Chapter 09

Change of Basis

00 · Symbol Glossary

$\mathcal{B}$script B — basis label

A calligraphic letter labelling a specific basis. $\mathcal{B} = \{\mathbf{b}_1, \mathbf{b}_2, \ldots, \mathbf{b}_n\}$ names a choice of $n$ linearly independent vectors that span the space. Used to distinguish one basis from another when multiple bases are in play at once.

$[\mathbf{v}]_\mathcal{B}$v in basis B — coordinate vector

The coordinate representation of $\mathbf{v}$ with respect to basis $\mathcal{B}$ . If $\mathbf{v} = c_1\mathbf{b}_1 + c_2\mathbf{b}_2 + \cdots + c_n\mathbf{b}_n$ , then $[\mathbf{v}]_\mathcal{B} = \begin{pmatrix}c_1\\\vdots\\c_n\end{pmatrix}$ . The same geometric vector $\mathbf{v}$ has different coordinate vectors in different bases — this is the core idea of the whole chapter.

$P_{\mathcal{B}}$P sub B — change-of-basis matrix

The matrix whose columns are the basis vectors of $\mathcal{B}$ written in standard coordinates. Converts $\mathcal{B}$ -coordinates to standard coordinates: $\mathbf{v} = P_\mathcal{B} [\mathbf{v}]_\mathcal{B}$ . Its inverse goes the other direction: $[\mathbf{v}]_\mathcal{B} = P_\mathcal{B}^{-1}\mathbf{v}$ .

$[T]_\mathcal{B}$T in basis B — matrix representation

The matrix of a linear transformation $T$ expressed relative to basis $\mathcal{B}$ . The same transformation looks like a different matrix when you change the basis — and choosing the right basis can make the matrix diagonal (Chapter 7 diagonalisation is a special case of this).

$P^{-1}AP$P inverse A P — similarity transform

A similarity transformation. Matrices $A$ and $P^{-1}AP$ represent the same linear transformation in different bases. They share eigenvalues, determinant, and trace — all invariants that do not depend on the choice of basis.

01 · What Changes of Basis Mean

You already know that a vector $\mathbf{v} \in \mathbb{R}^2$ has coordinates $(3, 5)$ — meaning $3$ steps in the $x$ -direction and $5$ steps in the $y$ -direction. But those coordinates depend entirely on the choice of axes. Rotate the axes 45°, and the same physical arrow has completely different coordinate numbers. The arrow does not change; the measuring frame does.

This is the heart of change of basis: one geometric object, infinitely many coordinate descriptions. The standard basis $\{\mathbf{e}_1, \mathbf{e}_2\}$ is just one choice. Any two linearly independent vectors form a valid alternative.

Definition — Coordinates in a Basis

Let $\mathcal{B} = \{\mathbf{b}_1, \ldots, \mathbf{b}_n\}$ be a basis for $\mathbb{R}^n$ . For any $\mathbf{v} \in \mathbb{R}^n$ there exist unique scalars $c_1, \ldots, c_n$ such that:

\mathbf{v} = c_1\mathbf{b}_1 + c_2\mathbf{b}_2 + \cdots + c_n\mathbf{b}_n

The coordinate vector of $\mathbf{v}$ in basis $\mathcal{B}$ is:

[\mathbf{v}]_\mathcal{B} = \begin{pmatrix}c_1\\c_2\\\vdots\\c_n\end{pmatrix}

$c_i$ — the scalar coefficient on the $i$ -th basis vector. Uniqueness is guaranteed because $\mathcal{B}$ is a basis (linearly independent and spanning).

✓ Example — Same Vector, Two Descriptions

Let $\mathcal{B} = \left\{\mathbf{b}_1 = \begin{pmatrix}1\\1\end{pmatrix},\; \mathbf{b}_2 = \begin{pmatrix}1\\-1\end{pmatrix}\right\}$ and $\mathbf{v} = \begin{pmatrix}3\\1\end{pmatrix}$ .

In standard coordinates: $\mathbf{v} = \begin{pmatrix}3\\1\end{pmatrix}$ .

In $\mathcal{B}$ -coordinates: solve $c_1\begin{pmatrix}1\\1\end{pmatrix} + c_2\begin{pmatrix}1\\-1\end{pmatrix} = \begin{pmatrix}3\\1\end{pmatrix}$ .

