Examples from Seen Questions

Stability of two discrete-time systems

Let

Consider the discrete-time systems

Here “stable” uses the convention from the discrete stability notes: every solution must tend to zero in the stated time direction. A bounded but non-decaying unit-modulus mode is a boundary case, not stable under this convention.

Select all true statements:

1. The first DTS is stable as $n\to\infty$.
2. The first DTS is stable as $n\to-\infty$.
3. The second DTS is stable as $n\to\infty$.
4. The second DTS is stable as $n\to-\infty$.

Solution

Forward behaviour depends on the powers $A^n$ and $B^n$. Eigenmodes with modulus less than $1$ decay as $n\to\infty$ and those with modulus greater than $1$ grow. A unit-modulus eigenvalue must be checked for Jordan-block factors such as $n$. For $n\to-\infty$, the same test applies to the inverse matrix.

For the first matrix, write

For $n\ge0$, the binomial theorem truncates because $N^2=0$:

For $n=-k<0$, $(I+N)^{-1}=I-N$, so the binomial theorem and $N^2=0$ give

For example,

This vector is unbounded both as $n\to\infty$ and as $n\to-\infty$. Thus the first DTS is not stable in either direction. The cause is not the unit eigenvalue alone: the non-trivial Jordan block creates the factor $n$.

Let

Their product is

Hence $Q=40P^{-1}$ and the displayed expression is the diagonalisation $B=PDP^{-1}$, with $D=\operatorname{diag}(1/2,1/3)$. Its eigenmodes are constant multiples of $(1/2)^n$ and $(1/3)^n$. Therefore

so the second DTS is stable as $n\to\infty$. In the reverse direction, the eigenvalues of $B^{-1}$ are $2$ and $3$. Writing $n=-k$ gives $B^n=(B^{-1})^k$ with $k\to\infty$, so non-zero inverse modes grow. The second DTS is therefore not stable backwards in time.

The only true statement is

This is the matrix version of the modulus test for discrete modes. Reversing time replaces each non-zero eigenvalue by its reciprocal, which is why the second system is stable forwards but not backwards.

A system with zero initial derivative

Consider

We wish to solve the system of differential equations

subject to the initial condition

The solutions $\mathbf{x}(t)$ of this system form a vector space $S$.

(a) Find $\dim(S)$.

(b) Find a basis for $S$.

Solution

The condition prescribes the initial derivative rather than the initial state, so $\mathbf{x}(0)$ may vary and the condition selects a subspace of the general solution.

Solve the triangular component equations from the bottom upwards, then impose the condition on the derivative at $t=0$. Multiplying out $A\mathbf{x}$ gives

The last equation gives

for a constant $c_3$. Substituting this into the second equation and integrating gives

Now substitute $x_2(t)$ and $x_3(t)$ into the first equation:

Integrating once more gives the general solution

Differentiate this expression and evaluate at $t=0$:

The initial condition $\mathbf{x}'(0)=\mathbf0$ therefore gives

The second equation forces $c_3=0$ and the first then forces $c_2=0$. The constant $c_1$ remains free, so

Consequently,

and one basis is

As a check, $A(1,0,0)^{\mathsf T}=\mathbf0$, so every function in the displayed family is constant and has zero derivative. Equivalently, the condition $\mathbf{x}'(0)=A\mathbf{x}(0)=\mathbf0$ requires the initial vector to lie in $\ker(A)=\operatorname{span}\{(1,0,0)^{\mathsf T}\}$, which gives the same one-dimensional solution space.

Diagonalisation of a stress matrix

Keegan and Hongyi are engineers studying the stress on a certain physical object. This stress can be expressed as a $3\times3$ matrix $A$ and Keegan and Hongyi are particularly interested in the diagonalisation of $A$. After some computations, Keegan discovers that

where

where $x>20$ is some unknown parameter.

(a) Which statements are true?

Which of the following are true about Keegan and Hongyi's matrices? Select all that apply.

1. $\displaystyle \lim_{n\to\infty}A^n=B^3.$
2. If $\lambda$ is an eigenvalue of $A$, then $$(\lambda+4)(\lambda+20)(\lambda+x)=0.$$
3. The vectors $$\begin{pmatrix}-6\\0\\0\end{pmatrix}, \qquad \begin{pmatrix}0\\2\\0\end{pmatrix}, \qquad \begin{pmatrix}0\\0\\7\end{pmatrix}$$ are eigenvectors of $D$.
4. The columns of $A$ are linearly independent.
5. $A=BDB^{-1}$.

