Jordan Normal Form and eigenvalue 0 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.
 A: 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:
$$J = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}$$
A: 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).
A: 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.
