Chapter 02

Matrix Operations

00 · Symbol Glossary

$A$uppercase A — matrix

An uppercase letter denoting a matrix. Matrices extend vectors: a vector is a single column of numbers; a matrix is a rectangular grid. Bold uppercase is sometimes written $\mathbf{A}$ , but plain uppercase $A$ is equally standard — context makes it clear.

$a_{ij}$a sub i j — matrix entry

The entry of matrix $A$ in row $i$ , column $j$ . Read aloud as "a-i-j." The first subscript is always the row, the second is always the column — row then column, always. $a_{23}$ is in row 2, column 3.

$m \times n$m by n — matrix dimensions

The dimensions of a matrix: $m$ rows and $n$ columns. An $m \times n$ matrix has $mn$ entries total. Read aloud as "m by n." When dimensions appear in a product, the inner dimensions must match.

$\mathbb{R}^{m \times n}$R-m-by-n — matrix space

The set of all $m \times n$ matrices with real entries. Extends $\mathbb{R}^n$ from Chapter 1: a vector in $\mathbb{R}^n$ is an element of $\mathbb{R}^{n \times 1}$ . Addition and scalar multiplication on $\mathbb{R}^{m \times n}$ satisfy the same 8 vector space axioms.

$A^T$A transpose — matrix transpose

The matrix formed by flipping $A$ across its main diagonal: rows become columns, columns become rows. $(A^T)_{ij} = a_{ji}$ . Chapter 1 defined transpose for vectors; this is the same operation extended to full matrices.

$AB$A B — matrix product

The product of matrices $A$ and $B$ , computed via row-by-column dot products. Not component-wise multiplication. The entry $(AB)_{ij}$ equals the dot product of row $i$ of $A$ with column $j$ of $B$ . Order matters — $AB \neq BA$ in general.

$I_n$I sub n — identity matrix

The $n \times n$ identity matrix: ones on the main diagonal, zeros everywhere else. $AI_n = A$ and $I_m A = A$ for any $m \times n$ matrix $A$ — multiplication by $I$ leaves the matrix unchanged, just as multiplication by 1 leaves a scalar unchanged.

$\text{tr}(A)$trace of A

The sum of the diagonal entries of a square matrix: $\text{tr}(A) = a_{11} + a_{22} + \cdots + a_{nn}$ . Only defined for square matrices. In Chapter 6 it will reappear as the sum of the eigenvalues — a fundamental connection between entries and structure.

01 · What is a Matrix

A vector organises a list of numbers into one column. A matrix organises numbers into a rectangular grid — $m$ rows and $n$ columns. Every quant operation on multi-dimensional data eventually becomes a matrix operation: a portfolio of $n$ assets observed over $m$ days is an $m \times n$ matrix, each row a snapshot and each column an asset's history.

Definition — Matrix

An $m \times n$ matrix $A$ over $\mathbb{R}$ is a rectangular array of real numbers with $m$ rows and $n$ columns:

A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix} \in \mathbb{R}^{m \times n}

$a_{ij}$ — the entry in row $i$ , column $j$ . First index is row, second is column.

$\ddots$ — diagonal dots meaning the pattern continues in both directions.

$m \times n$ — read "m by n." The number of rows comes first, always.

✓ Example — Portfolio Returns Matrix

A fund tracks 3 assets over 4 trading days. Returns (in percent) are stored as a $4 \times 3$ matrix — 4 days (rows), 3 assets (columns):

R = \begin{pmatrix} 1.2 & -0.5 & 2.1 \\ -0.3 & 1.8 & 0.4 \\ 0.7 & 0.9 & -1.2 \\ 2.3 & -1.1 & 0.8 \end{pmatrix} \in \mathbb{R}^{4 \times 3}

$r_{13} = 2.1$ — day 1, asset 3 returned $2.1\%$ . $r_{42} = -1.1$ — day 4, asset 2 returned $-1.1\%$ . The row index names the day, the column index names the asset.

