0
$\begingroup$

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:

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

But what about:

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

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?

$\endgroup$

1 Answer 1

2
$\begingroup$

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.
$\endgroup$
5
  • $\begingroup$ My third matrix was not meant to have a 1 in row 1 column 2, that was a typo and I fixed it. $\endgroup$ Commented Apr 1, 2017 at 21:37
  • 2
    $\begingroup$ Then its the same again. Jordan normal form is only defined up to switching the order of the blocks. But other than that it is unique. $\endgroup$
    – Simon
    Commented Apr 1, 2017 at 21:38
  • $\begingroup$ 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 $\mathbb{R}^4$? $\endgroup$ Commented Apr 1, 2017 at 21:50
  • $\begingroup$ This works only for dimension-preserving maps, because you want to do the same change of basis on both sides of the matrix. If you allow different bases on the left and right then you have much more freedom in modifying your matrix. In particular both expanding and collapsing cases are fine, and instead of Jordan-normal-form you can for example use the "singular value decomposition" (which is always diagonal. No off-diagonals at all). $\endgroup$
    – Simon
    Commented Apr 1, 2017 at 21:54
  • $\begingroup$ @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. $\endgroup$
    – amd
    Commented Apr 2, 2017 at 0:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .