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Let $$A \in M_{3 \times 3}^\Bbb R $$ be a matrix such that $$tr(A)=-3 $$

$$P_A=(A-I)^5(A+2I)^7=0$$

Also, $A^2$ is not diagonalizable.

Question: Find the A's Jordan Normal Form.

Since $A$ is a square matrix, from Cayley Hamilton theorem I can conclude that $P_A$ is $A$'s characteristic polynomial.

However, I find it hard to prove that the $minimal$ polynomial is, to find the Jordan Canonical Form.

Since the minimal polynomial has to divide the characteristic polynomial, the minimal polynomial must be: $$(t-1)^x(t+2)^y$$ where $$x+y\le 3$$ Since a Jordan Normal Form's rank can't be bigger than the original matrix's rank.

All in all, my options for the minimal polynomial are:

$$M_A=\begin {cases} (t-1) \\ (t+2) \\ (t-1)^2(t+2)\\(t-1)(t+2)^2\end {cases}$$

How do I choose between these options? And what does it have to do with the $trace$ of $A$?

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Actually the characteristic polynomial can't have a degree higher than 3 in this case, but we can say for sure that the characteristic polynomial must divide that larger polynomial so it must share the factors in some combination. So we still need to pick some configuration from $1$ and $-2$.

Trace is sum of eigenvalues. What combinations of sums of $1$ and $-2$ can give $-3$? The number of combinations is given by the coefficient for exponent $-3$ in $(x^1+x^{-2})^3$ and we easily see that we must pick $1$ once and $-2$ twice. The coefficient is 3 and corresponds to that we can place the $1$ eigenvalue on 3 distinct spots on the diagonal.

Further, if $A^2$ is not diagonalizable, then A can't either be diagonalizable ( as a Jordan block when squared remains the same size ). This means the possible configurations for generalized eigenspace sizes are (2,1) and 3. But it can't be 3 since the eigenvalue occuring the most times (-2) occurs only 2 times. So the only possibility left is $$\left[\begin{array}{rrr} 1&0&0\\0&-2&1\\0&0&-2 \end{array}\right]$$

Edit Actually we can't say the characteristic polynomial must divide that larger polynomial, but that the minimal polynomial must divide it ( since that is the definition of minimal polynomial ). And the minimal polynomial is related to the maximum Jordan block sizes.

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  • $\begingroup$ Thank you. So you're saying it's that easy? Just looking at the trace and picking the right combination for the Jordan Normal Form which fits the Minimal Polynomial (or vice verca)? $\endgroup$ Commented Oct 10, 2015 at 18:01
  • $\begingroup$ How can we say for sure that the characteristic polynomial must divide that larger polynomial? Is it from the Ciley - Hamilton theorem? $\endgroup$ Commented Oct 10, 2015 at 18:06
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    $\begingroup$ In this example there were not many combinations and we were also given the possible eigenvalues. If we had a large matrix the trace would only give us one piece of information and then it may be more difficult if we could only know which eigenvalues and not how many times each of them occured. $\endgroup$ Commented Oct 10, 2015 at 18:06

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