(b) A normalised eigenvector

Keegan and Hongyi are also interested in finding a normalised set of the eigenvectors of $A$. That is, they want eigenvectors of length $1$.

Suppose Hongyi writes $v_\lambda$ to denote an eigenvector of $A$ of length $1$ with the corresponding eigenvalue $\lambda$. Using Keegan's diagonalisation of $A$, find a valid choice for $v_4$.

Square roots, multiplication and exponents were to be written in Maple syntax; for example,

(c) A matrix power applied to a vector

Keegan and Hongyi are also interested in finding the vector $A^nB\mathbf y$, where $\mathbf y\in\mathbb R^3$, for all $n\in\mathbb Z$.

In one trial, suppose Hongyi considers

Use the above diagonalisation of $A$, or otherwise, to find the components of

(d) When does the nullity become non-zero?

A second pair of engineers, Max and Mira, uses Keegan's diagonalisation. For this part only, allow $x$ to range over all real values rather than imposing the restriction $x>20$ from parts (a)--(c). In Max's simulations, the program runs whenever $\operatorname{nullity}(A)=0$. There exists only one choice of $x\in\mathbb R$ for which the software fails and Mira is tasked with finding this value.

Which choice of $x$ gives $\operatorname{nullity}(A)\ne0$? In words, briefly explain your answer.

Solution

The equation $D=B^{-1}AB$ is a similarity relation. Multiplying it on the left by $B$ and on the right by $B^{-1}$ gives

This relation carries eigenvalues, eigenvectors, powers and invertibility between $A$ and the diagonal matrix $D$.

(a) Testing the statements

For positive integers $n$, diagonalisation gives

Since $x>20$, the diagonal entries $4^n,20^n,x^n$ do not converge to finite limits. If $A^n$ had a finite limit, then $B^{-1}A^nB=D^n$ would also have a finite limit, which is impossible. Therefore statement 1 is false.

The eigenvalues of $A$ are the diagonal entries of $D$, namely $4,20,x$. Thus the correct characteristic-root equation is

not the equation with plus signs in statement 2. Statement 2 is false.

The three vectors in statement 3 are non-zero scalar multiples of the standard basis vectors. Direct multiplication gives

and

Hence statement 3 is true.

Because $4$, $20$ and $x$ are all non-zero, $D$ is invertible. The matrix $B$ is invertible because $B^{-1}$ appears in the given relation, so $A=BDB^{-1}$ is invertible. An invertible square matrix has linearly independent columns, so statement 4 is true. Statement 5 is the relation derived above:

Therefore the true statements are

(b) Finding $v_4$

Let

Since $D\mathbf e_1=4\mathbf e_1$, the corresponding eigenvector of $A$ is $B\mathbf e_1$:

The first column of $B$ is

whose length is

Dividing by this length gives the valid normalised choice

Then $\|v_4\|=1$ and $Av_4=4v_4$. Its negative is an equally valid choice.

(c) Finding $w_1,w_2,w_3$

The diagonal entries $4$, $20$ and $x$ are non-zero, so $A$ is invertible and the diagonalisation power identity is valid for every $n\in\mathbb Z$, not only for non-negative $n$. Using $A^n=BD^nB^{-1}$,

For the given vector,

Multiplying by $B$ gives

Therefore

The factor $B$ cancels $B^{-1}$, so $\mathbf y$ acts directly as a vector of diagonal coordinates.

(d) Finding the value that makes $A$ singular

Similarity preserves the nullity, so

The diagonal matrix $D$ is singular exactly when one of its diagonal entries is zero. The entries $4$ and $20$ are non-zero, so

At $x=0$, $D$ has rank $2$ and nullity $1$ and so does $A$. The required value is

This value is admissible because part (d) lifts the restriction $x>20$ that applies to parts (a)--(c).

