# Finding possible Jordan forms of a matrix

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 :)

• 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. – Bernard May 3 at 19:34