# example of non-diagonalizable matrices

Some matrices with real entries which are not diagonalizable over $\mathbb{R}$ are diagonalizable over the complex numbers $\mathbb{C}$.

For instance,

 $A=\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}$

has $\lambda^{2}+1$ as characteristic polynomial. This polynomial doesn’t factor over the reals, but over $\mathbb{C}$ it does. Its roots are $\lambda=\pm i$.

Interpreting the matrix as a linear transformation $\mathbb{C}^{2}\to\mathbb{C}^{2}$, it has eigenvalues $i$ and $-i$ and linearly independent eigenvectors $(1,-i)$, $(-i,1)$. So we can diagonalize $A$:

 $A=\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}=\begin{pmatrix}1&-i\\ -i&1\end{pmatrix}\begin{pmatrix}i&0\\ 0&-i\end{pmatrix}\begin{pmatrix}.5&.5i\\ .5i&.5\end{pmatrix}$

But there exist real matrices which aren’t diagonalizable even if complex eigenvectors and eigenvalues are allowed.

For example,

 $B=\begin{pmatrix}0&1\\ 0&0\end{pmatrix}$

cannot be written as $UDU^{-1}$ with $D$ diagonal.

In fact, the characteristic polynomial is $\lambda^{2}$ and it has only one double root $\lambda=0$. However the eigenspace corresponding to the $0$ (kernel) eigenvalue has dimension 1.

$B\begin{pmatrix}v_{1}\\ v_{2}\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}\iff v_{2}=0$ and thus the eigenspace is $ker(B)=span_{\mathbb{C}}\left\{(1,0)^{T}\right\}$, with only one dimension.

There isn’t a change of basis where $B$ is diagonal.

Title example of non-diagonalizable matrices ExampleOfNondiagonalizableMatrices 2013-03-22 14:14:30 2013-03-22 14:14:30 cvalente (11260) cvalente (11260) 14 cvalente (11260) Example msc 15-00