Row 1: $c_1 + c_2 = 3$ . Row 2: $c_1 - c_2 = 1$ . Adding: $2c_1 = 4 \Rightarrow c_1 = 2$ . Then $c_2 = 1$ .

Therefore $[\mathbf{v}]_\mathcal{B} = \begin{pmatrix}2\\1\end{pmatrix}$ . The same arrow has coordinates $(3,1)$ in the standard frame and $(2,1)$ in the $\mathcal{B}$ -frame.

❌ Common Mistake — Coordinates Depend on Basis

Writing $[\mathbf{v}]_\mathcal{B} = \begin{pmatrix}3\\1\end{pmatrix}$ by simply copying the standard-coordinate entries. The standard coordinates of $\mathbf{v}$ are its coefficients with respect to $\{\mathbf{e}_1, \mathbf{e}_2\}$ , not $\mathcal{B}$ . Unless $\mathcal{B}$ happens to be the standard basis, $[\mathbf{v}]_\mathcal{B} \neq \mathbf{v}$ . You must solve the linear system $P_\mathcal{B} [\mathbf{v}]_\mathcal{B} = \mathbf{v}$ every time.

02 · The Change-of-Basis Matrix

Solving a linear system each time is slow. The change-of-basis matrix packages the conversion into a single matrix multiplication.

Definition — Change-of-Basis Matrix

Given a basis $\mathcal{B} = \{\mathbf{b}_1, \ldots, \mathbf{b}_n\}$ for $\mathbb{R}^n$ , the change-of-basis matrix is:

P_\mathcal{B} = \begin{pmatrix} \mathbf{b}_1 & \mathbf{b}_2 & \cdots & \mathbf{b}_n \end{pmatrix}

the $n \times n$ matrix whose columns are the basis vectors in standard coordinates.

To convert $\mathcal{B}$ -coordinates to standard: $\mathbf{v} = P_\mathcal{B}\, [\mathbf{v}]_\mathcal{B}$

To convert standard to $\mathcal{B}$ -coordinates: $[\mathbf{v}]_\mathcal{B} = P_\mathcal{B}^{-1}\, \mathbf{v}$

$P_\mathcal{B}$ is always invertible because the columns are linearly independent (they form a basis).

Step-by-step — Converting $\mathbf{v}=\begin{pmatrix}4\\2\end{pmatrix}$ to $\mathcal{B}$-coordinates where $\mathcal{B}=\left\{\begin{pmatrix}2\\1\end{pmatrix},\begin{pmatrix}0\\1\end{pmatrix}\right\}$

Write the change-of-basis matrix: place the basis vectors as columns.

P_\mathcal{B} = \begin{pmatrix}2 & 0 \\ 1 & 1\end{pmatrix}

Column 1 is $\mathbf{b}_1 = (2,1)^T$ , column 2 is $\mathbf{b}_2 = (0,1)^T$ .

Compute $P_\mathcal{B}^{-1}$ : for a $2\times 2$ matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ , the inverse is $\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ .

$\det(P_\mathcal{B}) = 2\cdot1 - 0\cdot1 = 2$ . Therefore:

P_\mathcal{B}^{-1} = \frac{1}{2}\begin{pmatrix}1 & 0 \\ -1 & 2\end{pmatrix}

The $\frac{1}{2}$ comes from dividing every entry by $\det = 2$ .

Multiply $P_\mathcal{B}^{-1}\mathbf{v}$ : convert $\mathbf{v}=\begin{pmatrix}4\\2\end{pmatrix}$ to $\mathcal{B}$ -coordinates.

[\mathbf{v}]_\mathcal{B} = \frac{1}{2}\begin{pmatrix}1 & 0 \\ -1 & 2\end{pmatrix}\begin{pmatrix}4\\2\end{pmatrix} = \frac{1}{2}\begin{pmatrix}4\\-4+4\end{pmatrix} = \frac{1}{2}\begin{pmatrix}4\\0\end{pmatrix} = \begin{pmatrix}2\\0\end{pmatrix}

Row 1: $1(4)+0(2)=4$ , then $\frac{4}{2}=2$ . Row 2: $-1(4)+2(2)=-4+4=0$ , then $\frac{0}{2}=0$ .