A linear map, bases and a discrete-time system

Consider the linear map $T:\mathbb R^3\to\mathbb R^3$ given by

The following eigenvalue computation is useful:

T = [ 103/126  -37/126  15/56
      -25/252  157/252  15/56
      -11/252  19/126  115/168 ];

e = eig(T)

e = 3x1

    0.4166
    0.7916
    0.9166

(a)

Suppose $B=\{\mathbf u,\mathbf v,\mathbf w\}\subset\mathbb R^3$ is a set of three vectors. Select all of the following sets for which $B$ is a basis for $\mathbb R^3$:

1. $\displaystyle B=\left\{ \begin{pmatrix}-5\\5\\4\end{pmatrix}, \begin{pmatrix}-4\\-2\\5\end{pmatrix}, \begin{pmatrix}-1\\5\\-3\end{pmatrix} \right\}$;
2. $\displaystyle B=\left\{ \begin{pmatrix}9\\0\\0\end{pmatrix}, \begin{pmatrix}9\\0\\9\end{pmatrix}, \begin{pmatrix}0\\9\\9\end{pmatrix} \right\}$;
3. $\displaystyle B=\left\{ \begin{pmatrix}14\\-11\\7\end{pmatrix}, \begin{pmatrix}-18\\3\\16\end{pmatrix}, \begin{pmatrix}-12\\-24\\69\end{pmatrix} \right\}$;
4. $\displaystyle B=\left\{ \begin{pmatrix}-5\\5\\4\end{pmatrix}, \begin{pmatrix}-4\\-2\\5\end{pmatrix}, \begin{pmatrix}31\\-7\\-32\end{pmatrix} \right\}$;
5. $\displaystyle B=\left\{ \begin{pmatrix}0\\0\\0\end{pmatrix}, \begin{pmatrix}2\\1\\0\end{pmatrix}, \begin{pmatrix}0\\0\\3\end{pmatrix} \right\}$;
6. $\displaystyle B=\left\{ \begin{pmatrix}10\\3\\-15\end{pmatrix}, \begin{pmatrix}10\\-19\\-19\end{pmatrix}, \begin{pmatrix}-40\\120\\84\end{pmatrix} \right\}$;
7. $\displaystyle B=\left\{ \begin{pmatrix}-9\\11\\0\end{pmatrix}, \begin{pmatrix}-9\\0\\11\end{pmatrix}, \begin{pmatrix}-18\\11\\11\end{pmatrix} \right\}$;
8. $\displaystyle B=\left\{ \begin{pmatrix}10\\3\\-15\end{pmatrix}, \begin{pmatrix}10\\-19\\-19\end{pmatrix}, \begin{pmatrix}6\\7\\-5\end{pmatrix} \right\}$;
9. $\displaystyle B=\left\{ \begin{pmatrix}14\\-11\\7\end{pmatrix}, \begin{pmatrix}-18\\3\\16\end{pmatrix}, \begin{pmatrix}-2\\6\\-2\end{pmatrix} \right\}$;
10. $\displaystyle B=\left\{ \begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\1\\0\end{pmatrix}, \begin{pmatrix}0\\0\\1\end{pmatrix} \right\}$.

(b)

Suppose $B$ is a basis for $\mathbb R^3$ and let $T_B$ denote the matrix of $T$ with respect to the basis $B$. Define the discrete-time system

Select all true statements about $T_B$ and this system:

• The eigenvalues of $T_B$ are not necessarily the same as those of $T$.
• There exist some $B$ such that the stability of the DTS depends on the choice of $x_0$.
• There exist some $B$ such that for all $x_0$, the DTS is unstable as $n\to-\infty$.
• The determinant of $T_B$ does not change depending on $B$.
• There exist some $B$ such that for all $x_0$, the DTS is stable as $n\to+\infty$.
• The DTS has undefined behaviour for some choices of $B$.

Solution

Put the vectors in each candidate basis into a matrix as columns. A set of three vectors in $\mathbb R^3$ is a basis exactly when this matrix has a pivot in every column, equivalently when its determinant is non-zero.

For the first, second, eighth, ninth and tenth candidates, elementary row reduction gives

The corresponding determinants are, respectively,

so these five sets are bases.

