Let A be a 7 x 7 matrix with characteristic polynomial $(t − 2)^4(3 − t)^3$. It is known that in the Jordan form of A, the largest blocks for both the eigenvalues are of order 2. Show that there are only two possible Jordan forms for A; and determine those Jordan forms.
Shouldn't there be 8 possible Jordan forms for the given matrix? The eigenvalue 3 will have one 2x2 block, and one 1x1 block. And the eigenvalue 2 can have two 2x2 blocks or, one 2x2 block and two 1x1 blocks, giving 8 possible arrangements between the two cases.
Or am I missing something? I also saw a similar question but the characteristic polynomial was $(t − 2)^4(t − 3)^3$ instead of $(t − 2)^4(3 − t)^3$. Is there any difference between these two?
Any help would be appreciated. Thanks
First of all, it is exactly the same as $(t-2)^4 (t-3)^3$ it is just a matter of convention $\det (XI_7 -A) $ or $\det(A - X I_7)$.
For your big question, it all depends on the numbers of ordered lists $(j_1 \leqslant \cdots \leqslant j_r)$ such that $\sum_{i=1}^r j_i= 4$ and $(k_1 \leqslant \cdots \leqslant k_s)$ such that $\sum_{i=1}^s k_i = 3$ (partitions).
Why? Because your Jordan decomposition will be given by diagonals $$\operatorname{diag} (J(j_1,2), \dots , J(j_r,2),J(k_1,2), \dots , J(k_r,2))$$ with $j_i \leqslant j_{i+1}$ and $k_i \leqslant k_{i+1}$ where $$J(n,\lambda) = \begin{pmatrix} \lambda & 1 & 0 & \cdots &0\\ 0 & \ddots & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & \ddots & 0 \\ \vdots & & \ddots & \ddots & 1 \\ 0 & \cdots & \cdots & 0& \lambda \end{pmatrix} \quad \text{of size } n \times n.$$ Analaysing the size of the Jordan Blocks gives you algonside with the kernel lemma gives $$\dim \ker((A-2I_7)^4) = 4 =j_1 + \dots + j_r$$ $$\dim \ker((A-3I_7)^3) = 3 =k_1 + \dots + k_s$$
But you have another assumption: the largest Jordan block's size is two for each eigenvalue, which means that $j_r = 2$ and $k_s = 2$.
This gives you for $(j_1,\dots ,j_r)$:
And for $(k_1,\dots ,k_s)$:
Which means you have $2 \times 1= 2$ possibilities! Which are $$\begin{pmatrix} 2&0&0&0&0&0&0 \\ 0&2&0&0&0&0&0 \\ 0&0&2&1&0&0&0 \\ 0&0&0&2&0&0&0 \\ 0&0&0&0&3&0&0 \\ 0&0&0&0&0&3&1 \\ 0&0&0&0&0&0&3 \end{pmatrix} \qquad \begin{pmatrix} 2&1&0&0&0&0&0 \\ 0&2&0&0&0&0&0 \\ 0&0&2&1&0&0&0 \\ 0&0&0&2&0&0&0 \\ 0&0&0&0&3&0&0 \\ 0&0&0&0&0&3&1 \\ 0&0&0&0&0&0&3 \end{pmatrix}$$