# what is the significance of Jordan blocks (for some matrix)?

I know that each Jordan block in the Jordan canonical form of a matrix has eigenvalues on its diagonal, but there is also other information there in the form of ones on the superdiagonal or other entries in each Jordan block matrix. Why is this form preferred to the diagonal matrix with eigenvalues on the diagonal?

For some strange reason my book doesn't have an explanation of Jordan canonical form and it is not even mentioned in the index.

Also, if I have the following eigenvalues:

$\lambda_i = 1, 2, 4$

For some $4 \times 4$ matrix, then the Jordan Canonical form is:

$A = \begin{bmatrix} 1 & 0 & 0 &0 \\ 0 & 2 & 0 & 0\\ 0 & 0 & 4 & 1 \\ 0 & 0 & 0 & 4\end{bmatrix}$

$A = \begin{bmatrix} 4 & 1 & 0 &0 \\ 0 & 4 & 0 & 0\\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}$

Or

$A = \begin{bmatrix} 2 & 0 & 0 &0 \\ 0 & 4 & 1 & 0\\ 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}$

Would those be wrong?

1 . It is impossible to transform an arbitray matrix to diagonal form. For example the matrix $$\begin{pmatrix}5&1\\0&5\end{pmatrix}$$ has clearly eigenvalue $5$ (and no other eigenvalue), and it can not be transformed into diagonal form. Jordan normal form is simply the best one can do in general.
1. To your edited example: $\begin{bmatrix} 1 & 0 & 0 &0 \\ 0 & 2 & 0 & 0\\ 0 & 0 & 4 & 1 \\ 0 & 0 & 0 & 4\end{bmatrix}$ and $\begin{bmatrix} 4 & 1 & 0 &0 \\ 0 & 4 & 0 & 0\\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}$ are equivalent, the order of the Jordan blocks does not matter. But $\begin{bmatrix} 2 & 1 & 0 &0 \\ 0 & 4 & 1 & 0\\ 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}$ is different. It is not even in Jordan-normal form.
• I see. It only works for dimension preserving linear transformations, and for dimension collapsing transformations we have to settle for the Jordan block form. What about dimension expanding transformations? Is there anything for those? Like if I go from $\mathbb{R}^2$ to $\mathbb{R}^4$? Commented Apr 1, 2017 at 21:50
• @SocraticaFan Not at all. $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ is dimension-collapsing, but is trivially diagonalizable. You need Jordan block form when the matrix is deficient—some eigenvalue has geometric multiplicity less than its algebraic multiplicity.