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Systems in matrix form

Coupled equations written as one matrix equation, solutions built from eigenvalues and eigenvectors, and a second-order equation seen as a system.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to write a pair of coupled linear equations as a single matrix equation, find the eigenvalues and eigenvectors of the coefficient matrix, assemble the general solution from the pairs, check your eigenvalues against the trace and determinant, and turn a second-order equation into an equivalent first-order system.

2. What you already have

You can find eigenvalues and eigenvectors of a two-by-two matrix, and you know that $y' = ky$ has solution $y_0e^{kt}$. This lesson puts those two facts together and gets the complete theory of linear systems out of them, which is a return on linear algebra that is worth noticing while it happens.

3. Words this lesson uses

4. One equation, with a matrix in it

Two coupled equations

$$x' = ax + by, \qquad y' = cx + dy$$

become a single equation $\mathbf{x}' = A\mathbf{x}$, which is the scalar $y' = ky$ with a matrix where the constant was. So try the same thing that worked there: look for a solution $\mathbf{x} = e^{\lambda t}\mathbf{v}$ with $\mathbf{v}$ a fixed vector.

Substituting gives $\lambda e^{\lambda t}\mathbf{v} = e^{\lambda t}A\mathbf{v}$, and cancelling the exponential leaves

$$A\mathbf{v} = \lambda\mathbf{v}.$$

That is the eigenvalue equation, and it arrived rather than being imposed. Every eigenvalue-eigenvector pair of $A$ is a solution of the system, and for two distinct eigenvalues the two solutions are independent, so

$$\mathbf{x} = c_1e^{\lambda_1 t}\mathbf{v}_1 + c_2e^{\lambda_2 t}\mathbf{v}_2$$

is the general solution, with the two constants fixed by the starting vector.

This also subsumes unit 2. Setting $u = y$ and $v = y'$ turns $y'' + by' + cy = 0$ into a system whose matrix has characteristic polynomial $\lambda^{2} + b\lambda + c$ — the very same characteristic equation, reached from the other direction.

5. Two free checks on an eigenvalue calculation

For a two-by-two matrix the two eigenvalues add to the trace and multiply to the determinant. Both are one line of arithmetic and between them they catch almost every slip.

The second check is on the eigenvector: $(A - \lambda I)$ must be singular, so its two rows have to be multiples of each other. If they are not, either the eigenvalue is wrong or the subtraction was. This is a genuinely useful signal, because the system $(A - \lambda I)\mathbf{v} = \mathbf{0}$ looks over-determined to anyone meeting it for the first time, and the resolution is that one of its equations is redundant — by construction.

An eigenvector is also only ever determined up to scale. $(1, 1)$ and $(3, 3)$ are the same direction and give the same solutions, the difference being absorbed into the arbitrary constant in front.

6. Where this goes wrong

Building the matrix by columns. Row one is equation one. Transposing $A$ changes the determinant not at all and the eigenvectors completely.

Stopping at the eigenvalues. $e^{\lambda t}$ is not a solution of a system; $e^{\lambda t}\mathbf{v}$ is, and without the direction there is nothing to combine.

Pairing an eigenvalue with the wrong eigenvector. Each exponential travels with its own direction, and swapping them produces something that solves no system at all.

Fitting the constants before combining. The starting vector is applied to the whole general solution, not to one term of it.

7. The method, step by step, and how to check it

A linear system $\mathbf{x}' = A\mathbf{x}$ with a constant two-by-two matrix is solved by the same six moves every time.

1. Build the matrix row by row. Row one holds the coefficients of the first equation, in the order the unknowns are listed.
2. Trace and determinant. Write both down before anything else; they are the sum and the product of the eigenvalues.
3. Eigenvalues from $\lambda^{2} - (\operatorname{tr}A)\lambda + \det A = 0$.
4. Eigenvectors. For each $\lambda$, write $A - \lambda I$; its two rows must be multiples of each other, and either one gives the direction.
5. General solution $\mathbf{x} = c_1e^{\lambda_1t}\mathbf{v}_1 + c_2e^{\lambda_2t}\mathbf{v}_2$, each exponential with its own vector.
6. Starting vector. Put $t = 0$; the exponentials become $1$ and the constants solve a two-by-two linear system.