❌ Confusion — Row Index vs Column Index

Reading $a_{32}$ as row 2, column 3 is wrong. The subscript order is row-then-column without exception. For the matrix $A = \begin{pmatrix}7&2\\5&9\\1&4\end{pmatrix}$ : $a_{32} = 4$ (row 3, column 2), not $a_{23} = 9$ (row 2, column 3). Swapping the indices retrieves a completely different entry.

02 · Matrix Addition and Scalar Multiplication

Matrix addition and scalar multiplication work entry-by-entry — the same pattern as vector operations from Chapter 1, now applied to every slot in the grid.

Definition — Matrix Addition

For $A, B \in \mathbb{R}^{m \times n}$ , their sum $A + B \in \mathbb{R}^{m \times n}$ is defined entry-by-entry:

(A + B)_{ij} = a_{ij} + b_{ij}

Both matrices must have identical dimensions. Adding an $m \times n$ matrix to a $p \times q$ matrix is undefined unless $m = p$ and $n = q$ .

Definition — Scalar Multiplication

For $A \in \mathbb{R}^{m \times n}$ and scalar $c \in \mathbb{R}$ :

(cA)_{ij} = c \cdot a_{ij}

Every entry is multiplied by $c$ . This preserves dimensions: $cA \in \mathbb{R}^{m \times n}$ .

Step-by-step — Computing $A + 2B$ where $A = \begin{pmatrix}3&-1\\0&4\end{pmatrix}$, $B = \begin{pmatrix}-2&5\\1&-3\end{pmatrix}$

Confirm dimensions match: $A \in \mathbb{R}^{2 \times 2}$ and $B \in \mathbb{R}^{2 \times 2}$ . Both $2 \times 2$ . ✓ Addition is defined.

Scale $B$ by 2: multiply every entry of $B$ by 2. $2B = \begin{pmatrix}2 \times (-2) & 2 \times 5\\ 2 \times 1 & 2 \times (-3)\end{pmatrix} = \begin{pmatrix}-4&10\\2&-6\end{pmatrix}$ . Each of the four entries of $B$ is doubled individually.

Add entry $(1,1)$ : $a_{11} + (2B)_{11} = 3 + (-4) = -1$ . Adding a negative is subtraction.

Add entry $(1,2)$ : $a_{12} + (2B)_{12} = (-1) + 10 = 9$ .

Add entry $(2,1)$ : $a_{21} + (2B)_{21} = 0 + 2 = 2$ .

Add entry $(2,2)$ : $a_{22} + (2B)_{22} = 4 + (-6) = -2$ .

Assemble the result:

A + 2B = \begin{pmatrix}-1 & 9 \\ 2 & -2\end{pmatrix}

❌ Failure — Adding Matrices of Different Dimensions

$A = \begin{pmatrix}1&2\\3&4\end{pmatrix} \in \mathbb{R}^{2 \times 2}$ and $B = \begin{pmatrix}5&6&7\\8&9&10\end{pmatrix} \in \mathbb{R}^{2 \times 3}$ . Adding them fails at slot $(1,3)$ : $A$ has no entry in column 3. The dimensions differ ( $2 \times 2$ vs $2 \times 3$ ). This is undefined — full stop. The number of columns (2 vs 3) must match, not just the number of rows.

03 · Matrix Multiplication

Matrix multiplication is the most important — and the most misunderstood — operation in linear algebra. It is not entry-by-entry. It is row-by-column dot products.

Definition — Matrix Multiplication

For $A \in \mathbb{R}^{m \times n}$ and $B \in \mathbb{R}^{n \times p}$ , the product $AB \in \mathbb{R}^{m \times p}$ has entries:

(AB)_{ij} = \sum_{k=1}^{n} a_{ik} b_{kj} = a_{i1}b_{1j} + a_{i2}b_{2j} + \cdots + a_{in}b_{nj}

$(AB)_{ij}$ — the dot product of row $i$ of $A$ with column $j$ of $B$ .