Verify: $P_\mathcal{B}[\mathbf{v}]_\mathcal{B} = \begin{pmatrix}2&0\\1&1\end{pmatrix}\begin{pmatrix}2\\0\end{pmatrix} = \begin{pmatrix}4\\2\end{pmatrix} = \mathbf{v}$ ✓. The coordinates $(2,0)$ mean $\mathbf{v} = 2\mathbf{b}_1 + 0\mathbf{b}_2$ — the vector $\mathbf{v}$ is simply twice $\mathbf{b}_1$ .

03 · Changing Between Two Non-Standard Bases

When both the source and target are non-standard bases, the conversion routes through standard coordinates.

Definition — Basis Change Between $\mathcal{B}$ and $\mathcal{C}$

Let $\mathcal{B}$ and $\mathcal{C}$ be two bases for $\mathbb{R}^n$ with change-of-basis matrices $P_\mathcal{B}$ and $P_\mathcal{C}$ . Then:

[\mathbf{v}]_\mathcal{C} = P_\mathcal{C}^{-1} P_\mathcal{B}\, [\mathbf{v}]_\mathcal{B}

The matrix $P_\mathcal{C}^{-1} P_\mathcal{B}$ is the change-of-basis matrix from $\mathcal{B}$ to $\mathcal{C}$ . Reading right to left: first convert $\mathcal{B}$ -coordinates to standard ( $P_\mathcal{B}$ ), then standard to $\mathcal{C}$ -coordinates ( $P_\mathcal{C}^{-1}$ ).

Standard Basis Is the Relay Station

Every coordinate conversion routes through standard coordinates. You never need a direct formula between arbitrary bases — compose the two matrices $P_\mathcal{C}^{-1}$ and $P_\mathcal{B}$ and the routing happens automatically.

04 · Matrices Under Change of Basis

When a linear transformation $T: \mathbb{R}^n \to \mathbb{R}^n$ has standard matrix $A$ , its matrix in a different basis $\mathcal{B}$ is related by a similarity transformation.

Definition — Similarity Transformation

Matrices $A$ and $B$ are similar if there exists an invertible matrix $P$ such that:

B = P^{-1}AP

$B$ is the matrix of the same linear transformation, but expressed in the basis whose vectors are the columns of $P$ . Similar matrices represent the same transformation and share: eigenvalues, determinant, trace, rank, and characteristic polynomial.

✓ Example — Diagonalisation Is Change of Basis

Diagonalising $A = PDP^{-1}$ is exactly a change of basis. In the basis of eigenvectors (columns of $P$ ), the transformation becomes the diagonal matrix $D = P^{-1}AP$ . The transformation scales each eigenvector by its eigenvalue — a much simpler picture than the off-diagonal mixing in $A$ .

This is why diagonalisation matters: it finds the natural coordinate system for $A$ .

Step-by-step — Finding $[T]_\mathcal{B}$ for $T(\mathbf{x})=A\mathbf{x}$ with $A=\begin{pmatrix}3&1\\0&2\end{pmatrix}$ in basis $\mathcal{B}=\left\{\begin{pmatrix}1\\0\end{pmatrix},\begin{pmatrix}1\\1\end{pmatrix}\right\}$

Write $P_\mathcal{B}$ : columns are the basis vectors in standard order.

P = P_\mathcal{B} = \begin{pmatrix}1&1\\0&1\end{pmatrix}

Compute $P^{-1}$ : $\det(P)=1\cdot1-1\cdot0=1$ , so $P^{-1} = \begin{pmatrix}1&-1\\0&1\end{pmatrix}$ .

Compute $AP$ : right-multiply $A$ by $P$ .

AP = \begin{pmatrix}3&1\\0&2\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix} = \begin{pmatrix}3&4\\0&2\end{pmatrix}

Entry $(1,1)$ : $3(1)+1(0)=3$ . Entry $(1,2)$ : $3(1)+1(1)=4$ . Entry $(2,1)$ : $0+2(0)=0$ . Entry $(2,2)$ : $0+2(1)=2$ .

Compute $[T]_\mathcal{B} = P^{-1}AP$ :

[T]_\mathcal{B} = \begin{pmatrix}1&-1\\0&1\end{pmatrix}\begin{pmatrix}3&4\\0&2\end{pmatrix} = \begin{pmatrix}3&2\\0&2\end{pmatrix}

Entry $(1,1)$ : $1(3)+(-1)(0)=3$ . Entry $(1,2)$ : $1(4)+(-1)(2)=2$ . Entry $(2,1)$ : $0$ . Entry $(2,2)$ : $1(2)=2$ .

