DLA Examples

🔗

Section 33.5 Examples

🔗

In this section.

🔗

Subsection 33.5.1 Determining the form indirectly

🔗

Here we will carry out the indirect analysis method described in Subsection 33.4.2 for a matrix

$A$ that is similar to a triangular-block nilpotent form matrix

$N$ as in Discovery 33.1.

🔗

Example 33.5.1.

Suppose $A$ is an $19 \times 19$ matrix that is similar to the triangular-block nilpotent matrix

$\begin{equation*} N = \begin{bmatrix} N_1 \\ \amp N_2 \\ \amp\amp N_3 \\ \amp\amp\amp N_4 \\ \amp\amp\amp\amp N_5 \\ \amp\amp\amp\amp\amp N_6 \end{bmatrix} \text{,} \end{equation*}$

where the elementary nilpotent blocks $N_1$ and $N_2$ are each $5 \times 5\text{,}$ blocks $N_3$ and $N_4$ are each $3 \times 3\text{,}$ block $N_5$ is $2 \times 2\text{,}$ and block $N_6$ is $1 \times 1\text{.}$

We will carry out our analysis as if we don't actually know the form matrix $N\text{,}$ but at the same time we will use our “secret” knowledge of $N$ to “know” the required properties of $A\text{.}$

We should first compute the powers of $A$ to determine the degree of nilpotency. We should find $A^5 = \zerovec$ while $A^4 \neq \zerovec\text{.}$ If we didn't already know the form $N\text{,}$ this computation would tell us that the largest block in $N$ is size $5 \times 5\text{,}$ because any larger block would not yet be zero at the $\nth[5]$ power, and smaller blocks become zero before the $\nth[5]$ power. After this we should compute $\rank A^4\text{,}$ which in this case we would find to be $2\text{.}$ When we back off the exponent by one like this, blocks that are smaller than $4 \times 4$ will still all be taken to zero blocks, but a single $1$ in the bottom left corner of each of the $5 \times 5$ blocks will have reappeared. So $\rank A^4 = 2$ tells us that we have two such “corner ones” in $N^4\text{,}$ one in each of two blocks of size $5 \times 5\text{.}$

We continue with rank computations of lower powers. Backing off the exponent by one again, in this case we should find $\rank A^3 = 4\text{.}$ Our expectation is that in $N^3\text{,}$ each of the two (now known) $5 \times 5$ blocks would have grown in rank from $1$ to $2$ as they each “re-grow” a subdiagonal of two $1$ s in the lower left, for a total rank contribution of $4$ from these two blocks. Since that is our total rank for $A^3\text{,}$ this tells us that there are no blocks of size $4 \times 4\text{,}$ since if there were they would have “reappeared” at the $\ird$ power with a “corner one” in the lower left of each, causing a larger jump in rank than expected.

However, calculating $\rank A^2$ should yield $8$ in this case, a larger rank than expected. The two $5 \times 5$ blocks in $N$ each gain one in rank, as their subdiagonals of $1$ s march higher up from the lower left toward the main diagonal. So their total contribution to the rank increases from $4$ to $6$ as we move from $A^3$ to $A^2\text{.}$ Since the calculated rank is $2$ larger than the expected rank of $6$ from our two known $5 \times 5$ blocks, we conclude that two new blocks have reappeared with a solitary $1$ in their respective lower-left corners. Since these blocks reappeared at the second power, they must have size $3 \times 3\text{.}$ So now we know about two $5 \times 5$ blocks and two $3 \times 3$ blocks.

Finally, with four known blocks, we would expect the rank to increase from $8$ to $12$ as we move from $A^2$ to $A^1\text{,}$ an increase by $1$ for each known block. But a calculation should reveal $\rank A = 13\text{,}$ which means that one new block has reappeared. And since it appeared at exponent one, it must have size $2 \times 2\text{.}$

From above, we have $\rank A = 13\text{,}$ so

$\begin{equation*} \nullity A = 19 - 13 = 6\text{.} \end{equation*}$

This means that $N$ should have six blocks (which we secretly know it does). And since we already have five blocks (two $5 \times 5$ blocks, two $3 \times 3$ blocks, and one $2 \times 2$ block), this means that there must be one remaining unknown block. It can't be size $2 \times 2\text{,}$ as the $\rank A^1$ calculation should have revealed all of the blocks of that size. So there must be one remaining $1 \times 1$ block, and we now have enough information to know the form matrix $N$ precisely.

