# Determine the JCF from $\operatorname{rank}(A-\lambda I)^j$

Given the following facts about a square complex matrix $C$:

• Characteristic polynomial of C is $x^5 (x-1)^4$
• $\operatorname{rank}(C)=7$
• $\operatorname{rank}(C-I)=6$
• $\operatorname{rank}(C^2)=5$

Find the Jordan Canonical Form of $C$.

My attempt so far to find the JCF of C:
From the characteristic polynomials I know that the eigenvalues are $0$ and $1$ with algebraic multiplicities $5$ and $4$ respectively.

For evalue $\lambda =0$:
From $\operatorname{rank}(C)=7$, I know that the geometric multilicity is $9-7=2$. So J-Blocks are either: $J_1(0) \oplus J_4(0)$ or $J_2(0) \oplus J_3(0)$.

For evalue $\lambda =1$:
From $\operatorname{rank}(C-I)=6$, I know that the geometric multilicity is $9-6=3$. So J-Blocks are: $J_2(1) \oplus J_1(1) \oplus J_1(1)$.

I'm confused on how to use the last fact that $\operatorname{rank}(C^2)=5$ to determine which of the to JNFs for $\lambda=0$ to choose.

I read the section on the Uniqueness of the Jordan Normal blocks in Wikipedia where it explained the significance of the $\operatorname{rank}(A-\lambda I)^j$ in determining the JNF form but I think I require to know $k_1$ in $(A-\lambda I)^{k_1}=0$ to do so.

## 1 Answer

Lets $C = MJM^{-1}$. Then $C^2 = MJ^2M^{-1}$, where $J$ is the Jordan form of $C$.

Now for the first case, $J_1(0) \oplus J_4(0)$

$$J_4^2(0) = \left[\begin{array}\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0 \end{array}\right]^2 = \left[\begin{array}\ 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0\\ 0&0&0&0 \end{array}\right]$$

For the second case $J_2(0) \oplus J_3(0)$ $$J_3^2(0 ) = \left[\begin{array}\ 0&1&0\\ 0&0&1\\ 0&0&0 \end{array}\right]^2 = \left[\begin{array}\ 0&0&1\\ 0&0&0\\ 0&0&0 \end{array}\right]$$

$$J_2^2(0) = \left[\begin{array}\ 0&1\\ 0&0 \end{array}\right]^2 = \left[\begin{array}\ 0&0\\ 0&0 \end{array}\right]$$

So if $J(0) = J_1(0) \oplus J_4(0)$, then rank($C^2$) = 6 and if $J(0) = J_2(0) \oplus J_3(0)$, rank($C^2$) = 5.

• How did you calculate $\operatorname{rank}(C)^2$? I know that $\operatorname{rank}(C)^2 = \operatorname{rank}(J)^2$ since $C$ and $J$ are similar, but I am confused on how you derived that without considering the $J$-blocks of $\lambda = 1$. – Nick Aug 21 '18 at 16:34
• @Nick, Just do the Kronecker product of Jordan blocks and check for yourself. Given the simplicity of these blocks, you can see that $J_4^2$ adds a single new row full of $0$s, and $J_1^2$ do not do this. Whereas $J_3^2$ and $J_2^2$ each add extra row of $0$s. J blocks of $\lambda = 1$ have all $1$s on the diagonal and do not extra zero rows when squared. So all in all, first case adds one row with all $0$s where as second case adds two rows with all $0$s. What I mean by adding row of $0$s is squaring it makes one row go all $0$s. If you want I can do the $9x9$ matrix multiplication for proof – artha Aug 21 '18 at 17:36
• and with each new row going all $0$s, rank is reduced by 1 – artha Aug 21 '18 at 17:38