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Find the Jordan forms of a matrix $A$ subject to the following conditions:

  1. the characteristic polynomial is $(x-1)^4(x+3)^5$.

  2. matrix $A-I$ has nullity $4$ and matrix $A+3I$ has nullity $1$.

Since $(x-1)^4(x+3)^5$ is the characteristic polynomial we get that a Jordan form of $A$ is

$$\begin{bmatrix}1&a_1\\&1&a_2\\&&1&a_3\\&&&1&0\\&&&&-3&a_4\\&&&&&-3&a_5\\&&&&&&-3&a_6&\\&&&&&&&-3&a_7\\&&&&&&&&-3\end{bmatrix},$$

where $a_i=0,1$ for all $i$. Now, since $A-I$ has nullity $4$, we deduce that

$$\begin{bmatrix}0&a_1\\&0&a_2\\&&0&a_3\\&&&0&0\\&&&&-4&a_4\\&&&&&-4&a_5\\&&&&&&-4&a_6&\\&&&&&&&-4&a_7\\&&&&&&&&-4\end{bmatrix},$$

does as well, hence $a_1,a_2,a_3=0$. Since $A+3I$ has nullity $1$, by similar logic, we can deduce that $a_4,a_5,a_6,a_7 = 1$, giving us the Jordan form of

$$\begin{bmatrix}1&0\\&1&0\\&&1&0\\&&&1&0\\&&&&-3&1\\&&&&&-3&1\\&&&&&&-3&1&\\&&&&&&&-3&1\\&&&&&&&&-3\end{bmatrix}.$$

Could anyone tell me if my logic for finding the Jordan form is correct? Thank you :)

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    $\begingroup$ Yes, in short, you should retain that the nullity of $A-\lambda I$, where $\lambda$ is an eigenvalue of $A$, is the number of Jordan blocks corresponding to this eigenvalue. $\endgroup$ – Bernard May 3 at 19:34

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