find the Jordan form of A?1

Question

Let A be the complex matrix

Let $$A =\begin{bmatrix} 2&0&0&0&0&0 \\1&2&0&0&0&0\\-1 &0&2&0&0&0\\0&1&0&2&0&0\\1&1&1&1&2&0\\0&0&0&0&1&-1\end{bmatrix}$$

Now find the Jordan form of A?

My attempt is the characteristic polynomial of $A$ =$( λ-2)^5$$(λ +1)$ and minimal polynomial is $(λ+1) (λ-2)^2$....after that I am not able to proceed,,,,

pliz help me....

Answer 1

With the given info about the characteristic and minimal polynomials, your options for JCF of that matrix are:

$$\begin{pmatrix} 2&1&0&0&0&0 \\0&2&0&0&0&0\\0 &0&2&0&0&0\\0&0&0&2&0&0\\0&0&0&0&2&0\\0&0&0&0&0&-1\end{pmatrix}\;,\;\;\begin{pmatrix} 2&1&0&0&0&0 \\0&2&0&0&0&0\\0 &0&2&1&0&0\\0&0&0&2&0&0\\0&0&0&0&2&0\\0&0&0&0&0&-1\end{pmatrix}$$

The first one: one block of size $2$, three of size $1$, and the second option is two blocks each of size $2$, one of size $\;1\;$ . These are the unique possible JCF's, up to order of blocks,for your matrix with the given info.

If you already calculate the dimension of the eigenspace $\;V_2\;$ corresponding to $\;\lambda=2\;$, thenL the first case above is for $\;\dim V_2=4\;$ , and the second case for $\;\dim V_2=3\;$

Answer 2

You have to consider the sequence of subspaces: $$\{0\}\subset\ker(A-2I)\subset\ker(A-2I)^2\subset\ker(A-2I)^3\subset\dots$$ and the corresponding sequence of their dimensions, $\;0\le d_1\le d_2\le d_3\le\dots$

These sequences are strictly increasing to begin with, then stabilise. A fundamental result is this:

With the above notations, for each $k\ge 1$, the number $n_k$ of Jordan blocks of size $\ge k$ is $$n_k=d_k-d_{k-1}.$$

Edit: This procedure is also the starting point to determine a Jordan basis.

Answer 3

$$ \frac{1}{81} \left( \begin{array}{cccccc} 10&6&9&9&-27&81 \\ -108&-81&-81&0&0&0 \\ -38&-39&-18&-18&-27&0 \\ -48&-45&-27&-27&0&0 \\ -18&-27&0&0&0&0 \\ -27&0&0&0&0&0 \\ \end{array} \right) \left( \begin{array}{cccccc} 2&0&0&0&0&0 \\ 1&2&0&0&0&0 \\ -1&0&2&0&0&0 \\ 0&1&0&2&0&0 \\ 1&1&1&1&2&0 \\ 0&0&0&0&1&-1 \\ \end{array} \right) \left( \begin{array}{cccccc} 0&0&0&0&0&-3 \\ 0&0&0&0&-3&2 \\ 0&-1&0&0&3&2 \\ 0&1&0&-3&2&0 \\ 0&0&-3&2&1&0 \\ 1&0&-1&1&0&0 \\ \end{array} \right) = \left( \begin{array}{cccccc} -1&0&0&0&0&0 \\ 0&2&0&0&0&0 \\ 0&0&2&1&0&0 \\ 0&0&0&2&1&0 \\ 0&0&0&0&2&1 \\ 0&0&0&0&0&2 \\ \end{array} \right) $$