# Find the Jordan forms of a matrix from just the ranks of its eigenspaces

Let $$C$$ be a $$10\times 10$$ matrix whose characteristic polynomial is $$(t+2)^5(t-3)^5$$. Suppose that $$rank((C+2I)^2)=6$$ and $$rank((C-3I))=8$$. What are the possible Jordan forms of $$C$$?

This is the first problem in which I have ever encountered Jordan forms, so when it says to find all possible Jordan forms of $$C$$, I take that to mean finding every combination of possible Jordan blocks. Since $$rank((C-3I))=8$$, I know by rank nullity theorem that the dimension of the eigenspace of $$\lambda=3$$ is $$2$$. This tells me that there will be two Jordan blocks for $$\lambda=3$$, but I don't know how to determine the respective sizes because I don't know what the ranks of higher powers of $$(C-3I)$$ are. Additionally, I don't know how to glean any information from $$rank(C+2I)^2=6$$.

• So you have that for $\lambda =3$ we have 2 Jordan blocks. The question doesn't ask which ones exactly. Instead it asks which blocks they might be. Can you find the 2 possible configurations? – Klaas van Aarsen Dec 4 '18 at 23:37
• Suppose $C$ is already in Jordan normal form. Now pick some configuration of the blocks, say, 1 big block for $\lambda=-2$. What will $(C+2I)^2$ look like? And what is its rank? – Klaas van Aarsen Dec 4 '18 at 23:37
• Hint: the number of Jordan blocks of an eigenvalue is equal to its geometric multiplicity. – amd Dec 5 '18 at 3:47

Throughout the post, I use $$J(c;m)$$ to denote the Jordan block with diagonal entries $$c$$ of size $$m\times m$$.

According to the char. polynomial, the Jordan form $$J$$ of $$C$$ has five $$-2$$ and five $$3$$ on the diagonal. For Jordan blocks with diagonal entries $$3$$, yes you have 2 blocks. For Jordan blocks with diag. entries $$-2$$, easy to see that $$\DeclareMathOperator\rank{rank} \rank (J(-2;m)+2I) = m-1$$ and $$\rank ((J(-2;m)+2I)^2) = m-2$$. Restricting to the blocks with diagonal entries $$-2$$, the total dimension is $$5$$ by the char. polynomial, so $$\rank(J(-2)^2)=1$$, then there exists at least $$1$$ block $$J(-2;3)$$, and the rest of the blocks could be 2 $$J(-2;1)$$ or 1 $$J(-2;2)$$.

Conclusion: up to permutation, the possible Jordan forms are $$\DeclareMathOperator\diag{diag} \diag(J(-2;3), J(-2;2), J(3;1), J(3;4))$$, $$\diag (J(-2;3), J(-2;2), J(3;2), J(3;3))$$, $$\diag(J(-2;3), -2, -2, 3, J(3;4))$$ and $$\diag (J(-2;3), -2, -2, J(3;2), J(3;3))$$.

### UPDATE

You could prove that $$\rank\begin{bmatrix} A & O \\O & B\end{bmatrix} = \rank A + \rank B$$ where the LHS is a block diagonal matrix.

Then, for example, $$J = \begin{bmatrix} -2 & 1 &&&\\ &-2&&&\\ &&3 & * & * \\ &&&3 &* \\ &&&&3\end{bmatrix} = \mathrm {diag}(J(-2), J(3)),$$ then $$(J+2I)^2= \begin{bmatrix} 0 & 0 &&&\\ &0&&&\\ &&25 & * & * \\ &&&25 &* \\ &&&&25\end{bmatrix} = \mathrm {diag}((J(-2)+2I)^2, (J(3)+2I)^2),$$ and $$\rank((J+2I)^2) = 3$$, so $$\rank((J(-2)+2I)^2) = 3 - \rank((J(3)+2I)^2) = 3-3 = 0.$$

• What does $J(-2)$ mean? Also, since the dimension for diagonal entries $-2$ is $5$, how does this tell you that $rank(J(-2)^2)=1$? I know that this must involve the fact that $rank(C+2I)^2=6$ but I don't see how. – Ryan Greyling Dec 5 '18 at 4:42
• @RyanGreyling $J(-2)$ is the biggest block with the diagonal entries $-2$. In this question, it is a upper triangular matrix with all diagonal entries $-2$. Example if the Jordan blocks are $J(3;4), J(3;5)$, then the $J(3) = \mathrm {diag}(J(3;4), J(3;5))$. – xbh Dec 5 '18 at 4:47
• @RyanGreyling For the 2nd question, note that for the Jordan form $J =\mathrm {diag}(J(-2), J(3))$, $J+2I = \mathrm {diag}(J(0), J(5))$, so the rank equals $\mathrm {rank}(J(0)) + \mathrm {rank}(J(5))$, where $J(5)$ is of full rank, i.e. 5. – xbh Dec 5 '18 at 4:51
• Okay, I got it. The way I understand it uses the same logic in your explanation but with notation that makes more sense for me personally. Since $rank(C+2I)^2=6$, by rank-nullity theorem the dimension of the solution space for $(C+2I)^2\vec{x}=0$ is $4$. $4<5$, the algebraic multiplicity for $\lambda=-2$, so when drawing the box diagram for the Jordan form, there will be at least one column of $3$ boxes (one row above and one below). This means one of the Jordan blocks is of size $3$. – Ryan Greyling Dec 5 '18 at 19:03