For the remaining candidates, the row reductions give

Each of these has only two pivots and the dependence relations are respectively

The third relation follows because candidate five contains the zero vector. Therefore the correct selections in part (a) are

For part (b), if $P$ is the matrix whose columns are the vectors in the ordered basis $B$, then

Thus $T_B$ is similar to $T$, so they have the same characteristic polynomial, eigenvalues and determinant. The decimal output suggests rational eigenvalues. For a monic factorisation, use the alternative convention $\det(\lambda I-T)=(-1)^3p_T(\lambda)$, which has the same roots as the course convention $p_T(\lambda)=\det(T-\lambda I)$. An exact calculation gives

Thus the exact eigenvalues are $5/12$, $19/24$ and $11/12$.

All three eigenvalues are distinct and satisfy $0<|\lambda_j|<1$. Therefore the eigenvectors form a basis and every forward trajectory has the form

for eigenvectors $z_j$ of $T_B$. Every term tends to zero as $n\to+\infty$, so the DTS is asymptotically stable for every basis $B$ and every initial state.

Because $T_B$ is invertible, negative times are defined. If $x_0\ne0$, at least one coefficient $c_j$ is non-zero. In the eigenbasis, the corresponding coordinate is

Since a fixed change of basis preserves boundedness, every non-zero initial state is unbounded backwards in time.

Finally,

which is independent of $B$. The eigenvalues also cannot change and no choice of basis makes the system undefined.

Thus the true statements are

Changing the basis changes coordinates but not the linear map, so the eigenvalues and determinant are the same for every $B$. With every eigenvalue strictly inside the unit circle, trajectories decay forwards and grow backwards, whatever the basis.

A three-state population model

Each year the populations of South Australia, Western Australia and Tasmania migrate as follows:

• four fifths of those in South Australia and four fifths of those in Tasmania move to Western Australia;
• one quarter of those in Western Australia move to South Australia and one half of those in Western Australia move to Tasmania;
• because of animosity between South Australia and Tasmania due to the infamous koala wars, no one ever moves directly between these two states.

Let $x_k,y_k,z_k$ be the populations of South Australia, Western Australia, and Tasmania, respectively, after $k$ years and let

Assume that the initial total population of the three states is $10$ million.

The following partially redacted computation supplies the eigenvalues and an eigenbasis once the transition matrix has been found.

>> B = [partially redacted]
B =
[partially redacted]
[partially redacted]
[partially redacted]

>> [V,D] = eig(B)
V =
   -1.0000    0.5000    0.5000
    0.0000   -1.5000    1.6000
    1.0000    1.0000    1.0000

D =
    0.2000         0         0
         0   -0.5500         0
         0         0    1.0000
>>

(a) Find the Markov matrix $B$ that describes the population dynamics

Write the matrix explicitly.

(b) State the eigenvalues of $B$

Give the eigenvalues as exact rational values.

(c) Find the eventual long-term population distribution

What is the eventual long-term population distribution of the three states after $k$ years as $k\to\infty$?

Give each limiting population in millions to at least two decimal places.

Solution

Use the column convention in $\mathbf v(k+1)=B\mathbf v(k)$. Each column describes where the people from one state go.

For people starting in South Australia, one fifth remain in South Australia and four fifths move to Western Australia. For people starting in Western Australia, one quarter move to South Australia, one quarter remain in Western Australia and one half move to Tasmania. For people starting in Tasmania, one fifth remain in Tasmania and four fifths move to Western Australia. Therefore

Equivalently,

[1/5, 1/4, 0; 4/5, 1/4, 4/5; 0, 1/2, 1/5]

The columns sum to $1$, so the total population is preserved:

To find the eigenvalues exactly, use the monic convention from the previous question:

Thus

The eigenvalues, in the requested row-vector format, are

For the long-term distribution, observe that the other two eigenvalues satisfy

while $1$ is a simple eigenvalue. The three distinct eigenvalues give an eigenbasis, so the two non-stationary modes decay to zero. Every initial population vector with total $10$ million therefore converges to the same stationary vector and the initial split between the states is not needed.

The stationary vector spans the eigenspace for $\lambda=1$. Let

and solve $B\mathbf p=\mathbf p$. The first and third coordinates give

and

Therefore the stationary vector has the ratio

Write it as

The total population is $10$ million, so

Hence the limiting populations are

and

The eigenvalue-$1$ mode carries the long-term distribution, while the two decaying modes record transient migration.