Editing Diagonalizable matrix (section)

=== 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.