Questions on Jordan forms

Question

I am studying Jordan form of a matrix from wiki. I am wondering how could two matrices have same eigenvalues with same multiplicities, but have different Jordan form? Also, if two matrices have different Jordan forms, does this mean that these two matrices are not similar? Thank you so much!

Answer 1

how could two matrices have same eigenvalues with same multiplicities, but have different Jordan form?

Zero matrix $\begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}$ and $\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$ have the only eigenvalue (zero) with algebraic multiplicity 2, but they are not similar. The zero matrix is not similar to any non-zero matrix.

In general, Jordan matrix consists of blocks with $\lambda_i$'s on the diagonal and $1$'s above the diagonal (if the block size is larger than 1); you can have more blocks with the same $\lambda_i$. The number and structure of all the blocks determine the similarity class. The zero matrix consists, in this language, of two diagonal blocks $1\times 1$, and the second matrix above of one block $2\times 2$.

Also, if two matrices have different Jordan forms, does this mean that these two matrices are not similar?

Yes, that's correct.

Answer 2

For both matrices below, the characteristic polynomial is $(x-3)^4$ and the minimal polynomial is $(x-3)^2.$ However, they are not similar, the Jordan normal forms are different. The exponents in the minimal polynomial tell us only the size of the largest Jordan block with a given eigenvalue. They do not tell us the full partition of Jordan block sizes. Below, for $A$ we have $2 + 2$ but for $B$ we have $1+1+2.$

$$ A \; = \; \left( \begin{array}{cccc} 3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 3 \end{array} \right). $$

$$ B \; = \; \left( \begin{array}{cccc} 3 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 3 \end{array} \right). $$

I do not see how to get such an example in dimension 3 or 2, so in those dimensions characteristic and minimal polynomials suffice. Note that these examples are not diagonalizable, they are already in Jordan Normal Form, these are the best that can be done.

Answer 3

There is something called the Jordan decomposition, (or Jordan-Chevalley decomposition) which says that every linear operator $T$ can be written uniquely as $T=S+N$ where $S$ is diagonalizable (over an algebraically closed field, over a general field just semisimple) and $N$ is nilpotent. Now the eigenvalues and their multiplicities determine and are determined by $S$. But there are still many possiblities for $N$. That is there may be vectors that are mapped into eigenspaces without being eigenvectors themselves.