1}
  &= \left[\begin{array}{rrr}
       1 & \,0 &          1 \\
       1 &   2 &          0 \\
       0 &   1 & \!\!\!\!-1
     \end{array}\right]
     \begin{bmatrix} 1^k & 0 & 0 \\ 0 & 1^k & 0 \\ 0 & 0 & 2^k \end{bmatrix}
     \left[\begin{array}{rrr}
       1 & \,0 &          1 \\
       1 &   2 &          0 \\
       0 &   1 & \!\!\!\!-1
     \end{array}\right]^{-1} \\[1em]
  &= \begin{bmatrix}
         2 - 2^k & -1 + 2^k &  2 - 2^{k + 1} \\
               0 &        1 &              0 \\
        -1 + 2^k &  1 - 2^k & -1 + 2^{k + 1}
      \end{bmatrix}.
\end{align}</math>

This approach can be generalized to [[matrix exponential]] and other [[matrix function]]s that can be defined as power series. For example, defining {{nowrap|<math display="inline">\exp(A) = I + A + \frac{1}{2!}A^2 + \frac{1}{3!}A^3 + \cdots</math>,}} we have:
: <math>\begin{align}
  \exp(A) = P \exp(D) P^{-1}
  &= \left[\begin{array}{rrr}
       1 & \,0 &          1 \\
       1 &   2 &          0 \\
       0 &   1 & \!\!\!\!-1
     \end{array}\right]
     \begin{bmatrix} e^1 & 0 & 0 \\ 0 & e^1 & 0 \\ 0 & 0 & e^2 \end{bmatrix}
     \left[\begin{array}{rrr}
       1 & \,0 & 1\\
       1 & 2 & 0\\
       0 & 1 & \!\!\!\!-1
     \end{array}\right]^{-1} \\[1em]
  &= \begin{bmatrix}
       2 e - e^2 & -e + e^2 & 2 e - 2 e^2 \\
               0 &        e &           0 \\
        -e + e^2 &  e - e^2 &  -e + 2 e^2
     \end{bmatrix}.
\end{align}</math>

This is particularly useful in finding closed form expressions for terms of [[linear recursive sequences]], such as the [[Fibonacci number#Matrix form|Fibonacci numbers]].

=== Particular application ===
For example, consider the following matrix:

:<math>M = \begin{bmatrix}a & b - a\\ 0 & b\end{bmatrix}.</math>

Calculating the various powers of <math>M</math> reveals a  surprising pattern:

:<math>
  M^2 = \begin{bmatrix}a^2 & b^2-a^2 \\ 0 &b^2 \end{bmatrix},\quad
  M^3 = \begin{bmatrix}a^3 & b^3-a^3 \\ 0 &b^3 \end{bmatrix},\quad
  M^4 = \begin{bmatrix}a^4 & b^4-a^4 \\ 0 &b^4 \end{bmatrix},\quad
  \ldots
</math>

The above phenomenon can be explained by diagonalizing {{nowrap|<math>M</math>.}}  To accomplish this, we need a basis of <math>\R^2</math> consisting of eigenvectors of {{nowrap|<math>M</math>.}}  One such eigenvector basis is given by

:<math>
  \mathbf{u} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \mathbf{e}_1,\quad
  \mathbf{v} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \mathbf{e}_1 + \mathbf{e}_2,
</math>

where '''e'''<sub>''i''</sub> denotes the standard basis of '''R'''<sup>''n''</sup>. The reverse change of basis is given by

:<math>\mathbf{e}_1 = \mathbf{u},\qquad \mathbf{e}_2 = \mathbf{v} - \mathbf{u}.</math>

Straightforward calculations show that

:<math>M\mathbf{u} = a\mathbf{u},\qquad M\mathbf{v} = b\mathbf{v}.</math>

Thus, ''a'' and ''b'' are the eigenvalues corresponding to '''u''' and '''v''', respectively. By linearity of matrix multiplication, we have that

:<math> M^n \mathbf{u} = a^n \mathbf{u},\qquad M^n \mathbf{v} = b^n \mathbf{v}.</math>

Switching back to the standard basis, we have

:<math>\begin{align}
  M^n \mathbf{e}_1 &= M^n \mathbf{u} = a^n \mathbf{e}_1, \\
  M^n \mathbf{e}_2 &= M^n \left(\mathbf{v} - \mathbf{u}\right) = b^n \mathbf{v} - a^n\mathbf{u} = \left(b^n - a^n\right) \mathbf{e}_1 + b^n\mathbf{e}_2.
\end{align}</math>

The preceding relations, expressed in matrix form, are

:<math>M^n = \begin{bmatrix} a^n & b^n - a^n \\ 0 & b^n \end{bmatrix}, </math>

thereby explaining the above phenomenon.