A-Level Further Mathematics – Chapter 4: Linear Transformations and Eigenvalues Further Pure
Contents
Linear transformations connect the algebraic world of matrices to the geometric world of shapes, rotations, and stretches. This chapter develops the theory systematically: first by understanding what matrices do to the plane and three-dimensional space, then by identifying the special directions (eigenvectors) that matrices merely scale rather than rotate, and finally by exploiting those directions to simplify matrix computations enormously.
4.1 Matrices as Linear Transformations
A linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ is a function satisfying $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$ and $T(k\mathbf{v}) = kT(\mathbf{v})$ for all vectors $\mathbf{u}, \mathbf{v}$ and scalars $k$. Every such transformation can be represented as multiplication by a matrix.
Definition: Matrix of a Linear Transformation
If $T: \mathbb{R}^2 \to \mathbb{R}^2$ is a linear transformation, then there exists a unique $2 \times 2$ matrix $A$ such that $T(\mathbf{x}) = A\mathbf{x}$ for all $\mathbf{x} \in \mathbb{R}^2$. The columns of $A$ are the images of the standard basis vectors: $A = \bigl[T(\mathbf{e}_1) \mid T(\mathbf{e}_2)\bigr]$.
The unit square with vertices at $(0,0)$, $(1,0)$, $(0,1)$, $(1,1)$ is a useful diagnostic tool. Under the transformation $A$, these four vertices map to $(0,0)$, the first column of $A$, the second column of $A$, and the sum of both columns. The area of the image parallelogram equals $|\det(A)|$ times the original area.
Example 4.1.1 — Image of the unit square
Let $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$. Find the image of the unit square and the area scale factor.
Step 1 Map the four vertices:
- $(0,0) \mapsto (0,0)$
- $(1,0) \mapsto A\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}2\\0\end{pmatrix}$
- $(0,1) \mapsto A\begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}1\\3\end{pmatrix}$
- $(1,1) \mapsto A\begin{pmatrix}1\\1\end{pmatrix} = \begin{pmatrix}3\\3\end{pmatrix}$
Step 2 The image is a parallelogram with vertices $(0,0)$, $(2,0)$, $(1,3)$, $(3,3)$.
Step 3 Area scale factor $= |\det(A)| = |2 \times 3 - 1 \times 0| = 6$. The unit square has area 1, so the image has area 6. ▮
Figure 4.1 — The unit square (blue) and its image (red) under $A = \begin{pmatrix}2&1\\0&3\end{pmatrix}$. The area ratio equals $|\det A| = 6$.
Example 4.1.2 — Composition of transformations
Transformation $T_1$ has matrix $A = \begin{pmatrix}0&-1\\1&0\end{pmatrix}$ (rotation by $90°$) and $T_2$ has matrix $B = \begin{pmatrix}2&0\\0&1\end{pmatrix}$ (horizontal stretch). Find the matrix of "$T_1$ followed by $T_2$".
Key rule The combined transformation is $BA$ (apply $A$ first, then $B$ acts on the result).
$$BA = \begin{pmatrix}2&0\\0&1\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix} = \begin{pmatrix}0&-2\\1&0\end{pmatrix}$$Note that $AB \neq BA$ in general — matrix multiplication is not commutative, and the order of composition is critical. ▮
Example 4.1.3 — Identifying the transformation from its matrix
Describe the geometric effect of $C = \begin{pmatrix}-1&0\\0&1\end{pmatrix}$.
The image of $(x, y)$ is $(-x, y)$. This reflects every point in the $y$-axis. The columns of $C$ confirm this: $\mathbf{e}_1 = (1,0)$ maps to $(-1,0)$ and $\mathbf{e}_2 = (0,1)$ is fixed. ▮
Exam Tip — Order of Matrix Multiplication
In A-Level Further Maths, "apply $A$ then $B$" corresponds to the product $BA\mathbf{x}$. The rightmost matrix acts first. Marks are frequently lost by reversing the order. Always check by applying to $\mathbf{e}_1$ and $\mathbf{e}_2$.
4.2 Standard 2D Transformations
The following matrices appear repeatedly in examinations. All should be derived from first principles rather than memorised blindly — the derivations are straightforward and show up as multi-mark questions.