In basis $\mathcal{B}$ , the transformation matrix is upper triangular — same eigenvalues $3$ and $2$ , but simpler off-diagonal structure.

❌ What Breaks — Matrix Change of Basis Is Not Transpose

A common error is computing $P^T A P$ instead of $P^{-1}AP$ . This equals the similarity transformation only when $P$ is orthogonal ( $P^T = P^{-1}$ ). For a general change-of-basis matrix, $P^T \neq P^{-1}$ , so $P^TAP$ produces a wrong result that does not represent the transformation in the new basis.

05 · Quant Application — PCA as Change of Basis

Principal Component Analysis is exactly a change of basis for a covariance matrix. Given a sample covariance matrix $\Sigma \in \mathbb{R}^{p \times p}$ (symmetric, positive semi-definite), the spectral theorem guarantees an orthogonal diagonalisation:

\Sigma = Q \Lambda Q^T

where $Q$ has orthonormal eigenvectors as columns and $\Lambda = \text{diag}(\lambda_1, \ldots, \lambda_p)$ with $\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_p \geq 0$ .

Changing to the eigenvector basis means $[\Sigma]_Q = Q^{-1}\Sigma Q = Q^T \Sigma Q = \Lambda$ . In the eigenvector coordinate system, all assets' risk contributions are diagonal — uncorrelated. The original covariance matrix had cross-correlation terms; in the PC basis those vanish.

The first column of $Q$ points in the direction of maximum variance ( $\lambda_1$ largest). Projecting the return data onto $Q$ gives the principal component scores — a new set of uncorrelated factors that explain variance in decreasing order.

06 · Practice Exercises

EXERCISE 9.1

Write the change-of-basis matrix $P_\mathcal{B}$ by placing $\mathbf{b}_1$ and $\mathbf{b}_2$ as columns. Then apply $P_\mathcal{B}^{-1}$ to $\mathbf{v}$ .

$P_\mathcal{B} = \begin{pmatrix}1&2\\2&3\end{pmatrix}$ .

$\det(P_\mathcal{B}) = 1(3)-2(2) = 3-4 = -1$ .

$P_\mathcal{B}^{-1} = \frac{1}{-1}\begin{pmatrix}3&-2\\-2&1\end{pmatrix} = \begin{pmatrix}-3&2\\2&-1\end{pmatrix}$ .

$[\mathbf{v}]_\mathcal{B} = \begin{pmatrix}-3&2\\2&-1\end{pmatrix}\begin{pmatrix}5\\7\end{pmatrix} = \begin{pmatrix}-15+14\\10-7\end{pmatrix} = \begin{pmatrix}-1\\3\end{pmatrix}$ .

Verify: $\begin{pmatrix}1&2\\2&3\end{pmatrix}\begin{pmatrix}-1\\3\end{pmatrix} = \begin{pmatrix}-1+6\\-2+9\end{pmatrix} = \begin{pmatrix}5\\7\end{pmatrix} = \mathbf{v}$ ✓.

Let $\mathcal{B} = \left\{\begin{pmatrix}1\\2\end{pmatrix}, \begin{pmatrix}2\\3\end{pmatrix}\right\}$ and $\mathbf{v} = \begin{pmatrix}5\\7\end{pmatrix}$ . Find $[\mathbf{v}]_\mathcal{B}$ .

EXERCISE 9.2

Compute the change-of-basis matrix from $\mathcal{B}$ to $\mathcal{C}$ as $P_\mathcal{C}^{-1} P_\mathcal{B}$ . Apply it to the given coordinates.

$P_\mathcal{B} = \begin{pmatrix}2&0\\0&1\end{pmatrix}$ and $P_\mathcal{C} = \begin{pmatrix}1&1\\0&2\end{pmatrix}$ .

$\det(P_\mathcal{C})=2$ , so $P_\mathcal{C}^{-1} = \frac{1}{2}\begin{pmatrix}2&-1\\0&1\end{pmatrix}$ .

$[\mathbf{v}]_\mathcal{C} = P_\mathcal{C}^{-1}P_\mathcal{B}[\mathbf{v}]_\mathcal{B} = \frac{1}{2}\begin{pmatrix}2&-1\\0&1\end{pmatrix}\begin{pmatrix}2&0\\0&1\end{pmatrix}\begin{pmatrix}3\\1\end{pmatrix}$ .

