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If A is a matrix, Find a Jordan Canonical matrix similar to A:

$ c(x)=\text{det}(xI-A)=(x-3)^{5}(x-2)^{4} $

The information given about A is:

$ \text{rank}(A-3I)=7 $

$ \text{rank}(A-3I)^{2}=5 $

$ \text{rank}(A-3I)^{3}=4 $

$ \text{rank}(A-3I)^{4}=4 $

$ \text{rank}(A-2I)=7 $

$ \text{rank}(A-2I)^{2}=5 $

$ \text{rank}(A-2I)^{3}=5 $

I cannot find an example that will show me how to apply the theorems in the textbook to help me answer this question. :(

The textbook is: "Matrices and Linear Transformations" by Cullen.

I was googling, and came across something on wikipedia. I'm not sure if this is correct or useful:

$ rank(A-3I)=7 \therefore \text{there are 7 Jordan blocks of size 2} $

is this correct?

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No, not correct. – Arturo Magidin Jul 17 '12 at 21:59
up vote 6 down vote accepted

The eigenvalues are $3$ (with multiplicity 5) and $2$ (with multiplicity 4).

The number of Jordan blocks associated to $3$ is equal to the dimension of the eigenspace of $3$, namely, to $\mathrm{nullity}(A-3I)$. By the Rank-Nullity Theorem, since $A$ is $9\times 9$ and the rank of $A-3I$ is $7$, the nullity is $2$, so there will be two Jordan blocks associated to $3$.

How many blocks are there of size at least 1? The nullity of $A-3I$. How many blocks are there of size at least 2? The initial vectors and second vectors of every block will be annihilated by $(A-3I)^2$; so you can figure out how many blocks have size at least $2$ by comparing the nullity of $A-3I$ and the nullity of $(A-3I)^2$. Every "extra" dimension you get in $(A-3I)^2$ gives you a block of size at least $2$. We have enough information to know that the nullity of $(A-3I)^2$ is 4; that means that there are 4 vectors in a Jordan canonical basis that are annihilated by $(A-3I)^2$. Two of them are the eigenvectors we already had, so that leaves $2$ vectors that correspond to blocks of size at least $2$. Continuing this way, we see that since the nullity of $(A-3I)^3$ is $5$, we get one more vector than before. That means that there is one block of size at least $3$ (four of the vectors were already accounted for, so you just add one more); the nullity of $(A-3I)^4$ is also $5$, so there are no new vectors added. There are no blocks of size at least $4$.

So we have two Jordan blocks associated with $3$: one block of size $2$, one block of size $3$.

Do something similar with $\lambda=2$.

This can all be visualized with the dot diagram of $A$ associated to $3$. This is an array of dots where there is a column for each block, and the number of rows in each column represents the length of the corresponding cycle.

If column $i$ has $p_i$ rows, then we will have $p_1\geq p_2\geq\cdots\geq p_k$ (where $k$ is the number of columns). Let $r_1\leq r_2\leq\cdots\leq r_{p_1}$ be the number of dots in each row. You can reconstruct the diagram on the basis of the $p_i$ or on the basis of the $r_j$. The following formulas give you the $r_i$:

$$\begin{align*} r_1 &= \dim(V) - \mathrm{rank}(T-\lambda I)\\ r_j &= \mathrm{rank}((T-\lambda I)^{j-1}) - \mathrm{rank}((T-\lambda I)^j)\quad\text{if }j\gt 1. \end{align*}$$ In your example, for $\lambda=3$, we would have $$\begin{align*} r_1 &= 9-7 = 2\\ r_2 &= 7-5 = 2\\ r_3 &= 5-4 = 1\\ r_4 &= 4-4 = 0. \end{align*}$$ So the dot diagram associated to $\lambda=3$ is: $$\begin{array}{ccc} \bullet &\quad & \bullet\\ \bullet & \quad & \bullet\\ \bullet\\ \end{array}$$ The first column tells you there is a block of length $3$; the second column that there is a block of length $2$; the lack of a third column tells you there are only two blocks.

The $\bullet$ correspond to vectors in Jordan canonical basis. If $v_1$ and $v_2$ are the end vectors of the Jordan cycles associated to the two blocks, the dots correspond as follows: $$\begin{array}{lcl} \bullet\ (A-3 I)^{3-1}(v_1) &\quad&\bullet\ (A-3 I)^{2-1}(v_2)\\ \bullet\ (A-3 I)^{3-2}(v_1) &\quad&\bullet\ v_2\\ \bullet\ v_1 \end{array}$$

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Modulo any conventions you may need to follow about the order of the Jordan blocks, The Jordan canonical form will be $$J=\begin{pmatrix} 3 & p_1 \\ & 3 & p_2 \\ & & 3 & p_3 \\ & & & 3 & p_4 \\ & & & & 3 \\ & & & & & 2 & q_1 \\ & & & & & & 2 & q_2 \\ & & & & & & & 2 & q_3 \\ & & & & & & & & 2 \end{pmatrix} $$ for $p_1,\ldots,q_3\in\{0,1\}$.

Since you know that $A-3I$ has rank 7, $J-3I$ also has rank 7, so exactly one of the $p_i$s must be 0; without loss of generality it must be $p_1$ or $p_2$.

Similarly, $J-2I$ has rank 7, so exactly one of the $q_i$s must be 0; without loss of generality let it be $q_1$ or $q_2$.

This gives you four possibilities to check for which ones get the other ranks right. (There's a fairly simple principle to what works here, too, but it is better to discover it oneself).

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