I understand the processes of putting a matrix into Jordan normal form and forming the transformation matrix associated to "diagonalizing" the matrix. So here's my question:

Why is it that when you have an eigenvalue x=0 with algebraic multiplicity greater than 1, that you don't put a 1 in the superdiagonal of the JNF matrix but when the eigenvalue is non-zero and satisfies the same properties, we put a 1 in the superdiagonal of the Jordan normal form?

My professor posted solutions to an assignment involving finding a matrix exponential, but the JNF of a matrix had eigenvalue x=0 with algebraic multiplicity of 3,yet had no entries of 1 along the superdiagonal.

In advance, I would like to thank you for your help.

  • $\begingroup$ Can you post the matrix? $\endgroup$ – Amzoti Oct 28 '13 at 3:41
  • $\begingroup$ I do not know how to post a visual using latex but the matrix is a 4x4 matrix with 1's in the first column and 0's everywhere else. $\endgroup$ – tamefoxes Oct 28 '13 at 3:45
  • $\begingroup$ the solutions can be found here: lalashan.mcmaster.ca/theobio/3F03/images/6/6e/3fa3s_2013.pdf The question is 3c $\endgroup$ – tamefoxes Oct 28 '13 at 3:48
  • $\begingroup$ Because we can find a null space the contains three linearly independent eigenvectors. $\endgroup$ – Amzoti Oct 28 '13 at 3:51
  • $\begingroup$ Thanks for the help but how does that explain a matrix in JNF with zero eigenvalues not have 1s above the diagonal with 0 entries. ie. JNF of matrix in 3c, why does it have only diagonal elements when eigenvalues with multiplicity >1 have 1's on the superdiagonal corresponding to eigenvalues with multiplicity >1 $\endgroup$ – tamefoxes Oct 28 '13 at 3:55

We have a single eigenvalue of $\lambda_1 = 1$ and a triple eigenvalue of $\lambda_{2,3,4} = 0$.

For $\lambda=0$, we need to find three linearly independent eigenvectors and can just use the null space of $A$ for this. We have:

$$NS(A) = NS \left(\begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix}\right)$$

This produces $v_{2,3,4} = (0,0,0,1), (0,0,1,0), (0,1,0,0)$ as three linearly independent eigenvectors, thus this matrix is diagonalizable and we can write the Jordan block using the eigenvalues down the main diagonal as:

$$J = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}$$

  • $\begingroup$ So this is true only if the eigenvectors associated to an eigenvalue with multiplicity >1 are linearly independent? $\endgroup$ – tamefoxes Oct 28 '13 at 4:05
  • 1
    $\begingroup$ Yes, recall that if you cannot find linearly independent eigenvectors, you need to us generalized ones and this will put $1's$ in the Jordan block. When we can find independent eigenvectors, we can diagonalize the matrix and have no Jordan blocks. You can see another example on the Wiki using the algebraic and geometric multiplicty language. Regards $\endgroup$ – Amzoti Oct 28 '13 at 4:07
  • $\begingroup$ Ohhhh ok, that makes sense. Thanks for all the help Amzoti, and I apologize if I seemed a bit hostile at first. In our linear algebra sequence, we were taught bilinear form rather than Jordan normal form so this differential equations course is the first time I'm encountering it. $\endgroup$ – tamefoxes Oct 28 '13 at 4:15
  • $\begingroup$ @tamefoxes: No problem! You are very welcome. Regards $\endgroup$ – Amzoti Oct 28 '13 at 4:17
  • $\begingroup$ Nice work, and great feedback, too. +1 $\endgroup$ – Namaste Oct 29 '13 at 1:12

All Jordan blocks do have their entries on the super-diagonal (if any) equal to$~1$, whether the eigenvalue of the block is$~0$ or not. What confuses you is that one can have multiple Jordan blocks for the same eigenvalue; then between adjacent Jordan blocks for the same $\lambda$ there is a super-diagonal entry that is not part of any Jordan block at all, and which therefore is$~0$. Don't make the error of thinking that taking these blocks together forms a new, larger, Jordan block with (exceptionally?) some entries$~0$ on the super-diagonal; they don't. If all Jordan blocks for have size$~1$, as is the case in the example, then none have any entries on the super-diagonal; therefore the entire super-diagonal will be zero in this case (and the Jordan form is a diagonal matrix).


I can make some counter-example about $1$.

Let's see two matrices: $A=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $B=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$.

In both cases eigenvalues are $1$ with multiplicity $2$, but $A$ has two eigenvectors such that $Ae=e (\{1,0\},\{0,1\})$ and $B$ has only one such eigenvector$(\{1,0\})$.

We put a $1$ into subdiagonal only then the corresponding block has only one eigenvector with such eigenvalue.

  • $\begingroup$ Thanks for the response but I'm still confused as to why JNF's with zero eigenvalues don't have 1's on the superdiagonal entries that are associated with those eigenvalues $\endgroup$ – tamefoxes Oct 28 '13 at 4:00
  • $\begingroup$ It depends on the matrix. You would put 1 on the superdiagonal if there were a block of dimension $\geq 2$ that had only one eigenvector. In your case any vector is eigenvector, thus we don't put 1. $\endgroup$ – Тимофей Ломоносов Oct 28 '13 at 11:13

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