Standard 2D Transformation Matrices
- Rotation by $\theta$ anticlockwise about origin: $R_\theta = \begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}$
- Reflection in the line $y = x\tan\alpha$: $M_\alpha = \begin{pmatrix}\cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha\end{pmatrix}$
- Stretch parallel to $x$-axis by factor $a$: $S_x = \begin{pmatrix}a & 0 \\ 0 & 1\end{pmatrix}$
- Enlargement (isotropic stretch) by factor $k$: $kI = \begin{pmatrix}k & 0 \\ 0 & k\end{pmatrix}$
- Shear parallel to $x$-axis by factor $k$: $H_x = \begin{pmatrix}1 & k \\ 0 & 1\end{pmatrix}$
Example 4.2.1 — Deriving the rotation matrix
Derive $R_{90°}$, the matrix for anticlockwise rotation by $90°$.
$\mathbf{e}_1 = (1,0)$ rotates to $(0,1)$; $\mathbf{e}_2 = (0,1)$ rotates to $(-1,0)$.
Columns of $R_{90°}$ are these images: $R_{90°} = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$.
Using the formula with $\theta = 90°$: $\cos 90° = 0$, $\sin 90° = 1$, giving $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$. ▮
Example 4.2.2 — Reflection in $y = x$
The line $y = x$ corresponds to angle $\alpha = 45°$ (since $\tan 45° = 1$). The reflection matrix is:
$$M_{45°} = \begin{pmatrix}\cos 90° & \sin 90° \\ \sin 90° & -\cos 90°\end{pmatrix} = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$$Check: $(3, 5) \mapsto (5, 3)$. Correct — reflection in $y = x$ swaps coordinates. ▮
Example 4.2.3 — Combined transformation
Find the single matrix for: reflect in the $x$-axis, then rotate anticlockwise by $60°$.
Reflection in the $x$-axis: $\alpha = 0°$, so $M_0 = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$.
Rotation by $60°$: $R_{60°} = \begin{pmatrix}1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & 1/2\end{pmatrix}$.
Combined (rotation after reflection): $R_{60°}M_0 = \begin{pmatrix}1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & 1/2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix} = \begin{pmatrix}1/2 & \sqrt{3}/2 \\ \sqrt{3}/2 & -1/2\end{pmatrix}$. ▮
Example 4.2.4 — Shear transformation
The matrix $H = \begin{pmatrix}1 & 3 \\ 0 & 1\end{pmatrix}$ represents a shear. Find the image of the triangle with vertices $(0,0)$, $(2,0)$, $(1,2)$.
$(0,0) \mapsto (0,0)$; $(2,0) \mapsto (2,0)$; $(1,2) \mapsto (1 + 6, 2) = (7, 2)$.
Note: $\det(H) = 1$, so area is preserved. The triangle with base 2 and height 2 has area 2 both before and after the shear. ▮
Exam Tip — Determinant and Area
For any $2\times2$ transformation matrix $A$, the area scale factor is $|\det A|$. If $\det A = 0$, the transformation is singular and maps the entire plane onto a line (or point), so area becomes zero. Always compute $\det A$ when asked about area.
4.3 3D Transformations
The extension to three dimensions follows the same column-of-images principle. The standard rotations about each coordinate axis are central to this topic.
Standard 3D Rotation Matrices (right-hand rule, anticlockwise when viewed along positive axis toward origin)
- Rotation by $\theta$ about the $x$-axis: $R_x(\theta) = \begin{pmatrix}1&0&0\\0&\cos\theta&-\sin\theta\\0&\sin\theta&\cos\theta\end{pmatrix}$
- Rotation by $\theta$ about the $y$-axis: $R_y(\theta) = \begin{pmatrix}\cos\theta&0&\sin\theta\\0&1&0\\-\sin\theta&0&\cos\theta\end{pmatrix}$
- Rotation by $\theta$ about the $z$-axis: $R_z(\theta) = \begin{pmatrix}\cos\theta&-\sin\theta&0\\\sin\theta&\cos\theta&0\\0&0&1\end{pmatrix}$
Example 4.3.1 — Rotation about the $z$-axis
Find the image of the point $P(1, 0, 2)$ under rotation by $90°$ anticlockwise about the $z$-axis.
$$R_z(90°)\begin{pmatrix}1\\0\\2\end{pmatrix} = \begin{pmatrix}0&-1&0\\1&0&0\\0&0&1\end{pmatrix}\begin{pmatrix}1\\0\\2\end{pmatrix} = \begin{pmatrix}0\\1\\2\end{pmatrix}$$The $z$-coordinate is unchanged (the axis of rotation), while $(x,y) = (1,0)$ rotates to $(0,1)$. ▮
Example 4.3.2 — Reflection in the $xy$-plane
The reflection in the $xy$-plane sends $(x, y, z) \mapsto (x, y, -z)$. Write down its matrix.