🔗

Remark 33.5.2.

At $\rank A\text{,}$ there is no further backwards one can go. A $1 \times 1$ elementary nilpotent block is just a $1 \times 1$ zero matrix, and makes no contribution to the rank at any exponent. So if our number of blocks hasn't reached $\nullity A$ and we haven't yet filled out the full size of the matrix at the end of the analysis (as in the previous example), we would conclude that the rest of the form matrix is filled out by $1 \times 1$ zero blocks.

🔗

Subsection 33.5.2 Determining a transition matrix

🔗

Here are two examples of determining a transition matrix

$P$ that will put a nilpotent matrix

$A$ into triangular-block nilpotent form. In the first we will carrying out Procedure 33.4.2. We use the matrix that you explored in both Discovery 33.2 and Discovery 33.5.

🔗

Example 33.5.3. A $9 \times 9$ “bottom-up” example.

Consider the matrix

$\begin{equation*} A = \left[\begin{array}{rrrrrrrrr} 0 \amp 1 \amp 0 \amp -1 \amp 0 \amp 2 \amp 0 \amp -5 \amp 0 \\ -1 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 2 \\ 0 \amp 1 \amp 0 \amp -1 \amp 0 \amp 2 \amp 0 \amp -5 \amp 0 \\ 2 \amp 0 \amp 1 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp -2 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \\ -1 \amp 0 \amp 0 \amp 0 \amp 1 \amp 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \\ 1 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 1 \amp 0 \amp -1 \\ 0 \amp 1 \amp 0 \amp -1 \amp 0 \amp 2 \amp 0 \amp -5 \amp 0 \end{array}\right]\text{.} \end{equation*}$

To carry out Procedure 33.4.2, we first need to determine the degree of nilpotency of $A\text{.}$ In Discovery 33.2, we found this degree to be $k = 4\text{.}$ So the largest block(s) in the triangular-block nilpotent form for $A$ will have size $4\text{,}$ and we need to look for the corresponding $4$ -dimensional $A$ -cyclic subspaces of $\R^9\text{.}$

🔗

Cyclic spaces of dimension $k = 4$ .

🔗

Compute a basis for the null space of

$A^{4-1} = A^3\text{.}$ With the aid of a computer algebra system (CAS), we obtain the basis vectors

$\begin{gather*} ( 0 , 1 , 0 , 0 , 0 , 0 , 0 , 0 , 0 ), \\ ( -1 , 0 , 10 , 0 , 0 , 0 , 0 , 0 , 0 ), \\ ( 0 , 0 , 0 , 1 , 0 , 0 , 0 , 0 , 0 ), \\ ( 1 , 0 , 0 , 0 , 5 , 0 , 0 , 0 , 0 ), \\ ( 0 , 0 , 0 , 0 , 0 , 1 , 0 , 0 , 0 ), \\ ( -1 , 0 , 0 , 0 , 0 , 0 , 2 , 0 , 0 ), \\ ( 0 , 0 , 0 , 0 , 0 , 0 , 0 , 1 , 0 ), \\ ( 11 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 10 ). \end{gather*}$

🔗

As the null space of

$A^4 = \zerovec$ is all of

$\R^9\text{,}$ and the null space of

$A^3$ has dimension

$8\text{,}$ we need to extend by only a single vector. This means that there is only a single

$4 \times 4$ block in the triangular-block nilpotent form for

$A\text{.}$

🔗

Hopefully it is clear that the first standard basis vector will be independent from the eight vectors above, and this choice gives us our

$4$ -dimensional cyclic space

$\begin{equation*} W_1 = \Span \{ \uvec{e}_1, A\uvec{e}_1, A^2 \uvec{e}_1, A^3 \uvec{e}_1 \} \text{.} \end{equation*}$

🔗

Cyclic spaces of dimension $k-1 = 3$ .

