LINEAR-ALGEBRA · Interactive Practice | Unit 22 · Video 4

From One Big Equation to a Matrix: The Companion Matrix Trick

IKey Formulas

Formula	Name	Description
$y'' + b y' + k y = 0$	Second-order linear ODE	The original scalar equation
$\mathbf{u} = \begin{pmatrix} y' \\ y \end{pmatrix}$	State vector	Highest derivative on top
$\mathbf{u}' = A\mathbf{u}, \quad A = \begin{pmatrix} -b & -k \\ 1 & 0 \end{pmatrix}$	Companion matrix system	Row 1 = physics, Row 2 = bookkeeping
$\det(A - \lambda I) = \lambda^2 + b\lambda + k$	Characteristic polynomial	Same as the ODE's characteristic equation

IIVisualization 1 — Building the Companion Matrix

The ODE's coefficients $b$ and $k$ drop into $A$ 's top row; its characteristic polynomial is $\lambda^2 + b\lambda + k$ .

b (coeff of y′) 3 k (coeff of y) 2

💡 Those x-intercepts are the eigenvalues of $A$ — computed a second way in the next visualization.

IIIVisualization 2 — Eigenvalues Are the Characteristic Roots

The eigenvalues of $A$ are exactly the roots of $\lambda^2 + b\lambda + k$ , whether real or complex.

k (coeff of y) 2 b (coeff of y′) 3

💡 For any $b, k$ the two eigenvalues sum to $-b$ (the trace of $A$ ) and multiply to $k$ (its determinant).

IVVisualization 3 — The Higher-Order Pattern

For an $n$ -th order ODE, one physics row sits atop a subdiagonal of bookkeeping $1$ 's.

ODE order n

💡 The eigenvalues of this matrix are the roots of the degree- $n$ characteristic polynomial — the same trick, one order higher.

VQuiz Questions

Question 1

Consider the second-order ODE $y'' + 4y' + 3y = 0.$

Using the convention $\mathbf{u} = (y', y)^T$ (highest derivative on top), what is the companion matrix $A$ such that $\mathbf{u}' = A\mathbf{u}$ ?

✅ Correct! The top row is the negated ODE coefficients $(-b, -k) = (-4, -3)$ , and the bottom row $(1, 0)$ encodes $y' = y'$ .

❌ Sign error. When you solve $y'' + 4y' + 3y = 0$ for $y''$ , you get $y'' = \mathbf{-}4 y' \mathbf{-} 3 y$ . The coefficients flip sign.

❌ Order mistake. With $\mathbf{u} = (y', y)^T$ , the top row acts on $y'$ first, then $y$ . So the entries are $(-b, -k) = (-4, -3)$ , not $(-k, -b)$ .

❌ That's the matrix you'd get from the reversed convention $\mathbf{u} = (y, y')^T$ . Our convention puts the highest derivative on top, which puts the physics row on top.

❌ Not quite. Try again — the hints above can help.

Show solution

Solution:

With $\mathbf{u} = (y', y)^T$ , we differentiate: $\mathbf{u}' = \begin{pmatrix} y'' \\ y' \end{pmatrix}.$

From the ODE: $y'' = -4y' - 3y$ . The bottom entry is just $y' = y'$ . So $\mathbf{u}' = \begin{pmatrix} -4 y' - 3 y \\ y' \end{pmatrix} = \begin{pmatrix} -4 & -3 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} y' \\ y \end{pmatrix}.$

Hence $A = \begin{pmatrix} -4 & -3 \\ 1 & 0 \end{pmatrix}$ .

Question 2

For the ODE $y'' + 3y' + 2y = 0$ , the companion matrix is $A = \begin{pmatrix} -3 & -2 \\ 1 & 0 \end{pmatrix}.$

What are the eigenvalues of $A$ ?

✅ Correct! The characteristic polynomial $\lambda^2 + 3\lambda + 2 = (\lambda+1)(\lambda+2)$ has roots $-1, -2$ .