Applying to the standard basis: $\mathbf{e}_1 \mapsto \mathbf{e}_1$, $\mathbf{e}_2 \mapsto \mathbf{e}_2$, $\mathbf{e}_3 \mapsto -\mathbf{e}_3$.
$$M_{xy} = \begin{pmatrix}1&0&0\\0&1&0\\0&0&-1\end{pmatrix}$$Note $\det(M_{xy}) = -1$, consistent with a reflection being orientation-reversing. ▮
Example 4.3.3 — Volume scale factor
A general $3\times3$ matrix $A$ scales volumes by $|\det A|$. For $A = \begin{pmatrix}2&0&0\\0&3&0\\0&0&5\end{pmatrix}$, a unit cube maps to a cuboid of volume $2 \times 3 \times 5 = 30$, confirming $\det A = 30$. ▮
4.4 Eigenvalues and Eigenvectors
For most vectors, multiplication by a matrix $A$ changes both the direction and magnitude. An eigenvector is a non-zero vector for which multiplication by $A$ only changes the magnitude (and possibly reverses direction). The scale factor is the corresponding eigenvalue.
Definition: Eigenvalue and Eigenvector
Let $A$ be an $n\times n$ matrix. A scalar $\lambda$ is an eigenvalue of $A$ if there exists a non-zero vector $\mathbf{v}$ such that:
$$A\mathbf{v} = \lambda\mathbf{v}$$The vector $\mathbf{v}$ is the corresponding eigenvector. The set of all eigenvectors corresponding to $\lambda$ (together with the zero vector) forms the eigenspace for $\lambda$.
Theorem: Characteristic Equation
The eigenvalues of $A$ are the solutions of the characteristic equation:
$$\det(A - \lambda I) = 0$$For a $2\times2$ matrix, this gives a quadratic; for a $3\times3$ matrix, a cubic. The characteristic polynomial of a $2\times2$ matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ is $\lambda^2 - (a+d)\lambda + (ad-bc) = \lambda^2 - \operatorname{tr}(A)\lambda + \det(A)$.
Example 4.4.1 — Eigenvalues of a $2\times2$ matrix
Find the eigenvalues of $A = \begin{pmatrix}3&1\\0&2\end{pmatrix}$.
Step 1 Form $\det(A - \lambda I) = 0$:
$$\det\begin{pmatrix}3-\lambda & 1 \\ 0 & 2-\lambda\end{pmatrix} = (3-\lambda)(2-\lambda) - 0 = 0$$Step 2 Solve: $(3-\lambda)(2-\lambda) = 0$, so $\lambda = 3$ or $\lambda = 2$. ▮
Example 4.4.2 — Eigenvectors of a $2\times2$ matrix
For $A = \begin{pmatrix}3&1\\0&2\end{pmatrix}$, find the eigenvectors.
For $\lambda = 3$: Solve $(A - 3I)\mathbf{v} = \mathbf{0}$:
$$\begin{pmatrix}0&1\\0&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix} \implies y = 0$$Eigenvector: $\mathbf{v}_1 = \begin{pmatrix}1\\0\end{pmatrix}$ (any non-zero multiple).
For $\lambda = 2$: Solve $(A - 2I)\mathbf{v} = \mathbf{0}$:
$$\begin{pmatrix}1&1\\0&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix} \implies x = -y$$Eigenvector: $\mathbf{v}_2 = \begin{pmatrix}1\\-1\end{pmatrix}$. ▮
Figure 4.2 — Eigenvectors of $A = \begin{pmatrix}3&1\\0&2\end{pmatrix}$: the red arrow (along the $x$-axis) is scaled by $\lambda=3$; the blue arrow is scaled by $\lambda=2$.