🔗

Now compute a basis for the null space of

$A^{3-1} = A^2\text{.}$ With the aid of our CAS, we obtain the basis vectors

$\begin{gather*} ( -1 , 0 , 10 , 0 , 0 , 0 , 0 , 0 , 0 ), \\ ( 0 , 1 , 0 , 1 , 0 , 0 , 0 , 0 , 0 ), \\ ( 1 , 0 , 0 , 0 , 5 , 0 , 0 , 0 , 0 ), \\ ( 0 , -2 , 0 , 0 , 0 , 1 , 0 , 0 , 0 ), \\ ( -1 , 0 , 0 , 0 , 0 , 0 , 2 , 0 , 0 ), \\ ( 0 , 5 , 0 , 0 , 0 , 0 , 0 , 1 , 0 ), \\ ( 11 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 10 ). \end{gather*}$

🔗

This null space has dimension

$7\text{,}$ and we already know from the previous step that the null space of

$A^3$ has dimension

$8\text{.}$ So we again need to extend by a single vector. However, we don't want to have any overlap with the

$4$ -dimensional cyclic space we've already obtained, so before we look for “new” vectors we should extend by old vectors. The “second” vector

$A \uvec{e}_1$ from the previous step is in the null space of

$A^3$ but not in the null space of

$A^2\text{.}$ This means that

$A \uvec{e}_1$ would generate a

$3$ -dimensional cyclic space, but it would not be independent from what we already have. On the other hand, including

$A \uvec{e}_1$ with our basis above gets us up to dimension

$8\text{,}$ so there is no further to go and no new independent vectors from the null space of

$A^3$ can be obtained.

🔗

We conclude that there are no blocks of size

$3$ in the triangular-block nilpotent form for

$A\text{,}$ as we already knew from Discovery 33.2.

🔗

Cyclic spaces of dimension $k-2 = 2$ .

🔗

Now compute a basis for the null space of

$A^{2-1} = A\text{.}$ With the aid of our CAS, we obtain the basis vectors

$\begin{gather*} ( 0 , 1 , 0 , 1 , 0 , 0 , 0 , 0 , 0 ), \\ ( 0 , -2 , 0 , 0 , 0 , 1 , 0 , 0 , 0 ), \\ ( 0 , 5 , 0 , 0 , 0 , 0 , 0 , 1 , 0 ), \\ ( 2 , 0 , -2 , 0 , 1 , 0 , -1 , 0 , 1 ). \end{gather*}$

🔗

This null space has dimension

$4\text{,}$ and we know from the previous step that the null space of

$A^2$ has dimension

$7\text{.}$ So we need to extend by three vectors, but again to avoid overlap with previously obtained cyclic spaces we should first extend by those cyclic basis vectors that are in the null space of

$A^2\text{.}$ As our first generator

$\uvec{e}_1$ was chosen from the null space of

$A^4\text{,}$ the vector

$A^2 \uvec{e}_1$ will be in the null space of

$A^2\text{.}$ Which brings us up to five vectors in our quest for a basis for the null space of

$A^2\text{:}$

$\begin{gather*} ( 0 , 1 , 0 , 1 , 0 , 0 , 0 , 0 , 0 ), \\ ( 0 , -2 , 0 , 0 , 0 , 1 , 0 , 0 , 0 ), \\ ( 0 , 5 , 0 , 0 , 0 , 0 , 0 , 1 , 0 ), \\ ( 2 , 0 , -2 , 0 , 1 , 0 , -1 , 0 , 1 ), \\ ( -10 , 0 , -10 , 0 , 0 , 0 , 0 , 0 , -10 ). \end{gather*}$

🔗

We also already have a basis for the null space of

$A^2$ from our previous step, so it is just a matter of trial-and-error to choose two of those old basis vectors to include with our five current basis vectors, with the criteria being independence with our five vectors. By inspection, we choose

$\begin{gather*} \uvec{u} = ( 1 , 0 , 0 , 0 , 5 , 0 , 0 , 0 , 0 ), \\ \uvec{v} = ( -1 , 0 , 0 , 0 , 0 , 0 , 2 , 0 , 0 ), \end{gather*}$

🔗

and with our CAS we confirm that all seven vectors together are linearly independent.

