Editing Diagonalizable matrix (section)

=== 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.