Example 4.4.3 — Eigenvalues of a $3\times3$ matrix
Find the eigenvalues of $B = \begin{pmatrix}2&0&0\\1&3&0\\0&2&4\end{pmatrix}$ (lower triangular).
For any triangular matrix, the eigenvalues are the entries on the main diagonal. Therefore $\lambda = 2, 3, 4$.
Verification: $\det(B - \lambda I) = (2-\lambda)(3-\lambda)(4-\lambda)$ since the matrix remains lower triangular after subtracting $\lambda I$. ▮
Example 4.4.4 — Repeated eigenvalue
Find all eigenvalues and eigenvectors of $C = \begin{pmatrix}5&-4\\1&1\end{pmatrix}$.
$\det(C - \lambda I) = (5-\lambda)(1-\lambda) + 4 = \lambda^2 - 6\lambda + 9 = (\lambda - 3)^2 = 0$.
Repeated eigenvalue $\lambda = 3$. Solve $(C - 3I)\mathbf{v} = \mathbf{0}$:
$$\begin{pmatrix}2&-4\\1&-2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \mathbf{0} \implies x = 2y$$Only one linearly independent eigenvector: $\mathbf{v} = \begin{pmatrix}2\\1\end{pmatrix}$. This matrix is not diagonalisable over $\mathbb{R}$. ▮
Exam Tip — Eigenvectors are Directions, not Points
Any non-zero scalar multiple of an eigenvector is also an eigenvector for the same eigenvalue. In exam questions you may write any convenient multiple, but check: the eigenvector must be non-zero, and you must show the working that leads to the eigenspace equation.
4.5 Diagonalisation
If a matrix $A$ has $n$ linearly independent eigenvectors, it can be diagonalised: written as $A = PDP^{-1}$, where $D$ is diagonal (eigenvalues on the diagonal) and $P$ has the corresponding eigenvectors as columns. Diagonalisation makes it trivial to compute powers of $A$.
Theorem: Diagonalisation
Suppose $n \times n$ matrix $A$ has eigenvalues $\lambda_1, \ldots, \lambda_n$ with corresponding linearly independent eigenvectors $\mathbf{v}_1, \ldots, \mathbf{v}_n$. Let $P = [\mathbf{v}_1 \mid \cdots \mid \mathbf{v}_n]$ and $D = \operatorname{diag}(\lambda_1, \ldots, \lambda_n)$. Then:
$$A = PDP^{-1} \qquad \text{and} \qquad A^n = PD^nP^{-1}$$where $D^n = \operatorname{diag}(\lambda_1^n, \ldots, \lambda_n^n)$.
Example 4.5.1 — Diagonalising a $2\times2$ matrix
Diagonalise $A = \begin{pmatrix}3&1\\0&2\end{pmatrix}$ (eigenvalues $\lambda_1 = 3$, $\lambda_2 = 2$; eigenvectors $\mathbf{v}_1 = \begin{pmatrix}1\\0\end{pmatrix}$, $\mathbf{v}_2 = \begin{pmatrix}1\\-1\end{pmatrix}$).
Form $P = \begin{pmatrix}1&1\\0&-1\end{pmatrix}$, $D = \begin{pmatrix}3&0\\0&2\end{pmatrix}$.
Find $P^{-1}$: $\det P = -1$, so $P^{-1} = \frac{1}{-1}\begin{pmatrix}-1&-1\\0&1\end{pmatrix} = \begin{pmatrix}1&1\\0&-1\end{pmatrix}$.
Verify: $PDP^{-1} = \begin{pmatrix}1&1\\0&-1\end{pmatrix}\begin{pmatrix}3&0\\0&2\end{pmatrix}\begin{pmatrix}1&1\\0&-1\end{pmatrix} = \begin{pmatrix}3&1\\0&2\end{pmatrix} = A$. ▮
Example 4.5.2 — Computing $A^{10}$ by diagonalisation
Using the diagonalisation above, compute $A^{10}$.