🔗

Each of these two new vectors generates a

$2$ -dimensional

$A$ -cyclic space,

$\begin{align*} W_2 \amp = \Span \{ \uvec{u}, A \uvec{u} \}, \amp W_3 \amp = \Span \{ \uvec{v}, A \uvec{u} \}, \end{align*}$

🔗

and so there are two

$2 \times 2$ blocks in the triangular-block nilpotent form for

$A\text{.}$

🔗

Cyclic spaces of dimension $k-3 = 1$ .

🔗

A one-dimensional

$A$ -cyclic space of the kind we seek will need to be generated by a vector in the null space of

$A\text{.}$ As we have already filled eight of the nine columns of our transition matrix

$P\text{,}$ we only need on vector from the null space of

$A$ that is independent from the eight vectors we already have.

🔗

With the aid of our CAS, we find the null space of

$A$ to be spanned by the vectors

$\begin{gather*} ( 0 , 1 , 0 , 1 , 0 , 0 , 0 , 0 , 0 ), \\ ( 0 , -2 , 0 , 0 , 0 , 1 , 0 , 0 , 0 ), \\ ( 0 , 5 , 0 , 0 , 0 , 0 , 0 , 1 , 0 ), \\ ( 2 , 0 , -2 , 0 , 1 , 0 , -1 , 0 , 1 ). \end{gather*}$

🔗

By trial-and-error, the only one of these four vectors that works is the last one, so set

$\begin{equation*} \uvec{w} = ( 2 , 0 , -2 , 0 , 1 , 0 , -1 , 0 , 1 ) \text{.} \end{equation*}$

🔗

This vector will generate a one-dimensional

$A$ -cyclic space

$\begin{equation*} W_4 = \Span \{ \uvec{w} \} \text{.} \end{equation*}$

🔗

The transition and form matrices.

🔗

Our transition matrix

$P$ is made up of the cyclic bases for our four cyclic spaces, taken all together:

$\begin{align*} P \amp = \left[ \begin{array}{cccc|cc|cc|c} \uvec{e}_1 \amp A \uvec{e}_1 \amp A^2 \uvec{e}_1 \amp \uvec{u} \amp A \uvec{u} \amp \uvec{v} \amp A \uvec{v} \amp \uvec{w} \end{array}\right]\\ \amp = \left[ \begin{array}{rrrr|rr|rr|r} 1 \amp 0 \amp -10 \amp 0 \amp 1 \amp 0 \amp -1 \amp 0 \amp 2 \\ 0 \amp -1 \amp 0 \amp -10 \amp 0 \amp -1 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp -10 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp -2 \\ 0 \amp 2 \amp 0 \amp -10 \amp 0 \amp 2 \amp 0 \amp -2 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \amp 5 \amp 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp -1 \amp 0 \amp 0 \amp 0 \amp 4 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 2 \amp 0 \amp -1 \\ 0 \amp 1 \amp 0 \amp 0 \amp 0 \amp 1 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp -10 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 1 \end{array}\right]\text{.} \end{align*}$

🔗

Finally, without computing

$\inv{P} A P$ we know the form matrix will be

$\begin{equation*} \inv{P}AP = \begin{bmatrix} \boxed{ \begin{matrix} 0 \\ 1 \amp 0 \\ \amp 1 \amp 0 \\ \amp \amp 1 \amp 0 \end{matrix} } \\ \amp \boxed{ \begin{matrix} 0 \\ 1 \amp 0 \end{matrix} } \\ \amp \amp \boxed{ \begin{matrix} 0 \\ 1 \amp 0 \end{matrix} } \\ \amp \amp \amp \boxed{ \begin{matrix} 0 \end{matrix} } \end{bmatrix}\text{.} \end{equation*}$

🔗

Now we will do an example of carrying out Procedure 33.4.3. Since this procedure is less efficient, we will use a smaller example matrix.