❌ Sign error. The polynomial $\lambda^2 + 3\lambda + 2 = 0$ factors as $(\lambda+1)(\lambda+2) = 0$ , giving $\lambda = -1, -2$ , not $+1, +2$ .

❌ You read off the diagonal entries of $A$ . That doesn't give eigenvalues — only the trace ( $-3$ ) and the determinant ( $2$ ). You need to solve $\lambda^2 + 3\lambda + 2 = 0$ .

❌ Not quite. These are the absolute values of the ODE coefficients, not the eigenvalues. Solve $\lambda^2 + 3\lambda + 2 = 0$ to get $\lambda = -1, -2$ .

❌ Not quite. Try again — the hints above can help.

Show solution

Solution:

Compute the characteristic polynomial: $\det(A - \lambda I) = \det\begin{pmatrix} -3 - \lambda & -2 \\ 1 & -\lambda \end{pmatrix} = (-3-\lambda)(-\lambda) - (-2)(1) = \lambda^2 + 3\lambda + 2.$

This is exactly the characteristic polynomial of the ODE. Factor: $\lambda^2 + 3\lambda + 2 = (\lambda + 1)(\lambda + 2) = 0.$

So $\lambda = -1$ and $\lambda = -2$ . Note these are the same exponents you'd get from the ansatz $y = e^{\lambda t}$ .

Question 3

True or False: The companion-matrix trick adds new information to the original ODE. The system $\mathbf{u}' = A\mathbf{u}$ has more solutions than the scalar equation $y'' + b y' + k y = 0$ .

✅ Correct! The bookkeeping rows are trivial identities — they don't add constraints. The matrix system has exactly the same solution set as the scalar ODE.

❌ Not quite. The companion-matrix trick is purely a repackaging. The trivial rows like $y' = y'$ add no new constraints; they only unpack derivatives into named variables.

❌ Not quite. Try again — the hints above can help.

Show solution

Solution:

False. The video stresses this point: the companion-matrix trick adds no new information. The bookkeeping rows of $A$ encode trivial identities like $y' = y'$ , which impose no new constraints. They only rename the higher derivatives as new state variables.

The set of solutions to $\mathbf{u}' = A\mathbf{u}$ is in one-to-one correspondence with the set of solutions to $y'' + b y' + k y = 0$ . Same solutions, new clothing.

Question 4

Consider the fourth-order ODE $y^{(4)} + 5 y''' + 6 y'' + 2 y' + 7 y = 0.$

Using the convention $\mathbf{u} = (y''', y'', y', y)^T$ , what is the top row of the $4 \times 4$ companion matrix $A$ ?

✅ Correct! The top row is the ODE coefficients negated and listed in the same order as the entries of $\mathbf{u}$ — from $y'''$ down to $y$ .

❌ Sign error. When you solve for $y^{(4)}$ , every coefficient flips sign: $y^{(4)} = -5 y''' - 6 y'' - 2 y' - 7 y$ .

❌ Order is reversed. Since $\mathbf{u}$ lists derivatives from highest to lowest, the row entries must match: coefficient of $y'''$ first, then $y''$ , $y'$ , $y$ — giving $(-5, -6, -2, -7)$ .

❌ That's a bookkeeping row, not the physics row. The top row carries the ODE coefficients; the $1$ 's appear on the subdiagonal below.

❌ Not quite. Try again — the hints above can help.

Show solution

Solution:

Solve the ODE for the highest derivative: $y^{(4)} = -5 y''' - 6 y'' - 2 y' - 7 y.$

The top entry of $\mathbf{u}'$ is $y^{(4)}$ . Since $\mathbf{u} = (y''', y'', y', y)^T$ , the top row of $A$ must produce this linear combination, reading off the coefficients in the same order: $\text{Row 1} = (-5, \; -6, \; -2, \; -7).$

The remaining three rows are the bookkeeping subdiagonal of $1$ 's, giving $A = \begin{pmatrix} -5 & -6 & -2 & -7 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}.$

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