$k$ — the shared index, stepping through $1, 2, \ldots, n$ .

The inner dimensions must match: $A$ must have $n$ columns and $B$ must have $n$ rows. The result has the outer dimensions: $m$ rows (from $A$ ) and $p$ columns (from $B$ ).

Dimension Check Before Multiplying

Write the dimensions side by side: $(m \times n)(n \times p)$ . The inner two numbers must match — both must equal $n$ . The result takes the outer two: $m \times p$ . Example: $(3 \times 4)(4 \times 2) \to 3 \times 2$ ✓. Example: $(3 \times 4)(3 \times 2)$ — inner pair is $4, 3$ — mismatch. ✗ Undefined.

Step-by-step — Computing $AB$ where $A = \begin{pmatrix}1&2&3\\4&5&6\end{pmatrix}$, $B = \begin{pmatrix}7&8\\9&10\\11&12\end{pmatrix}$

Check dimensions: $A \in \mathbb{R}^{2 \times 3}$ , $B \in \mathbb{R}^{3 \times 2}$ . Inner dimensions both $3$ . ✓ Result will be $2 \times 2$ .

Compute entry $(1,1)$ : dot product of row 1 of $A$ with column 1 of $B$ .

Row 1 of $A$ : $(1, 2, 3)$ . Column 1 of $B$ : $(7, 9, 11)$ .

$(1)(7) + (2)(9) + (3)(11) = 7 + 18 + 33 = 58$ . The $7$ comes from $a_{11}b_{11} = 1 \times 7$ , the $18$ from $a_{12}b_{21} = 2 \times 9$ , the $33$ from $a_{13}b_{31} = 3 \times 11$ .

Compute entry $(1,2)$ : dot product of row 1 of $A$ with column 2 of $B$ .

Column 2 of $B$ : $(8, 10, 12)$ .

$(1)(8) + (2)(10) + (3)(12) = 8 + 20 + 36 = 64$ . The $8$ from $1 \times 8$ , the $20$ from $2 \times 10$ , the $36$ from $3 \times 12$ .

Compute entry $(2,1)$ : dot product of row 2 of $A$ with column 1 of $B$ .

Row 2 of $A$ : $(4, 5, 6)$ . Column 1 of $B$ : $(7, 9, 11)$ .

$(4)(7) + (5)(9) + (6)(11) = 28 + 45 + 66 = 139$ . The $28$ from $4 \times 7$ , the $45$ from $5 \times 9$ , the $66$ from $6 \times 11$ .

Compute entry $(2,2)$ : dot product of row 2 of $A$ with column 2 of $B$ .

$(4)(8) + (5)(10) + (6)(12) = 32 + 50 + 72 = 154$ . The $32$ from $4 \times 8$ , the $50$ from $5 \times 10$ , the $72$ from $6 \times 12$ .

Assemble the result:

AB = \begin{pmatrix}58 & 64 \\ 139 & 154\end{pmatrix} \in \mathbb{R}^{2 \times 2}

❌ Failure — Matrix Multiplication Is Not Commutative

For $A = \begin{pmatrix}1&2\\3&4\end{pmatrix}$ and $B = \begin{pmatrix}0&1\\1&0\end{pmatrix}$ :

$AB = \begin{pmatrix}1\cdot0+2\cdot1 & 1\cdot1+2\cdot0\\ 3\cdot0+4\cdot1 & 3\cdot1+4\cdot0\end{pmatrix} = \begin{pmatrix}2&1\\4&3\end{pmatrix}$ .

$BA = \begin{pmatrix}0\cdot1+1\cdot3 & 0\cdot2+1\cdot4\\ 1\cdot1+0\cdot3 & 1\cdot2+0\cdot4\end{pmatrix} = \begin{pmatrix}3&4\\1&2\end{pmatrix}$ .

$AB \neq BA$ . Swapping the order gives a different matrix. This is not a quirk of this specific pair — matrix multiplication is non-commutative in general. Never assume $AB = BA$ .

