Jordan canonical form of a 5 x 5 matrix

Question

If A is a complex 5 × 5 matrix with characteristic polynomial f = (x − 2)$^3$ (x + 7)$^2$ and minimal polynomial p = (x − 2)$^2$ (x + 7), what is the Jordan form for A?

I have got an answer very close to the correct one of \begin{bmatrix} 2 & 0 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & -7 & 0 \\ 0 & 0 & 0 & 0 & -7 \\ \end{bmatrix} However I was wondering why the answer is not \begin{bmatrix} 2 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & -7 & 0 \\ 0 & 0 & 0 & 0 & -7 \\ \end{bmatrix} Where does the 1 come from in the second row?

Thanks

Answer 1

A matrix is diagonalizable iff its minimal polynomial factors as a product of different linear factors, so for your matrix to be similar to the lower matrix its minimal polynomial would have to be $\;(x-2)(x+7)\;$.

The one in the second row of the correct matrix is what "breaks" the diagonalizability of your matrix, and it comes from something called "generalized eigenvector", meaning: there are not three linearly independet eigenvectors related to the eigenvalue $\;2\;$, only two of them. The "third one" is the so-called generalized eigenvector.

Answer 2

Check your notes and review the minimal polynomial. Note that a matrix will be diagonalizable if and only if every (distinct) factor in the minimal polynomial has exponent $1$.

In general, the exponent of a factor in the minimal polynomial tells you the size of the largest associated Jordan block.

Answer 3

We see that $\lambda = 2$ is a an eigenvalue with algebraic multiplicity $3$ since $(x-2)^3$ shows up in the characteristic polynomial but the geometric multiplicity if only $2$ since the minimal polynomial doesn't split into distinct linear factors. Thus the matrix is not diagonalizable and the eigenvalue $\lambda= 2$ has a degenerative eigenspace. That $1$ off of the diagonal expresses these qualities.

Answer 4

For each eigenvalue$~\lambda$, you may have one or more Jordan blocks, and the may have different sizes. The exponent of $(x-\lambda)$ in the characteristic polynomial gives the sum of all sizes of Jordan blocks for$~\lambda$, while the exponent of $(x-\lambda)$ in the minimal polynomial gives the size of the largest Jordan block for$~\lambda$.

So here for $\lambda=-7$ all blocks have size$~1$ (meaning "this part" is diagonalisable) and there are two such blocks. For $\lambda=2$ the largest block must have size$~2$ (so that you must have at least one nonzero off-diagonal coefficient, and this part is not diagonalisable), and the sum of the block sizes must be $3$, which is only possible with one block of size$~2$ and another block of size$~1$. This gives you all Jordan block sizes, and (up to permutation of the Jordan blocks) the Jordan normal form.