Matrices — Diagnostic Tests

Unit Tests

Tests edge cases, boundary conditions, and common misconceptions for matrices.

UT-1: Non-Commutativity and the Commutator Equation

Question:

Let $A = \begin{pmatrix} 1 & 2 \\ 0 & -1 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$.

(a) Show that $AB \neq BA$.

(b) Find all $2 \times 2$ matrices $X$ such that $AX = XA$Where $A$ is as defined above.

(c) A student claims that since $A$ and $B$ do not commute, no matrix can commute with $A$. Explain the error.

[Difficulty: hard. Tests understanding of matrix non-commutativity and solving the commutator equation.]

Solution:

(a)

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

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

Since $AB = \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix} \neq \begin{pmatrix} 0 & -1 \\ 1 & 2 \end{pmatrix} = BA$The matrices do not commute.

(b) Let $X = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Then:

$AX = \begin{pmatrix} a + 2c & b + 2d \\ -c & -d \end{pmatrix}$

$XA = \begin{pmatrix} a & 2a + b \\ c & 2c + d \end{pmatrix}$

Setting $AX = XA$:

• $(1,1)$: $a + 2c = a \implies c = 0$
• $(2,1)$: $-c = c \implies c = 0$ (consistent)
• $(1,2)$: $b + 2d = 2a + b \implies 2d = 2a \implies d = a$
• $(2,2)$: $-d = 2c + d \implies -d = d \implies d = 0$And therefore $a = 0$.

So $X = \begin{pmatrix} 0 & b \\ 0 & 0 \end{pmatrix}$ for any $b \in \mathbb{R}$.

(c) The student”s error is a logical fallacy: the fact that $B$ does not commute with $A$ does not imply that no matrix commutes with $A$. The identity matrix $I$The zero matrix $O$And all scalar multiples of $I$ commute with every matrix. Part (b) shows that there is in fact a one-parameter family of matrices commuting with $A$.

UT-2: 3x3 Eigenvalues with Complex Roots

Question:

Let $M = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$.

(a) Find the eigenvalues and eigenvectors of $M$.

(b) A student claims that since $M$ is triangular, the eigenvalues are the diagonal entries, so there should be three linearly independent eigenvectors. Is this correct?

[Difficulty: hard. Tests eigenvalue computation for a matrix with a repeated eigenvalue and the concept of defective matrices.]

Solution:

(a) The characteristic equation is $\det(M - \lambda I) = 0$:

$\det\begin{pmatrix} 2 - \lambda & 1 & 0 \\ 0 & 2 - \lambda & 0 \\ 0 & 0 & 3 - \lambda \end{pmatrix} = (2 - \lambda)^2(3 - \lambda) = 0$

Eigenvalues: $\lambda_1 = 2$ (repeated, algebraic multiplicity 2), $\lambda_2 = 3$.

For $\lambda_1 = 2$: $(M - 2I)\mathbf{v} = \mathbf{0}$:

$\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

This gives $y = 0$ and $z = 0$With $x$ free. So the eigenvectors are $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ times any scalar. There is only one linearly independent eigenvector for $\lambda = 2$.

For $\lambda_2 = 3$: $(M - 3I)\mathbf{v} = \mathbf{0}$:

$\begin{pmatrix} -1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

This gives $y = 0$, $-x = 0$ so $x = 0$With $z$ free. Eigenvector: $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.

(b) The student is incorrect. While it is true that the eigenvalues of a triangular matrix are the diagonal entries, the number of linearly independent eigenvectors is not necessarily equal to the number of eigenvalues (counting multiplicity). Here $\lambda = 2$ has algebraic multiplicity $2$ but geometric multiplicity $1$. The matrix is defective — it cannot be diagonalised.

Integration Tests

Tests synthesis of matrices with other topics.

IT-1: Diagonalisation for Computing Matrix Powers

Question:

Let $A = \begin{pmatrix} 5 & -6 \\ 2 & -2 \end{pmatrix}$.

(a) Find the eigenvalues and eigenvectors of $A$.

(b) Write $A = PDP^{-1}$ where $D$ is a diagonal matrix.

(c) Hence find $A^5$.

(d) A student claims that since $A$ has two distinct eigenvalues, $A^n$ can always be computed as $PD^nP^{-1}$ for any positive integer $n$. Is this correct? What conditions must be verified?

[Difficulty: hard. Combines eigenvalue computation, diagonalisation, and matrix exponentiation.]

Solution:

(a) Characteristic equation: $\det(A - \lambda I) = (5 - \lambda)(-2 - \lambda) + 12 = 0$

$\lambda^2 - 3\lambda - 10 + 12 = \lambda^2 - 3\lambda + 2 = 0$

$(\lambda - 1)(\lambda - 2) = 0 \implies \lambda_1 = 1, \; \lambda_2 = 2$

For $\lambda_1 = 1$: $(A - I)\mathbf{v} = \mathbf{0}$:

$\begin{pmatrix} 4 & -6 \\ 2 & -3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies 4x - 6y = 0 \implies 2x = 3y$

Eigenvector: $\mathbf{v}_1 = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$.

For $\lambda_2 = 2$: $(A - 2I)\mathbf{v} = \mathbf{0}$:

$\begin{pmatrix} 3 & -6 \\ 2 & -4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies x - 2y = 0$

Eigenvector: $\mathbf{v}_2 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$.

(b)

$P = \begin{pmatrix} 3 & 2 \\ 2 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$

$\det P = 3 - 4 = -1$

$P^{-1} = -\begin{pmatrix} 1 & -2 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ 2 & -3 \end{pmatrix}$

Verify: $PDP^{-1} = \begin{pmatrix} 3 & 2 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} -1 & 2 \\ 2 & -3 \end{pmatrix} = \begin{pmatrix} 3 & 4 \\ 2 & 2 \end{pmatrix}\begin{pmatrix} -1 & 2 \\ 2 & -3 \end{pmatrix}$

$= \begin{pmatrix} -3 + 8 & 6 - 12 \\ -2 + 4 & 4 - 6 \end{pmatrix} = \begin{pmatrix} 5 & -6 \\ 2 & -2 \end{pmatrix} = A$

(c)

$D^5 = \begin{pmatrix} 1^5 & 0 \\ 0 & 2^5 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 32 \end{pmatrix}$

$A^5 = PD^5P^{-1} = \begin{pmatrix} 3 & 2 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 32 \end{pmatrix}\begin{pmatrix} -1 & 2 \\ 2 & -3 \end{pmatrix}$

$= \begin{pmatrix} 3 & 64 \\ 2 & 32 \end{pmatrix}\begin{pmatrix} -1 & 2 \\ 2 & -3 \end{pmatrix} = \begin{pmatrix} -3 + 128 & 6 - 192 \\ -2 + 64 & 4 - 96 \end{pmatrix}$

$= \begin{pmatrix} 125 & -186 \\ 62 & -92 \end{pmatrix}$

(d) The student’s claim is correct with the caveat that the matrix must be diagonalisable. Two distinct eigenvalues are a sufficient condition for diagonalisability, since each eigenvalue contributes at least one eigenvector and eigenvectors corresponding to distinct eigenvalues are linearly independent. So with two distinct eigenvalues for a $2 \times 2$ matrix, we are guaranteed two linearly independent eigenvectors, meaning $P$ is invertible.

However, the student should verify that $P$ is indeed invertible (i.e., $\det P \neq 0$) before using this method. If the eigenvectors were accidentally computed incorrectly (e.g., picking a dependent pair), $P^{-1}$ would not exist.