✓ Example — Portfolio Weights Applied to Returns

A $2 \times 3$ matrix $R$ holds daily returns for 3 assets (columns) over 2 days (rows). A $3 \times 1$ weight vector $\mathbf{w}$ holds the portfolio allocation for each asset. The product $R\mathbf{w}$ gives the portfolio return on each day:

\begin{pmatrix}0.8 & -0.3 & 1.2 \\ 1.5 & 0.6 & -0.4\end{pmatrix} \begin{pmatrix}0.5\\0.3\\0.2\end{pmatrix} = \begin{pmatrix}0.8(0.5)+(-0.3)(0.3)+1.2(0.2)\\ 1.5(0.5)+0.6(0.3)+(-0.4)(0.2)\end{pmatrix} = \begin{pmatrix}0.4-0.09+0.24\\ 0.75+0.18-0.08\end{pmatrix} = \begin{pmatrix}0.55\\0.85\end{pmatrix}

Day 1 portfolio return: $0.55\%$ . Day 2: $0.85\%$ . Each row of $R$ is dotted with $\mathbf{w}$ — exactly the row-by-column pattern of matrix-vector multiplication.

04 · The Transpose

The transpose $A^T$ of a matrix $A$ is formed by reflecting it across the main diagonal: row $i$ of $A$ becomes column $i$ of $A^T$ , and column $j$ of $A$ becomes row $j$ of $A^T$ .

Definition — Matrix Transpose

For $A \in \mathbb{R}^{m \times n}$ , the transpose $A^T \in \mathbb{R}^{n \times m}$ is defined by:

(A^T)_{ij} = a_{ji}

The row and column indices swap. An $m \times n$ matrix transposes to an $n \times m$ matrix — dimensions flip too.

Step-by-step — Transposing $A = \begin{pmatrix}1&2&3\\4&5&6\end{pmatrix}$

Identify dimensions: $A \in \mathbb{R}^{2 \times 3}$ . The transpose will be $\mathbb{R}^{3 \times 2}$ — dimensions flip.

Write row 1 of $A$ as column 1 of $A^T$ : row 1 is $(1, 2, 3)$ , so column 1 of $A^T$ is $\begin{pmatrix}1\\2\\3\end{pmatrix}$ .

Write row 2 of $A$ as column 2 of $A^T$ : row 2 is $(4, 5, 6)$ , so column 2 of $A^T$ is $\begin{pmatrix}4\\5\\6\end{pmatrix}$ .

Assemble:

A^T = \begin{pmatrix}1&4\\2&5\\3&6\end{pmatrix} \in \mathbb{R}^{3 \times 2}

Verify:

(A^T)_{12} = a_{21} = 4

✓.

(A^T)_{31} = a_{13} = 3

✓.

Two transpose properties are critical — one is obvious, one trips people up:

Properties of the Transpose

(A^T)^T = A \qquad (AB)^T = B^T A^T

$(A^T)^T = A$ — transposing twice returns the original matrix.

$(AB)^T = B^T A^T$ — the order reverses when transposing a product. The reversed order is not optional; it is forced by the dimension math.

❌ Failure — Wrong Order in Product Transpose

$(AB)^T \neq A^T B^T$ . For $A \in \mathbb{R}^{2 \times 3}$ and $B \in \mathbb{R}^{3 \times 4}$ : $AB \in \mathbb{R}^{2 \times 4}$ , so $(AB)^T \in \mathbb{R}^{4 \times 2}$ .

$A^T \in \mathbb{R}^{3 \times 2}$ and $B^T \in \mathbb{R}^{4 \times 3}$ . The product $A^T B^T$ would need the inner dimensions to match: $3 \times 2$ times $4 \times 3$ — inner pair is $2, 4$ . Mismatch. $A^T B^T$ is not even defined here.

$B^T A^T$ : $4 \times 3$ times $3 \times 2$ — inner pair is $3, 3$ ✓ — gives $4 \times 2$ . Correct.