$$A^{10} = PD^{10}P^{-1} = \begin{pmatrix}1&1\\0&-1\end{pmatrix}\begin{pmatrix}3^{10}&0\\0&2^{10}\end{pmatrix}\begin{pmatrix}1&1\\0&-1\end{pmatrix}$$ $$= \begin{pmatrix}3^{10}&2^{10}\\0&-2^{10}\end{pmatrix}\begin{pmatrix}1&1\\0&-1\end{pmatrix} = \begin{pmatrix}3^{10}&3^{10}-2^{10}\\0&2^{10}\end{pmatrix}$$Numerically: $3^{10} = 59049$, $2^{10} = 1024$, so $A^{10} = \begin{pmatrix}59049 & 58025 \\ 0 & 1024\end{pmatrix}$. ▮
Example 4.5.3 — Conditions for diagonalisability
A matrix is diagonalisable over $\mathbb{R}$ if and only if it has $n$ linearly independent real eigenvectors. Key cases:
- If all $n$ eigenvalues are distinct, the matrix is automatically diagonalisable.
- If eigenvalue $\lambda$ has algebraic multiplicity $m$ (appears $m$ times in characteristic equation), it must also have geometric multiplicity $m$ (dimension of eigenspace = $m$) for diagonalisability.
- Rotation matrices (with $\theta \neq 0°, 180°$) have complex eigenvalues and are not diagonalisable over $\mathbb{R}$.
4.6 The Cayley-Hamilton Theorem
Theorem: Cayley-Hamilton
Every square matrix satisfies its own characteristic equation. That is, if $p(\lambda) = \det(A - \lambda I)$ is the characteristic polynomial of $A$, then $p(A) = O$ (the zero matrix).
For a $2\times2$ matrix with characteristic equation $\lambda^2 - (\operatorname{tr} A)\lambda + \det A = 0$:
$$A^2 - (\operatorname{tr} A)A + (\det A)I = O$$Example 4.6.1 — Verification of Cayley-Hamilton
Verify Cayley-Hamilton for $A = \begin{pmatrix}3&1\\0&2\end{pmatrix}$.
Characteristic equation: $\lambda^2 - 5\lambda + 6 = 0$ (since $\operatorname{tr} A = 5$, $\det A = 6$).
Compute $A^2 = \begin{pmatrix}3&1\\0&2\end{pmatrix}\begin{pmatrix}3&1\\0&2\end{pmatrix} = \begin{pmatrix}9&5\\0&4\end{pmatrix}$.
Check: $A^2 - 5A + 6I = \begin{pmatrix}9&5\\0&4\end{pmatrix} - \begin{pmatrix}15&5\\0&10\end{pmatrix} + \begin{pmatrix}6&0\\0&6\end{pmatrix} = \begin{pmatrix}0&0\\0&0\end{pmatrix}$. ▮
Example 4.6.2 — Using Cayley-Hamilton to find $A^{-1}$
For $A = \begin{pmatrix}4&1\\2&3\end{pmatrix}$, find $A^{-1}$ using Cayley-Hamilton.
$\det A = 12 - 2 = 10$, $\operatorname{tr} A = 7$. Characteristic equation: $A^2 - 7A + 10I = O$.
Rearranging: $A(A - 7I) = -10I$, so $A\cdot\frac{-(A-7I)}{10} = I$.
$$A^{-1} = \frac{7I - A}{10} = \frac{1}{10}\begin{pmatrix}3&-1\\-2&4\end{pmatrix}$$This is simply the standard formula $A^{-1} = \frac{1}{\det A}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$, confirmed by Cayley-Hamilton. ▮
Example 4.6.3 — Reducing higher matrix powers
Use Cayley-Hamilton to express $A^4$ in terms of $A$ and $I$, where $A^2 = 7A - 10I$.
$$A^3 = A \cdot A^2 = A(7A - 10I) = 7A^2 - 10A = 7(7A-10I) - 10A = 39A - 70I$$ $$A^4 = A \cdot A^3 = A(39A - 70I) = 39A^2 - 70A = 39(7A-10I) - 70A = 203A - 390I$$This technique reduces any polynomial in $A$ to a linear combination of $A$ and $I$. ▮
Exam Tip — Cayley-Hamilton Shortcuts
The Cayley-Hamilton theorem is tested in two ways: (1) verify it holds for a given matrix, and (2) use it to express $A^n$, $A^{-1}$, or a polynomial in $A$ more simply. Always start by writing the characteristic equation, then substitute $A$ for $\lambda$ and $O$ for zero.