Why $e^{\lambda t}\mathbf{v}$ is a solution. Its derivative is $\lambda e^{\lambda t}\mathbf{v}$, and $A$ applied to it is $e^{\lambda t}A\mathbf{v}$. These agree exactly when $A\mathbf{v} = \lambda\mathbf{v}$, which is the definition of an eigenvector. So the search for straight-line solutions is the eigenvalue problem, nothing more.

When the eigenvalues are complex, $\alpha \pm \beta i$, one complex solution is enough: its real and imaginary parts are two real solutions, each built from $e^{\alpha t}\cos\beta t$ and $e^{\alpha t}\sin\beta t$. The same thing happened for second-order equations, and for the same reason: a real problem has real solutions hiding inside every complex one.

How to check the answer. Put $t = 0$ into your solution and confirm the starting vector. Then differentiate one component and compare it with the right side of its equation: for $x' = x + 2y$, compute $x'$ from your formula and check it equals $x + 2y$ from your formulas. The eigenvalue checks are the trace and determinant, and an eigenvector is checked by multiplying: $A\mathbf{v}$ must be $\lambda\mathbf{v}$.

Reading the long run. Once the eigenvalues are known the behaviour is known. If both are negative, every solution decays to the origin. If one is positive, almost every solution grows along that eigenvalue's eigenvector, because its exponential eventually swamps the other term. Only a start exactly on the other eigenvector escapes that growth. The next lesson turns this reading into a picture.

8. In the world: a drug moving between blood and tissue

Pharmacologists model a drug as moving between compartments: the blood, where it is injected and cleared, and the body's tissues, which exchange it with the blood. With $x$ the amount in the blood and $y$ the amount in the tissue, a typical model is

$$x' = -3x + y, \qquad y' = x - 3y,$$

in which each compartment passes some of its drug to the other and loses some altogether. The matrix $\begin{pmatrix} -3 & 1 \\ 1 & -3 \end{pmatrix}$ has trace $-6$ and determinant $8$; its eigenvalues solve $\lambda^{2} + 6\lambda + 8 = 0$, so $\lambda = -2$ with eigenvector $(1, 1)$ and $\lambda = -4$ with eigenvector $(1, -1)$.

The two eigenvalues are the two phases a clinician sees in blood samples. The $(1, -1)$ mode, blood and tissue out of balance, decays fast, like $e^{-4t}$: the drug first spreads from the blood into the tissue until the two are level. The $(1, 1)$ mode, blood and tissue together, decays like $e^{-2t}$: the slower phase in which the body eliminates the drug from both. Injected with $10$ units into the blood, $x(0) = 10$ and $y(0) = 0$, the constants are $c_1 = c_2 = 5$, and $x = 5e^{-2t} + 5e^{-4t}$, $y = 5e^{-2t} - 5e^{-4t}$: the tissue level rises, peaks and falls, as the eigenvectors predicted.

9. In the world: two coupled pendulums

Two identical pendulums joined by a light spring swing in two normal modes, the eigenvectors of their system. In the first, $(1, 1)$, they swing together and the spring never stretches, so they swing at the pendulums' own frequency. In the second, $(1, -1)$, they swing in opposite directions and the spring adds to the restoring force, so they swing faster. Any motion is a combination of the two.

Start one pendulum swinging and hold the other still, and the combination produces something striking: the two modes, at slightly different frequencies, drift in and out of step, and the swinging passes entirely to the second pendulum and back again, over and over. The time for one transfer is set by the difference between the two eigen-frequencies. The same mathematics describes energy passing between the atoms of a molecule, which is how infrared spectroscopy identifies chemicals by the frequencies of their normal modes.