A symmetric matrix satisfies $A = A^T$ — it is unchanged by transpose. Covariance matrices in finance are always symmetric, which makes them tractable for eigenvalue analysis in Chapter 6.

05 · Special Matrices

Four matrices appear so often they have names. Recognising them on sight saves work.

Definition — Identity Matrix $I_n$

The $n \times n$ identity matrix $I_n$ has ones on the main diagonal and zeros elsewhere:

I_n = \begin{pmatrix}1&0&\cdots&0\\0&1&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&1\end{pmatrix}

For any $A \in \mathbb{R}^{m \times n}$ : $AI_n = A$ and $I_m A = A$ . Multiplying by $I$ is the matrix analogue of multiplying a number by 1.

Definition — Diagonal Matrix

A square matrix $D \in \mathbb{R}^{n \times n}$ is diagonal if $d_{ij} = 0$ whenever $i \neq j$ — all entries off the main diagonal are zero:

D = \begin{pmatrix}d_1&0&\cdots&0\\0&d_2&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&d_n\end{pmatrix}

Products of diagonal matrices are easy: $(D_1 D_2)_{ii} = (d_1)_i (d_2)_i$ — just multiply the diagonal entries. No row-by-column calculation needed.

Definition — Symmetric Matrix

A square matrix $A \in \mathbb{R}^{n \times n}$ is symmetric if $A = A^T$ , i.e., $a_{ij} = a_{ji}$ for all $i, j$ .

\text{Symmetric} \iff a_{ij} = a_{ji} \text{ for all } i, j

Symmetric matrices are their own transpose. Every covariance matrix is symmetric because $\text{Cov}(X_i, X_j) = \text{Cov}(X_j, X_i)$ .

✓ Example — Symmetric Covariance Matrix

For two assets with variances $\sigma_1^2 = 4$ , $\sigma_2^2 = 9$ , and covariance $\sigma_{12} = 3$ :

\Sigma = \begin{pmatrix}4 & 3 \\ 3 & 9\end{pmatrix}

$\Sigma = \Sigma^T$ ✓ — the off-diagonal entries are equal. This symmetry is not accidental; it holds for every covariance matrix by definition of covariance.

06 · The Trace

The trace of a square matrix is the sum of its diagonal entries. It is the first of several invariants — numbers that capture something essential about the matrix regardless of how it is written.

Definition — Trace

For $A \in \mathbb{R}^{n \times n}$ :

\text{tr}(A) = \sum_{i=1}^{n} a_{ii} = a_{11} + a_{22} + \cdots + a_{nn}

Only defined for square matrices. The trace is a single real number.

Step-by-step — Computing $\text{tr}(A)$ for $A = \begin{pmatrix}5&-2&1\\3&8&0\\-1&4&-3\end{pmatrix}$

Identify the main diagonal entries: $a_{11} = 5$ , $a_{22} = 8$ , $a_{33} = -3$ . The main diagonal runs from top-left to bottom-right.

Sum the diagonal entries: $\text{tr}(A) = 5 + 8 + (-3) = 10$ . The $5$ is from position $(1,1)$ , the $8$ from $(2,2)$ , the $-3$ from $(3,3)$ .

Two trace properties are useful:

Properties of the Trace

\text{tr}(A + B) = \text{tr}(A) + \text{tr}(B) \qquad \text{tr}(AB) = \text{tr}(BA)

$\text{tr}(AB) = \text{tr}(BA)$ — trace is cyclic under products. Note: this does not mean $AB = BA$ . The matrices may be completely different while their traces of products coincide.

✓ Example — Total Variance via Trace

For a covariance matrix $\Sigma = \begin{pmatrix}4&3\\3&9\end{pmatrix}$ , the total variance across all assets equals the trace:

\text{tr}(\Sigma) = 4 + 9 = 13

The $4$ is the variance of asset 1, the $9$ is the variance of asset 2. In PCA, the trace of the covariance matrix equals the total variance to be explained across all principal components.