Practice Problems
Problem 1
The matrix $A = \begin{pmatrix}1 & -1 \\ 1 & 1\end{pmatrix}$ represents a transformation of the plane. Describe the transformation geometrically and find the area scale factor.
Show solution
$\mathbf{e}_1 = (1,0) \mapsto (1,1)$ and $\mathbf{e}_2 = (0,1) \mapsto (-1,1)$. The angle that $(1,1)$ makes with the $x$-axis is $45°$, and its magnitude is $\sqrt{2}$. So $A$ represents an anticlockwise rotation by $45°$ combined with an enlargement by scale factor $\sqrt{2}$.
Area scale factor $= |\det A| = |1 \cdot 1 - (-1) \cdot 1| = 2$.
Problem 2
Find the matrix that represents reflection in the line $y = x\tan 30°$ (i.e., the line at $30°$ to the positive $x$-axis).
Show solution
Using $M_\alpha = \begin{pmatrix}\cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha\end{pmatrix}$ with $\alpha = 30°$:
$$M_{30°} = \begin{pmatrix}\cos 60° & \sin 60° \\ \sin 60° & -\cos 60°\end{pmatrix} = \begin{pmatrix}1/2 & \sqrt{3}/2 \\ \sqrt{3}/2 & -1/2\end{pmatrix}$$Problem 3
Find the eigenvalues and eigenvectors of $B = \begin{pmatrix}5 & 2 \\ 2 & 5\end{pmatrix}$.
Show solution
Characteristic equation: $(5-\lambda)^2 - 4 = 0 \Rightarrow \lambda^2 - 10\lambda + 21 = 0 \Rightarrow (\lambda-7)(\lambda-3) = 0$.
Eigenvalues: $\lambda = 7$ and $\lambda = 3$.
For $\lambda = 7$: $(B - 7I)\mathbf{v} = \mathbf{0} \Rightarrow \begin{pmatrix}-2&2\\2&-2\end{pmatrix}\mathbf{v} = \mathbf{0} \Rightarrow x = y$. Eigenvector: $\begin{pmatrix}1\\1\end{pmatrix}$.
For $\lambda = 3$: $(B - 3I)\mathbf{v} = \mathbf{0} \Rightarrow \begin{pmatrix}2&2\\2&2\end{pmatrix}\mathbf{v} = \mathbf{0} \Rightarrow x = -y$. Eigenvector: $\begin{pmatrix}1\\-1\end{pmatrix}$.
Problem 4
Diagonalise the matrix $B = \begin{pmatrix}5&2\\2&5\end{pmatrix}$ from Problem 3, and hence find $B^6$.
Show solution
$P = \begin{pmatrix}1&1\\1&-1\end{pmatrix}$, $D = \begin{pmatrix}7&0\\0&3\end{pmatrix}$, $P^{-1} = \frac{1}{-2}\begin{pmatrix}-1&-1\\-1&1\end{pmatrix} = \frac{1}{2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$.
$B^6 = PD^6P^{-1} = \begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}7^6&0\\0&3^6\end{pmatrix}\cdot\frac{1}{2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$.
$7^6 = 117649$, $3^6 = 729$. Result: $B^6 = \frac{1}{2}\begin{pmatrix}117649+729 & 117649-729 \\ 117649-729 & 117649+729\end{pmatrix} = \begin{pmatrix}59189 & 58460 \\ 58460 & 59189\end{pmatrix}$.
Problem 5
Find the eigenvalues of $C = \begin{pmatrix}1&0&0\\0&2&3\\0&3&2\end{pmatrix}$ and state whether $C$ is diagonalisable.
Show solution
The characteristic equation: expand along the first row. $\det(C - \lambda I) = (1-\lambda)\det\begin{pmatrix}2-\lambda&3\\3&2-\lambda\end{pmatrix} = (1-\lambda)[(2-\lambda)^2 - 9]$.
$(2-\lambda)^2 - 9 = 0 \Rightarrow 2 - \lambda = \pm 3 \Rightarrow \lambda = -1$ or $\lambda = 5$.
So eigenvalues are $\lambda = 1, -1, 5$ — three distinct real eigenvalues. $C$ is diagonalisable.
Problem 6
Verify the Cayley-Hamilton theorem for $A = \begin{pmatrix}2 & 3 \\ 1 & 4\end{pmatrix}$.