$P_\mathcal{C}^{-1}P_\mathcal{B} = \frac{1}{2}\begin{pmatrix}4&-1\\0&1\end{pmatrix}$ .

$\frac{1}{2}\begin{pmatrix}4&-1\\0&1\end{pmatrix}\begin{pmatrix}3\\1\end{pmatrix} = \frac{1}{2}\begin{pmatrix}12-1\\1\end{pmatrix} = \begin{pmatrix}11/2\\1/2\end{pmatrix}$ .

Let $\mathcal{B} = \left\{\begin{pmatrix}2\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix}\right\}$ and $\mathcal{C} = \left\{\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}1\\2\end{pmatrix}\right\}$ . Given $[\mathbf{v}]_\mathcal{B} = \begin{pmatrix}3\\1\end{pmatrix}$ , find $[\mathbf{v}]_\mathcal{C}$ .

EXERCISE 9.3

Form $P$ from the eigenvectors and compute $P^{-1}AP$ . The result should be diagonal.

Eigenvalues of $A=\begin{pmatrix}4&2\\1&3\end{pmatrix}$ : $p(\lambda)=(4-\lambda)(3-\lambda)-2=\lambda^2-7\lambda+10=(\lambda-5)(\lambda-2)$ , so $\lambda_1=5$ , $\lambda_2=2$ .

For $\lambda=5$ : $(A-5I)=\begin{pmatrix}-1&2\\1&-2\end{pmatrix}$ ; eigenvector $\mathbf{v}_1=\begin{pmatrix}2\\1\end{pmatrix}$ .

For $\lambda=2$ : $(A-2I)=\begin{pmatrix}2&2\\1&1\end{pmatrix}$ ; eigenvector $\mathbf{v}_2=\begin{pmatrix}-1\\1\end{pmatrix}$ .

$P=\begin{pmatrix}2&-1\\1&1\end{pmatrix}$ , $\det(P)=3$ , $P^{-1}=\frac{1}{3}\begin{pmatrix}1&1\\-1&2\end{pmatrix}$ .

$P^{-1}AP = \begin{pmatrix}5&0\\0&2\end{pmatrix}$ ✓ — diagonal, eigenvalues on diagonal.

For $A = \begin{pmatrix}4&2\\1&3\end{pmatrix}$ , find the change-of-basis matrix $P$ to the eigenvector basis and compute $[T]_P = P^{-1}AP$ . Verify the result is diagonal.

EXERCISE 9.4

Similar matrices satisfy $B = P^{-1}AP$ . They share eigenvalues because the characteristic polynomial is invariant: $\det(B-\lambda I) = \det(P^{-1}(A-\lambda I)P) = \det(A-\lambda I)$ .

Shared eigenvalues: $\det(P^{-1}AP - \lambda I) = \det(P^{-1}(A-\lambda I)P) = \det(P^{-1})\det(A-\lambda I)\det(P) = \det(A-\lambda I)$ since $\det(P^{-1})\det(P)=1$ . Same characteristic polynomial $\Rightarrow$ same eigenvalues.

Shared determinant: $\det(P^{-1}AP) = \det(P^{-1})\det(A)\det(P) = \det(A)$ .

Shared trace: $\text{tr}(P^{-1}AP) = \text{tr}(APP^{-1}) = \text{tr}(A)$ by the cyclic property of trace.

Shared rank: $P^{-1}AP$ and $A$ have the same null space dimension (multiplying by invertible matrices preserves rank).

These quantities are basis-independent invariants of the linear transformation itself.

Prove that similar matrices $A$ and $B = P^{-1}AP$ share the same eigenvalues, determinant, trace, and rank. For each, state which property of $\det$ or $\text{tr}$ makes the proof work.

EXERCISE 9.5

A matrix $A$ is diagonalisable $\iff$ it has $n$ linearly independent eigenvectors. The change-of-basis matrix $P$ then diagonalises it. If eigenvectors span $\mathbb{R}^n$ , write them as columns of $P$ .

$A=\begin{pmatrix}2&1\\0&2\end{pmatrix}$ has repeated eigenvalue $\lambda=2$ (algebraic multiplicity 2). $(A-2I)=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ ; $\ker(A-2I) = \text{span}\left\{\begin{pmatrix}1\\0\end{pmatrix}\right\}$ , geometric multiplicity 1.