10. In the world: why eigenvalues are the numbers engineers ask for

A system of equations describing a building, an aircraft or a power grid can have hundreds of unknowns, and no one writes its general solution. What engineers compute is its eigenvalues, because each one is a mode with its own rate: a negative real eigenvalue is a motion that dies away at that rate, a complex pair is a vibration, and a single eigenvalue with a positive real part is a failure mode that will grow. The two-by-two calculations of this lesson are the smallest case of that analysis, and every idea in it, from the trace and determinant checks to reading behaviour from the signs, scales up unchanged.

11. In the world: two tanks connected by pipes

Two tanks of brine exchanging water through pipes, with fresh water flowing into the first and mixed water draining from the second, hold salt amounts $x$ and $y$ that obey a linear system: each tank's salt changes by what flows in minus what flows out, and each flow carries its tank's concentration. With $100$ litres in each tank, $4$ litres a minute of fresh water in, $6$ litres a minute from the first tank to the second, $2$ back, and $4$ out, the system is $x' = -0.06x + 0.02y$, $y' = 0.06x - 0.06y$. Its matrix has trace $-0.12$ and determinant $0.0036 - 0.0012 = 0.0024$, so the eigenvalues solve $\lambda^{2} + 0.12\lambda + 0.0024 = 0$, giving $\lambda \approx -0.025$ and $\lambda \approx -0.095$ per minute. Both negative: all the salt eventually leaves, and the slow eigenvalue says it takes about $\frac{1}{0.025} = 40$ minutes to fall by a factor of $e$. Environmental engineers model chains of lakes, and pharmacologists chains of organs, in exactly this way.

12. An eigenvalue is not a solution without its direction

The scalar case makes it look as though the eigenvalues are the answer: for $y' = ky$ the number $k$ really is the whole story. In a system it is half of one. The substitution that produced the eigenvalue equation assumed a solution of the form $e^{\lambda t}\mathbf{v}$, and the eigenvalue only tells you how fast something grows along a particular direction — different directions grow at different rates, which is the entire content of having a matrix rather than a number. The practical symptom is an answer written as $c_1e^{\lambda_1 t} + c_2e^{\lambda_2 t}$ with no vectors in it, which is not a point in the plane and therefore cannot be a state of the system at all.

13. A system with two real eigenvalues

1. Take $x' = x + 2y$, $y' = 2x + y$. Write the matrix, its trace and determinant.

   $A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}, \quad \operatorname{tr}A = 1 + 1 = 2, \quad \det A = 1 \cdot 1 - 2 \cdot 2 = -3$

   Two invariants before any solving.

2. Solve the characteristic equation.

   $(1 - \lambda)^{2} - 4 = 0 \;\Rightarrow\; 1 - \lambda = \pm 2 \;\Rightarrow\; \lambda = -1,\ 3$

   A difference of squares. Check: $3 + (-1) = 2$ and $3 \times (-1) = -3$.

3. Find the eigenvector for $\lambda = 3$.

   $(A - 3I)\mathbf{v} = \begin{pmatrix} -2 & 2 \\ 2 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0} \;\Rightarrow\; v_1 = v_2, \quad \mathbf{v} = (1, 1)$

   Both rows say the same thing, as they must for an eigenvalue.

4. Find the eigenvector for $\lambda = -1$.

   $(A + I)\mathbf{v} = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix}\mathbf{v} = \mathbf{0} \;\Rightarrow\; v_1 = -v_2, \quad \mathbf{v} = (1, -1)$

   Again one independent equation.

5. Write the general solution.

   $\mathbf{x} = c_1e^{3t}\begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2e^{-t}\begin{pmatrix} 1 \\ -1 \end{pmatrix}$

   Each exponential travels with its own direction.

14. A second-order equation as a system

1. Take $y'' + 3y' + 2y = 0$. Name $u = y$ and $v = y'$.

   $u' = y' = v$

   The first equation of the system is just the definition of $v$.