🔗

Example 33.5.4. A $5 \times 5$ “top-down” example.

Consider the $5 \times 5$ matrix

$\begin{align*} A = \left[\begin{array}{rrrrr} 1 \amp -2 \amp -1 \amp -1 \amp 0 \\ -1 \amp -2 \amp -2 \amp 0 \amp 1 \\ 0 \amp 1 \amp 1 \amp 0 \amp 0 \\ 3 \amp 1 \amp 2 \amp -1 \amp -2 \\ 0 \amp -3 \amp -2 \amp -1 \amp 1 \end{array}\right]\text{.} \end{align*}$

Compute the powers of $A\text{:}$

$\begin{align*} A^2 \amp= \left[\begin{array}{rrrrr} 0 \amp 0 \amp 0 \amp 0 \amp 0 \\ 1 \amp 1 \amp 1 \amp 0 \amp -1 \\ -1 \amp -1 \amp -1 \amp 0 \amp 1 \\ -1 \amp -1 \amp -1 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \end{array}\right] \amp A^3 \amp= \zerovec\text{.} \end{align*}$

So $A$ is nilpotent of degree $3\text{.}$

The standard basis for $\R^5$ has five vectors in it, so we begin with five $A$ -cyclic spaces:

$\begin{gather*} W_1 = \Span \{ \uvec{e}_1, A\uvec{e}_1, A^2\uvec{e}_1, \dotsc \}, \quad W_2 = \Span \{ \uvec{e}_2, A\uvec{e}_2, A^2\uvec{e}_2, \dotsc \},\\ W_3 = \Span \{ \uvec{e}_3, A\uvec{e}_3, A^2\uvec{e}_3, \dotsc \}, \quad W_4 = \Span \{ \uvec{e}_4, A\uvec{e}_4, A^2\uvec{e}_4, \dotsc \},\\ W_5 = \Span \{ \uvec{e}_5, A\uvec{e}_5, A^2\uvec{e}_5, \dotsc \}. \end{gather*}$

But recall that for any matrix $B\text{,}$ the product $B \uvec{e}_j$ is precisely the $\nth[j]$ column of $B\text{.}$ So, by examining when the columns of powers of $A$ first become zero, we can use Theorem 32.6.3 to obtain a cyclic basis for each of these spaces:

$\begin{gather*} W_1 = \Span \{ \uvec{e}_1, A\uvec{e}_1, A^2\uvec{e}_1 \}, \quad W_2 = \Span \{ \uvec{e}_2, A\uvec{e}_2, A^2\uvec{e}_2 \},\\ W_3 = \Span \{ \uvec{e}_3, A\uvec{e}_3, A^2\uvec{e}_3 \}, \quad W_4 = \Span \{ \uvec{e}_4, A\uvec{e}_4 \},\\ W_5 = \Span \{ \uvec{e}_5, A\uvec{e}_5, A^2\uvec{e}_5 \}. \end{gather*}$

We don't have any dimension $1$ cyclic spaces, so let's move on to considering the collection of final vectors from each basis:

$\begin{equation*} \{ A^2\uvec{e}_1, A^2\uvec{e}_2, A^2\uvec{e}_3, A\uvec{e}_4, A^2\uvec{e}_5 \}. \end{equation*}$

Examining the columns of $A^2\text{,}$ notice that the first three vectors are all equal to each other, and the fifth vector is the negative of the those first three. Let's begin with the dependence relation

$\begin{equation*} A^2\uvec{e}_1 - A^2\uvec{e}_2 = \zerovec \text{.} \end{equation*}$