07 · Practice Exercises

EXERCISE 2.1

Add entry by entry in the same position. Confirm both matrices have the same dimensions before starting.

$A + B$ : add each entry at the same position.

$(1,1)$ : $2 + (-1) = 1$ . $(1,2)$ : $-3 + 4 = 1$ . $(2,1)$ : $5 + 2 = 7$ . $(2,2)$ : $0 + (-6) = -6$ . $(3,1)$ : $-1 + 3 = 2$ . $(3,2)$ : $4 + 0 = 4$ .

$A + B = \begin{pmatrix}1 & 1 \\ 7 & -6 \\ 2 & 4\end{pmatrix}$ .

$3A$ : multiply every entry of $A$ by 3. $(1,1)$ : $3 \times 2 = 6$ . $(1,2)$ : $3 \times (-3) = -9$ . $(2,1)$ : $3 \times 5 = 15$ . $(2,2)$ : $3 \times 0 = 0$ . $(3,1)$ : $3 \times (-1) = -3$ . $(3,2)$ : $3 \times 4 = 12$ . Therefore $3A = \begin{pmatrix}6&-9\\15&0\\-3&12\end{pmatrix}$ .

Let $A = \begin{pmatrix}2&-3\\5&0\\-1&4\end{pmatrix}$ and $B = \begin{pmatrix}-1&4\\2&-6\\3&0\end{pmatrix}$ , both in $\mathbb{R}^{3 \times 2}$ . Compute $A + B$ and $3A$ , showing every entry.

EXERCISE 2.2

Check inner dimensions first: $(2 \times 3)(3 \times 2)$ gives a $2 \times 2$ result. Each entry $(i,j)$ is the dot product of row $i$ of $A$ with column $j$ of $B$ .

$A \in \mathbb{R}^{2 \times 3}$ , $B \in \mathbb{R}^{3 \times 2}$ . Inner dimensions both 3. Result is $2 \times 2$ .

$(AB)_{11}$ : row 1 of $A$ = $(2, 0, -1)$ , col 1 of $B$ = $(1, 3, -2)$ . $2(1) + 0(3) + (-1)(-2) = 2 + 0 + 2 = 4$ .

$(AB)_{12}$ : row 1 of $A$ = $(2, 0, -1)$ , col 2 of $B$ = $(-1, 4, 5)$ . $2(-1) + 0(4) + (-1)(5) = -2 + 0 - 5 = -7$ .

$(AB)_{21}$ : row 2 of $A$ = $(1, -2, 3)$ , col 1 of $B$ = $(1, 3, -2)$ . $1(1) + (-2)(3) + 3(-2) = 1 - 6 - 6 = -11$ .

$(AB)_{22}$ : row 2 of $A$ = $(1, -2, 3)$ , col 2 of $B$ = $(-1, 4, 5)$ . $1(-1) + (-2)(4) + 3(5) = -1 - 8 + 15 = 6$ .

Therefore $AB = \begin{pmatrix}4 & -7 \\ -11 & 6\end{pmatrix}$ .

Compute $AB$ where $A = \begin{pmatrix}2&0&-1\\1&-2&3\end{pmatrix}$ and $B = \begin{pmatrix}1&-1\\3&4\\-2&5\end{pmatrix}$ . Show every entry computation in full.

EXERCISE 2.3

Transpose by writing each row of $A$ as the corresponding column of $A^T$ . Then check $(AB)^T = B^T A^T$ by computing both sides.

$A = \begin{pmatrix}1&2\\3&4\\5&6\end{pmatrix} \in \mathbb{R}^{3 \times 2}$ , so $A^T \in \mathbb{R}^{2 \times 3}$ : row 1 of $A$ becomes col 1 of $A^T$ , row 2 becomes col 2, row 3 becomes col 3. $A^T = \begin{pmatrix}1&3&5\\2&4&6\end{pmatrix}$ .

