What is the difference between the jordan normal forms of the two matrices

Question

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

$$B = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$

Do these matrices have the same Jordan form?

My confusion arises because a jordan block has the form

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

But I don't know what exactly dictates the size of our $J$ i.e. how many elements there are on the diagonals.

Is there a fundamental difference between $A$ and $B$ reflected in their jordan blocks?

Answer 1

The first is a classical Jordan form of $2\times2$, and the second diagonal, both with $0$ as the eigenvalue. For $A$ eigenvalue multiplicity is $2$. For $B$ multiplicity $1$.

Consider also, both the characteristic and minimal polynomials of them: $$\chi_A(x)=\mu_A(x)=x^2$$ and $$\chi_B(x)=x^2\quad,\quad \mu_B(x)=x$$

Answer 2

This comes down to the difference between the minimal polynomial vs the characteristic polynomial. They determinant of xI-A and xI-B are both x^2, but look at their minimal polynomials. That governs how the Jordan blocks break up.

Answer 3

The ones and zeros in the super-diagonal decide the blocks. The 1 in the $A_{1,2}$ decides that 1 and 2 are in the same block and therefore A has a $2 \times 2$ block. The zero in $B_{1,2}$ decide we have a "block breaker" between 1 and 2.

More complicated example to try and clarify:

$$A = \left[\begin{array}{ccccccc} 0&1&0&0&0&0&0\\ 0&0&1&0&0&0&0\\ 0&0&0&1&0&0&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&0&1&0\\ 0&0&0&0&0&0&1\\ 0&0&0&0&0&0&0 \end{array}\right]$$

Counting the ones we have "connections" between (1,2), (2,3), (3,4), (5,6),(6,7). Calculating $\sum_k A^k$ will then "propagate" these connections in one direction. If you do this by hand you will see that the 0 actually is kind of a "block breaker":

$$\sum_{k=1}^7 A^k = \left[\begin{array}{ccccccc} 0&1&1&1&0&0&0\\ 0&0&1&1&0&0&0\\ 0&0&0&1&0&0&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&0&1&1\\ 0&0&0&0&0&0&1\\ 0&0&0&0&0&0&0 \end{array}\right]$$

We can also "propagate" in the other direction:

$$\sum_{k=1}^7 (A^T)^k = \left[\begin{array}{ccccccc} 0&0&0&0&0&0&0\\ 1&0&0&0&0&0&0\\ 1&1&0&0&0&0&0\\ 1&1&1&0&0&0&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&1&0&0\\ 0&0&0&0&1&1&0 \end{array}\right] $$

These summed together with the identity matrix become:

$$I + \sum_{k=1}^7 (A^T)^k + A^k = \left[\begin{array}{ccccccc} 1&1&1&1&0&0&0\\ 1&1&1&1&0&0&0\\ 1&1&1&1&0&0&0\\ 1&1&1&1&0&0&0\\ 0&0&0&0&1&1&1\\ 0&0&0&0&1&1&1\\ 0&0&0&0&1&1&1 \end{array}\right]$$

And now the block structure is really visible.