Show solution
$\operatorname{tr} A = 6$, $\det A = 8 - 3 = 5$. Characteristic equation: $\lambda^2 - 6\lambda + 5 = 0$.
$A^2 = \begin{pmatrix}2&3\\1&4\end{pmatrix}\begin{pmatrix}2&3\\1&4\end{pmatrix} = \begin{pmatrix}7&18\\6&19\end{pmatrix}$.
$A^2 - 6A + 5I = \begin{pmatrix}7&18\\6&19\end{pmatrix} - \begin{pmatrix}12&18\\6&24\end{pmatrix} + \begin{pmatrix}5&0\\0&5\end{pmatrix} = \begin{pmatrix}0&0\\0&0\end{pmatrix}$. ✓
Problem 7
Use Cayley-Hamilton to find $A^{-1}$ for $A = \begin{pmatrix}3 & 5 \\ 1 & 2\end{pmatrix}$.
Show solution
$\operatorname{tr} A = 5$, $\det A = 6 - 5 = 1$. Characteristic equation: $A^2 - 5A + I = O$.
So $A(A - 5I) = -I$, giving $A^{-1} = 5I - A = \begin{pmatrix}5&0\\0&5\end{pmatrix} - \begin{pmatrix}3&5\\1&2\end{pmatrix} = \begin{pmatrix}2&-5\\-1&3\end{pmatrix}$.
Check: $AA^{-1} = \begin{pmatrix}3&5\\1&2\end{pmatrix}\begin{pmatrix}2&-5\\-1&3\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix}$. ✓
Problem 8
The transformation $T$ has matrix $A = \begin{pmatrix}0&-1&0\\1&0&0\\0&0&1\end{pmatrix}$. Describe $T$ geometrically and find $A^4$.
Show solution
$A$ is a rotation by $90°$ anticlockwise about the $z$-axis (compare with $R_z(90°)$).
$A^2$ is rotation by $180°$, $A^3$ by $270°$, and $A^4$ by $360°$ = identity. So $A^4 = I$.
Problem 9
A matrix $A$ has eigenvalues $2$ and $5$ with eigenvectors $\begin{pmatrix}3\\1\end{pmatrix}$ and $\begin{pmatrix}1\\1\end{pmatrix}$ respectively. Write down $A$.
Show solution
$P = \begin{pmatrix}3&1\\1&1\end{pmatrix}$, $D = \begin{pmatrix}2&0\\0&5\end{pmatrix}$. $\det P = 2$, $P^{-1} = \frac{1}{2}\begin{pmatrix}1&-1\\-1&3\end{pmatrix}$.
$A = PDP^{-1} = \begin{pmatrix}3&1\\1&1\end{pmatrix}\begin{pmatrix}2&0\\0&5\end{pmatrix}\cdot\frac{1}{2}\begin{pmatrix}1&-1\\-1&3\end{pmatrix} = \begin{pmatrix}6&5\\2&5\end{pmatrix}\cdot\frac{1}{2}\begin{pmatrix}1&-1\\-1&3\end{pmatrix}$.
$= \frac{1}{2}\begin{pmatrix}1&9\\-3&7\end{pmatrix} = \begin{pmatrix}1/2 & 9/2 \\ -3/2 & 7/2\end{pmatrix}$.
Or equivalently: $A = \frac{1}{2}\begin{pmatrix}1&9\\-3&7\end{pmatrix}$.
Problem 10
The shear matrix $H = \begin{pmatrix}1&k\\0&1\end{pmatrix}$ has a repeated eigenvalue. Find it, and explain why $H$ is not diagonalisable for $k \neq 0$.
Show solution
$\det(H - \lambda I) = (1-\lambda)^2 = 0$, so the only eigenvalue is $\lambda = 1$ with algebraic multiplicity 2.
$(H - I)\mathbf{v} = \begin{pmatrix}0&k\\0&0\end{pmatrix}\mathbf{v} = \mathbf{0}$. For $k \neq 0$, this forces $y = 0$, so all eigenvectors are multiples of $\begin{pmatrix}1\\0\end{pmatrix}$. The eigenspace has dimension 1 (geometric multiplicity 1), which is less than the algebraic multiplicity 2. Hence $H$ is not diagonalisable over $\mathbb{R}$ when $k \neq 0$.