Since geometric multiplicity (1) $<$ algebraic multiplicity (2), the matrix is not diagonalisable — there is no basis of eigenvectors. No change-of-basis matrix $P$ can produce a diagonal matrix $P^{-1}AP$ .

The best achievable is the Jordan form $\begin{pmatrix}2&1\\0&2\end{pmatrix}$ — which is already the Jordan normal form of $A$ .

The matrix $A = \begin{pmatrix}2&1\\0&2\end{pmatrix}$ has eigenvalue $\lambda=2$ with algebraic multiplicity 2. Determine whether $A$ is diagonalisable (i.e., whether a change-of-basis matrix to a diagonal representation exists) and explain why or why not.

EXERCISE 9.6

The covariance matrix $\Sigma$ is symmetric, so its eigenvectors are orthogonal. The change of basis $Q^T \Sigma Q = \Lambda$ transforms to principal component coordinates. Portfolio variance in PC coordinates is $\mathbf{w}_{PC}^T \Lambda \mathbf{w}_{PC} = \sum_i \lambda_i w_i^2$ .

$\Sigma = \begin{pmatrix}4&2\\2&1\end{pmatrix}$ .

Eigenvalues: $p(\lambda)=(4-\lambda)(1-\lambda)-4=\lambda^2-5\lambda=\lambda(\lambda-5)$ , so $\lambda_1=5$ , $\lambda_2=0$ .

For $\lambda=5$ : $(Σ-5I)\mathbf{v}=0$ gives $\begin{pmatrix}-1&2\\2&-4\end{pmatrix}\mathbf{v}=0$ ; eigenvector $\mathbf{v}_1=\frac{1}{\sqrt{5}}\begin{pmatrix}2\\1\end{pmatrix}$ .

For $\lambda=0$ : $\Sigma\mathbf{v}=0$ gives $\mathbf{v}_2=\frac{1}{\sqrt{5}}\begin{pmatrix}-1\\2\end{pmatrix}$ .

$Q = \frac{1}{\sqrt{5}}\begin{pmatrix}2&-1\\1&2\end{pmatrix}$ , $\Lambda = \begin{pmatrix}5&0\\0&0\end{pmatrix}$ .

$\lambda_2=0$ means the portfolio has a zero-variance direction — the assets are perfectly collinear. In the PC basis, $\mathbf{w}_{PC} = Q^T\mathbf{w}$ ; all portfolio variance concentrates in PC1. Holding the $\lambda_2=0$ direction incurs no variance — it is the risk-free combination of these two assets.

A two-asset portfolio has covariance matrix $\Sigma = \begin{pmatrix}4&2\\2&1\end{pmatrix}$ . Find the orthogonal change-of-basis matrix $Q$ that diagonalises $\Sigma$ (i.e., $Q^T\Sigma Q = \Lambda$ ). Interpret the result: what does $\lambda_2 = 0$ mean for portfolio construction?

07 · Chapter Summary

Concept	Formula / Rule
Coordinate vector	$[\mathbf{v}]_\mathcal{B} = (c_1,\ldots,c_n)^T$ where $\mathbf{v}=\sum c_i\mathbf{b}_i$
Change-of-basis matrix	$P_\mathcal{B}$ = columns are basis vectors in standard coords
Standard from $\mathcal{B}$	$\mathbf{v} = P_\mathcal{B}[\mathbf{v}]_\mathcal{B}$
$\mathcal{B}$ from standard	$[\mathbf{v}]_\mathcal{B} = P_\mathcal{B}^{-1}\mathbf{v}$
$\mathcal{B}$ to $\mathcal{C}$	$[\mathbf{v}]_\mathcal{C} = P_\mathcal{C}^{-1}P_\mathcal{B}[\mathbf{v}]_\mathcal{B}$
Similarity transform	$[T]_\mathcal{B} = P^{-1}AP$ ; same eigenvalues, det, trace
Diagonalisation	Special case: $P$ = eigenvector matrix; $[T]_P = \Lambda$
PCA	Covariance $\Sigma = Q\Lambda Q^T$ ; PC basis decorrelates risk

Next: Chapter 10 — Inner Products & Orthogonality generalises the dot product to abstract vector spaces and establishes the precise meaning of "angle" and "perpendicularity" in any dimension.