1
$\begingroup$

I'm solving a homework question that asks me to do the following:

"List the five upper Jordan canonical forms for a $4\times 4$ matrix $A$ with a real eigenvalue $\lambda$ of multiplicity $4$ and give the corresponding deficiency indices in each case."

I can't seem to understand what they mean by "$5$ upper Jordan canonical forms"? Isn't the answer unique and straightforward with the canonical form:

$$ J_= \left[ {\begin{array}{cccc} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 1 & 0 \\ 0 & 0 &\lambda & 1 \\ 0 & 0 & 0 & \lambda \end{array} } \right] $$

And what do they mean by deficiency indices? Are those the "$1$'s" that appear on top of each $\lambda$?

$\endgroup$
3
$\begingroup$

The other four Jordan canonical forms are $$ \begin{bmatrix} \lambda & 0 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \end{bmatrix} \quad \begin{bmatrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \end{bmatrix} \quad \begin{bmatrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 1 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \end{bmatrix} \quad \begin{bmatrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda \end{bmatrix} $$ The deficiency indices of a matrix $A\in\mathcal{M}_{n\times n}(\mathbb{C})$ are the numbers $$ n_{\pm}(A)=\dim\ker(A^*\mp i\cdot I_n) $$ Since $\lambda$ is assumed to be real, computing these numbers should be straight forward.

$\endgroup$
  • $\begingroup$ Do you know why for a $2x2$ matrix there are only 2 deficiency indices, for a $3x3$ matrix there are only 3 deficiency indices, yet for a $4x4$ there are 5? Perhaps the formula above holds the answer? $\endgroup$ – Arturo Feb 7 '14 at 7:31
  • $\begingroup$ There are always only two deficiency indices. $\endgroup$ – Brian Fitzpatrick Feb 7 '14 at 7:32
  • $\begingroup$ Why can't $ J_= \left[ {\begin{array}{cccc} \lambda & 0 & 0 & 0 \\ 0 & \lambda & 1 & 0 \\ 0 & 0 &\lambda & 1 \\ 0 & 0 & 0 & \lambda \end{array} } \right] $ be a Jordan canonical form? $\endgroup$ – Arturo Feb 7 '14 at 8:24
  • 1
    $\begingroup$ It can be but it's the same as the third matrix I've listed up to reordering of Jordan blocks. $\endgroup$ – Brian Fitzpatrick Feb 7 '14 at 8:27
  • $\begingroup$ Oh ok, this makes sense, I can now see why $ J_= \left[ {\begin{array}{cccc} \lambda & 0 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 &\lambda & 1 \\ 0 & 0 & 0 & \lambda \end{array} } \right] $ wouldn't produce an extra form since it is also a reordering of Jordan blocks in the second matrix! $\endgroup$ – Arturo Feb 7 '14 at 8:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.