2. Solve the original equation for $y''$ to get the second equation.

   $v' = y'' = -2y - 3y' = -2u - 3v$

   Subtract $3y' + 2y$ from both sides of the original, then rename.

3. Write the matrix, row by row.

   $A = \begin{pmatrix} 0 & 1 \\ -2 & -3 \end{pmatrix}$

   Row one is $u' = 0u + 1v$; row two is $v' = -2u - 3v$.

4. Find its characteristic polynomial.

   $\det(A - \lambda I) = (-\lambda)(-3 - \lambda) - (1)(-2) = \lambda^{2} + 3\lambda + 2$

   Exactly the characteristic equation of the second-order method.

5. Solve it and recover $y$.

   $(\lambda + 1)(\lambda + 2) = 0 \;\Rightarrow\; \lambda = -1,\ -2 \;\Rightarrow\; y = u = c_1e^{-t} + c_2e^{-2t}$

   The two theories are one; numerical solvers use this reduction for every higher-order equation.

6. Check the second component is the derivative of the first.

   $v = y' = -c_1e^{-t} - 2c_2e^{-2t}$

   Each eigenvector of $A$ has the form $(1, \lambda)$, so the second component is always $\lambda$ times the first: exactly what $v = y'$ requires.

15. Complex eigenvalues: motion round a circle

1. Solve $x' = -y$, $y' = x$ with $x(0) = 1$, $y(0) = 0$. Write the matrix, trace and determinant.

   $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \qquad \operatorname{tr}A = 0, \qquad \det A = 0 - (-1)(1) = 1$

   Trace $0$ and a positive determinant already point to a centre.

2. Solve the characteristic equation.

   $\lambda^{2} - 0\lambda + 1 = 0 \quad\Rightarrow\quad \lambda = \pm i$

   For a two-by-two matrix the equation is $\lambda^{2} - (\operatorname{tr}A)\lambda + \det A = 0$.

3. Find an eigenvector for $\lambda = i$.

   $(A - iI)\mathbf{v} = \begin{pmatrix} -i & -1 \\ 1 & -i \end{pmatrix}\begin{pmatrix} 1 \\ -i \end{pmatrix} = \begin{pmatrix} -i + i \\ 1 + i^{2} \end{pmatrix} = \mathbf{0}$

   Row one gives $v_2 = -iv_1$; take $v_1 = 1$. The second row checks, because $i^{2} = -1$.

4. Expand the complex solution with Euler's formula.

   $e^{it}\begin{pmatrix} 1 \\ -i \end{pmatrix} = \begin{pmatrix} \cos t + i\sin t \\ \sin t - i\cos t \end{pmatrix}$

   $e^{it} = \cos t + i\sin t$, and $-i(\cos t + i\sin t) = \sin t - i\cos t$.

5. Take the real and imaginary parts as two real solutions.

   $\begin{pmatrix} \cos t \\ \sin t \end{pmatrix}, \qquad \begin{pmatrix} \sin t \\ -\cos t \end{pmatrix}$

   The matrix is real, so the real and imaginary parts of a complex solution are each solutions.

6. Fit the starting vector.

   $c_1\begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2\begin{pmatrix} 0 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad\Rightarrow\quad c_1 = 1, \ c_2 = 0$

   At $t = 0$ the two solutions are $(1, 0)$ and $(0, -1)$.

7. Write the solution and check it.

   $x = \cos t, \quad y = \sin t: \qquad x' = -\sin t = -y, \quad y' = \cos t = x$

   Both equations hold, and $x^{2} + y^{2} = 1$: the point goes round the unit circle for ever, which is what a centre means.

16. Your turn: solve $x' = 3x + y$, $y' = x + 3y$ in general

1. Write the matrix, its trace and determinant.

   $A = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}, \quad \operatorname{tr}A = 6, \quad \det A = 9 - 1 = 8$

   Matrix first, then the two checks to come.

