DLA Discovery guide

Discovery guide 33.3 Discovery guide

Discovery 33.1.

Consider $19 \times 19$ matrix

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

in triangular-block nilpotent form, 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 $1 \times 1\text{.}$

Suppose $A$ is another $19 \times 19$ matrix that is similar to $N\text{.}$

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(a)

What is the rank of $A\text{?}$

What is nullity of $A$ (i.e. the dimension of the null space of $A$ )?

Describe how and why this number determines the number of blocks in $N\text{.}$

Hint

For the rank, consider Corollary 26.5.6.

For the nullity, see the Rank-Nullity Theorem.

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(b)

What is the first power of $A$ that is zero? Describe how and why this number determines the size of the largest block in $N\text{.}$

Hint

Consider Proposition 31.5.2, and recall that powers of $N$ can be computed by taking powers of the blocks.

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(c)

Let $k$ represent the degree of nilpotency of $A$ from Task b.

What is the rank of $A^{k-1}\text{?}$ Describe how and why this number determines the number of blocks in $N$ of the largest size.

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(d)

Continue working backwards, computing $\rank A^{k-2}\text{,}$ $\rank A^{k-3}\text{,}$ etc.. Describe how and why the jumps in these numbers determine the size and number of blocks in $N$ of second largest size, then of third largest size, etc..

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In Discovery 33.1 we worked backwards from the form matrix to uncover some patterns that must hold for any matrix that is similar to that specific form matrix. Now let's take what we've learned and work forwards, from a matrix to the form matrix.

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Discovery 33.2.

Consider the $9 \times 9$ matrix

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

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

Because of the size of $A\text{,}$ it will be more efficient for you to use your favourite computer algebra system (CAS) to perform the necessary calculations. If you are familiar with Sage, below is a Sage cell that you can use for your computations, with our $A$ matrix already set up.

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A = matrix([
    [  0, 1, 0, -1, 0, 2, 0, -5,  0 ],
    [ -1, 0, 0,  0, 0, 0, 0,  0,  2 ],
    [  0, 1, 0, -1, 0, 2, 0, -5,  0 ],
    [  2, 0, 1,  0, 0, 0, 0,  0, -2 ],
    [  0, 0, 0,  0, 0, 0, 0,  0,  0 ],
    [ -1, 0, 0,  0, 1, 0, 0,  0,  1 ],
    [  0, 0, 0,  0, 0, 0, 0,  0,  0 ],
    [  1, 0, 0,  0, 0, 0, 1,  0, -1 ],
    [  0, 1, 0, -1, 0, 2, 0, -5,  0 ]
]);
print(A);

Confirm that $A$ is nilpotent, and determine the degree of nilpotency. Compute the rank of $A\text{,}$ and from that determine the nullity of $A\text{.}$ Finally, compute the ranks of the powers of $A\text{,}$ up to the degree of nilpotency.

From all of this information, you should be able to determine the exact triangular-block nilpotent form matrix to which $A$ must be similar. (However, note that this information will not allow us to determine a transition matrix $P\text{.}$ )

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Discovery 33.3.

From Chapter 28 we know that we need a complete set of independent, $A$ -invariant subspaces to produce a block-diagonal form matrix. If each block is to be in elementary nilpotent form, then from Chapter 32 we know that for each block we need a corresponding $A$ -cyclic subspace of $\R^n$ (or $\C^n\text{,}$ as appropriate).

Suppose $W = \Span \{ \uvec{w}, A \uvec{w}, A^2 \uvec{w}, \dotsc \}$ is one the spaces in this collection.

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(a)

How is the dimension of $W$ reflected in the corresponding elementary nilpotent block of the triangular-block nilpotent form matrix to which $A$ is similar?

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(b)

Let $k$ represent $\dim W\text{.}$ What must be true about $A^k \uvec{w}\text{?}$

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Discovery 33.4.

Let's address the independence requirement of the $A$ -cyclic subspaces we need to achieve triangular-block nilpotent form, at least in the case of a set of two such subspaces. For this, it is not necessary to actually assume that $A$ is nilpotent.

Suppose

$\begin{align*} U \amp = \Span \{ \uvec{u}, A \uvec{u}, A^2 \uvec{u} \} \text{,} \amp W \amp = \Span \{ \uvec{w}, A \uvec{w} \} \text{,} \end{align*}$

where the provided spanning sets are assumed to independent, and generating vectors $\uvec{u}$ and $\uvec{w}$ each satisfy the condition you determined in Task b of Discovery 33.3. Further assume that the two final vectors, $A^2 \uvec{u}$ and $A \uvec{w} \text{,}$ are independent.

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(a)

Form the vector equation that begins the Test for Linear Dependence/Independence for the combined collection

$\begin{equation*} \{ \uvec{u}, A \uvec{u}, A^2 \uvec{u}, \uvec{w}, A \uvec{w} \} \text{.} \end{equation*}$

As there are five vectors in this collection, your vector equation will involve five unknown scalars.

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(b)

Multiply your vector equation from Task a through by $A^2\text{.}$ After applying that condition from Task b of Discovery 33.3 and simplifying, you should immediately be able to conclude that one of the five unknown scalars is zero.

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(c)

Use the knowledge from Task b that one of the scalars is zero to simplify your initial vector equation (before having multiplied by $A^2$ ) from Task a. Multiply this simplified vector equation through by $A\text{,}$ and again use that condition from Task b of Discovery 33.3 to simplify. Now you should be able to use the assumption that $A^2 \uvec{u}$ and $A \uvec{w}$ are independent to conclude that two more of the unknown scalars must be zero.

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(d)

Use the knowledge from Task c that two more of the scalars are zero to simplify your independence vector equation even further. You should be able to again use the assumption that $A^2 \uvec{u}$ and $A \uvec{w}$ are independent to conclude that the last two of the unknown scalars must be zero.

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(e)

Think about how this argument would unfold if the two spaces we started with had larger dimensions:

$\begin{align*} U \amp = \Span \{ \uvec{u}, A \uvec{u}, A^2 \uvec{u}, \dotsc, A^{k - 1} \uvec{u} \} \text{,} \amp W \amp = \Span \{ \uvec{w}, A \uvec{w}, A^2 \uvec{w}, \dotsc, A^{\ell - 1} \uvec{w} \} \text{.} \end{align*}$

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Discovery 33.5.

In Subsection 33.4.3, two different procedures for constructing a transition matrix $P$ that will put a given nilpotent matrix $A$ into triangular-block nilpotent form are given. If you've made it this far and have time to spare, see how far you can make it into applying the first of those two procedures (Procedure 33.4.2) to the $9 \times 9$ matrix from Discovery 33.2.

Once again, here is a Sage cell you can use to perform your calculations, if you wish.

xxxxxxxxxx
 
A = matrix([
    [  0, 1, 0, -1, 0, 2, 0, -5,  0 ],
    [ -1, 0, 0,  0, 0, 0, 0,  0,  2 ],
    [  0, 1, 0, -1, 0, 2, 0, -5,  0 ],
    [  2, 0, 1,  0, 0, 0, 0,  0, -2 ],
    [  0, 0, 0,  0, 0, 0, 0,  0,  0 ],
    [ -1, 0, 0,  0, 1, 0, 0,  0,  1 ],
    [  0, 0, 0,  0, 0, 0, 0,  0,  0 ],
    [  1, 0, 0,  0, 0, 0, 1,  0, -1 ],
    [  0, 1, 0, -1, 0, 2, 0, -5,  0 ]
]);
print(A);