Factoring out the $A^2\text{,}$ this turns into

$\begin{equation*} A^2(\uvec{e}_1 - \uvec{e}_2) = \zerovec. \end{equation*}$

We form the new cyclic space generated by $\uvec{f}_1 = \uvec{e}_1 - \uvec{e}_2\text{,}$ and call it $U_1\text{.}$ The above relation says that $A^2 \uvec{f}_1 = \zerovec\text{,}$ so we have

$\begin{equation*} U_1 = \Span \{ \uvec{f}_1, A \uvec{f}_1 \} \text{.} \end{equation*}$

Our dependence relation involved the final vectors from $W_1$ and $W_2\text{.}$ These two spaces have equal dimension, so we can use the new space $U_1$ to replace either of $W_1$ or $W_2\text{.}$ If we decide to replace $W_1\text{,}$ our new collection of cyclic subspaces is

$\begin{gather*} U_1 = \Span \{ \uvec{f}_1, A\uvec{f}_1 \}, \quad W_2 = \Span \{ \uvec{e}_2, A\uvec{e}_2, A^2\uvec{e}_2 \},\\ W_3 = \Span \{ \uvec{e}_3, A\uvec{e}_3, A^2\uvec{e}_3 \}, \quad W_4 = \Span \{ \uvec{e}_4, A\uvec{e}_4 \},\\ W_5 = \Span \{ \uvec{e}_5, A\uvec{e}_5, A^2\uvec{e}_5 \}. \end{gather*}$

We will work more quickly through this process now.

From the dependence relation
$\begin{equation*} A^2\uvec{e}_2 - A^2\uvec{e}_3 = \zerovec \end{equation*}$
obtain new cyclic vector $\uvec{f}_2 = \uvec{e}_2 - \uvec{e}_3\text{,}$ and use it to replace $W_2\text{:}$
$\begin{gather*} U_1 = \Span \{ \uvec{f}_1, A\uvec{f}_1 \}, \quad U_2 = \Span \{ \uvec{f}_2, A\uvec{f}_2 \}, \quad W_3 = \Span \{ \uvec{e}_3, A\uvec{e}_3, A^2\uvec{e}_3 \},\\ W_4 = \Span \{ \uvec{e}_4, A\uvec{e}_4 \}, \quad W_5 = \Span \{ \uvec{e}_5, A\uvec{e}_5, A^2\uvec{e}_5 \}. \end{gather*}$
From the dependence relation
$\begin{equation*} A^2\uvec{e}_3 + A^2\uvec{e}_5 = \zerovec \end{equation*}$
obtain new cyclic vector $\uvec{f}_3 = \uvec{e}_3 + \uvec{e}_5\text{,}$ and use it to replace $W_3\text{:}$
$\begin{gather*} U_1 = \Span \{ \uvec{f}_1, A\uvec{f}_1 \}, \quad U_2 = \Span \{ \uvec{f}_2, A\uvec{f}_2 \}, \quad U_3 = \Span \{ \uvec{f}_3, A\uvec{f}_3 \},\\ W_4 = \Span \{ \uvec{e}_4, A\uvec{e}_4 \}, \quad W_5 = \Span \{ \uvec{e}_5, A\uvec{e}_5, A^2\uvec{e}_5 \}. \end{gather*}$
From the dependence relation
$\begin{equation*} A\uvec{f}_2 - A\uvec{e}_4 = \zerovec \end{equation*}$
obtain new cyclic vector $\uvec{f}_4 = \uvec{f}_2 - \uvec{e}_4\text{,}$ and use it to replace $U_2\text{.}$ However, $\uvec{f}_4$ only generates a cyclic space of dimension $1\text{,}$ and is the negative of $A^2 \uvec{e}_5\text{.}$ So discard $\uvec{f}_4\text{,}$ bringing us down to four cyclic spaces:
$\begin{gather*} U_1 = \Span \{ \uvec{f}_1, A\uvec{f}_1 \}, \quad U_3 = \Span \{ \uvec{f}_3, A\uvec{f}_3 \},\\ W_4 = \Span \{ \uvec{e}_4, A\uvec{e}_4 \}, \quad W_5 = \Span \{ \uvec{e}_5, A\uvec{e}_5, A^2\uvec{e}_5 \}. \end{gather*}$
From the dependence relation
$\begin{equation*} A \uvec{f}_1 + A \uvec{f}_3 + 2 A \uvec{e}_4 = \zerovec \end{equation*}$
obtain new cyclic vector $\uvec{f}_5 = \uvec{f}_1 + \uvec{f}_3 + 2 \uvec{e}_4 \text{,}$ and use it to replace $U_1\text{.}$ However, $\uvec{f}_5$ only generates a cyclic space of dimension $1\text{,}$ and is equal to $A^2 \uvec{e}_5 - A \uvec{e}_4 \text{.}$ So discard $\uvec{f}_5\text{,}$ bringing us down to three cyclic spaces:
$\begin{equation*} U_3 = \Span \{ \uvec{f}_3, A\uvec{f}_3 \}, \quad W_4 = \Span \{ \uvec{e}_4, A\uvec{e}_4 \}, \quad W_5 = \Span \{ \uvec{e}_5, A\uvec{e}_5, A^2\uvec{e}_5 \}. \end{equation*}$
From the dependence relation
$\begin{equation*} A \uvec{f}_3 - A \uvec{e}_4 - A^2 \uvec{e}_5 = \zerovec \end{equation*}$
obtain new cyclic vector $\uvec{f}_6 = \uvec{f}_3 - \uvec{e}_4 - A \uvec{e}_5 \text{,}$ and use it to replace $U_3\text{.}$ However, $\uvec{f}_6$ only generates a cyclic space of dimension $1\text{,}$ and is equal to $A^2 \uvec{e}_5 \text{.}$ So discard $\uvec{f}_5\text{,}$ bringing us down to two cyclic spaces:
$\begin{equation*} W_4 = \Span \{ \uvec{e}_4, A\uvec{e}_4 \}, \quad W_5 = \Span \{ \uvec{e}_5, A\uvec{e}_5, A^2\uvec{e}_5 \}. \end{equation*}$