2. Your turn: work this step out. Its working is at the end of the packet.

   Solve the characteristic equation.

3. Your turn: work this step out. Its working is at the end of the packet.

   Find the eigenvectors and write the solution.

17. Guided practice

Which of these is an eigenvector of $\begin{pmatrix} 3 & 5 \\ 5 & 3 \end{pmatrix}$?

• $(1, 1)$
• $(1, 0)$
• $(3, 5)$
• $(8, 0)$

18. Guided practice

Complete the worked solution: the eigenvalues of $A = \begin{pmatrix} 4 & 5 \\ 5 & 4 \end{pmatrix}$.

1. Subtract $\lambda$ from each diagonal entry and take the determinant.

   $\det(A - \lambda I) = \begin{vmatrix} 4 - \lambda & 5 \\ 5 & 4 - \lambda \end{vmatrix} = (4 - \lambda)^{2} - 5^{2}$

   The eigenvalues are where $A - \lambda I$ is singular.

2. Set it to zero, move the square across, and take square roots of both sides.

   $(4 - \lambda)^{2} = 5^{2} \quad\Rightarrow\quad 4 - \lambda = \pm 5$

   A square equals $5^{2}$ when its base is $5$ or $-5$.

3. Take $4 - \lambda = -5$ and solve for $\lambda$.

   $\lambda_1 = 4 + 5 =$ _a

   Add $\lambda$ and $5$ to both sides.

4. Take $4 - \lambda = 5$ and solve for $\lambda$.

   $\lambda_2 = 4 - 5 =$ _b

   Add $\lambda$ and subtract $5$ on both sides.

5. Check: the eigenvalues must add to the trace.

   $\lambda_1 + \lambda_2 = \operatorname{tr}A = 4 + 4 =$ _t

   The sum of the eigenvalues is always the sum of the diagonal.

19. Guided practice

Write the coefficient matrix of the system $x' = 5x + y$, $y' = 6x + 2y$.

This task has no paper form; do it on a device.

20. Practice

For the system with matrix $\begin{pmatrix} 2 & 3 \\ 3 & 2 \end{pmatrix}$, fill in the trace, the determinant and the two eigenvalues.

21. Practice

What is the larger eigenvalue of $\begin{pmatrix} 7 & 5 \\ 5 & 7 \end{pmatrix}$?

Answer:

22. Practice

Two species help each other: each population, in hundreds, grows at its own rate and gains in proportion to the other's size, so after $t$ years they obey the system below. Solve $x' = 4x + y$, $y' = x + 4y$ with $x(0) = 3$ and $y(0) = 1$. Write $x(t)$ as a formula in $t$ (type each exponential as e^(...)).

Answer:

23. Somewhere new

Put the steps of solving $\mathbf{x}' = A\mathbf{x}$ with $A = \begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix}$ and a given starting vector into the order they must be done.

Number the steps in order (write the number in the box):

• Write the system as $\mathbf{x}' = A\mathbf{x}$
• Solve the characteristic equation for the eigenvalues
• Find an eigenvector for each eigenvalue
• Write the general solution as a combination
• Fit the constants to the starting vector

24. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

25. Test question

Two species help each other: each population, in hundreds, grows at its own rate and gains in proportion to the other's size, so after $t$ years they obey the system below. Solve $x' = 2x + y$, $y' = x + 2y$ with $x(0) = 5$ and $y(0) = 3$. Write $x(t)$ as a formula in $t$ (type each exponential as e^(...)).

Answer:

26. What you can do now

You can write a system in matrix form and build its general solution from eigenvalues and eigenvectors. Say in your own words why an eigenvalue on its own is not a solution of a system.

Working for the steps left to you

16. Your turn: solve $x' = 3x + y$, $y' = x + 3y$ in general, step 2

16. Your turn: solve $x' = 3x + y$, $y' = x + 3y$ in general, step 3