Since $A$ does not have maximum degree of nilpotency, we know it is not similar to an elementary nilpotent form matrix. In other words, it is not possible for us to reduce down to a single $A$ -invariant space, and the fact that $A \uvec{e}_4$ and $A^2 \uvec{e}_5$ are independent confirm that we are finished at this point. All that is left is to form the transition matrix

$\begin{equation*} P = \begin{bmatrix} | \amp | \amp | \amp | \amp | \\ \uvec{e}_5 \amp A \uvec{e}_5 \amp A^2 \uvec{e}_5 \amp \uvec{e}_4 \amp A \uvec{e}_4 \\ | \amp | \amp | \amp | \amp | \end{bmatrix} = \left[\begin{array}{rrrrr} 0 \amp 0 \amp 0 \amp 0 \amp -1 \\ 0 \amp 1 \amp -1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp -2 \amp 1 \amp 1 \amp -1 \\ 1 \amp 1 \amp 0 \amp 0 \amp -1 \end{array}\right]\text{,} \end{equation*}$

where we have ordered the bases for our $A$ -cyclic spaces in the columns of $P$ according to their dimensions to ensure that the elementary nilpotent blocks appear in order of decreasing size in the form matrix $\inv{P} A P\text{:}$

$\begin{equation*} \inv{P}AP = \begin{bmatrix} \boxed{ \begin{matrix} 0 \\ 1 \amp 0 \\ \amp 1 \amp 0 \end{matrix} } \\ \amp \boxed{ \begin{matrix} 0 \\ 1 \amp 0 \end{matrix} } \end{bmatrix}\text{.} \end{equation*}$

🔗

Remark 33.5.5.

Once again, in either example of this subsection, it is not necessary to compute $\inv{P}$ to know the form of $\inv{P}A P \text{.}$ We know that each block is elementary nilpotent, and the number and sizes of the blocks is determined by the number and dimensions of the obtained $A$ -cyclic spaces.