$B = \begin{pmatrix}0&-1\\2&3\\-1&1\end{pmatrix} \in \mathbb{R}^{3 \times 2}$ . $B^T = \begin{pmatrix}0&2&-1\\-1&3&1\end{pmatrix}$ .

$AB$ : dimensions $(3 \times 2)(3 \times 2)$ — inner pair is $2, 3$ . Mismatch. $AB$ is not defined. The product $A^T B$ is defined: $(2 \times 3)(3 \times 2) = 2 \times 2$ .

$(A^T B)_{11} = 1(0)+3(2)+5(-1) = 0+6-5 = 1$ . $(A^T B)_{12} = 1(-1)+3(3)+5(1) = -1+9+5 = 13$ . $(A^T B)_{21} = 2(0)+4(2)+6(-1) = 0+8-6 = 2$ . $(A^T B)_{22} = 2(-1)+4(3)+6(1) = -2+12+6 = 16$ . $A^T B = \begin{pmatrix}1&13\\2&16\end{pmatrix}$ .

For $A = \begin{pmatrix}1&2\\3&4\\5&6\end{pmatrix}$ and $B = \begin{pmatrix}0&-1\\2&3\\-1&1\end{pmatrix}$ : (a) Write down $A^T$ and $B^T$ . (b) Determine whether $AB$ is defined. (c) Compute $A^T B$ , showing all four entries.

EXERCISE 2.4

Compute both $AB$ and $BA$ explicitly. If they differ in even one entry, that proves non-commutativity for these matrices.

$A = \begin{pmatrix}1&1\\0&1\end{pmatrix}$ , $B = \begin{pmatrix}1&0\\1&1\end{pmatrix}$ .

$AB$ : $(1,1)$ : $1(1)+1(1)=2$ . $(1,2)$ : $1(0)+1(1)=1$ . $(2,1)$ : $0(1)+1(1)=1$ . $(2,2)$ : $0(0)+1(1)=1$ . $AB = \begin{pmatrix}2&1\\1&1\end{pmatrix}$ .

$BA$ : $(1,1)$ : $1(1)+0(0)=1$ . $(1,2)$ : $1(1)+0(1)=1$ . $(2,1)$ : $1(1)+1(0)=1$ . $(2,2)$ : $1(1)+1(1)=2$ . $BA = \begin{pmatrix}1&1\\1&2\end{pmatrix}$ .

$AB = \begin{pmatrix}2&1\\1&1\end{pmatrix} \neq \begin{pmatrix}1&1\\1&2\end{pmatrix} = BA$ . Therefore $AB \neq BA$ — matrix multiplication is not commutative for these matrices.

For $A = \begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $B = \begin{pmatrix}1&0\\1&1\end{pmatrix}$ , compute both $AB$ and $BA$ . Confirm or refute: $AB = BA$ .

EXERCISE 2.5

The trace is the sum of diagonal entries. For a symmetric matrix, check that $a_{ij} = a_{ji}$ for every off-diagonal pair.

$A = \begin{pmatrix}3&1&-2\\1&5&4\\-2&4&7\end{pmatrix}$ .

$\text{tr}(A) = a_{11} + a_{22} + a_{33} = 3 + 5 + 7 = 15$ . The 3 comes from position $(1,1)$ , the 5 from $(2,2)$ , the 7 from $(3,3)$ .

Symmetry check: $a_{12} = 1 = a_{21}$ ✓. $a_{13} = -2 = a_{31}$ ✓. $a_{23} = 4 = a_{32}$ ✓. All off-diagonal pairs match. Therefore $A$ is symmetric.

$\text{tr}(A) = 15$ and $A$ is symmetric.

For $A = \begin{pmatrix}3&1&-2\\1&5&4\\-2&4&7\end{pmatrix}$ : (a) Compute $\text{tr}(A)$ . (b) Determine whether $A$ is symmetric. Check every off-diagonal pair.

EXERCISE 2.6

The portfolio variance is $\mathbf{w}^T \Sigma \mathbf{w}$ . Compute $\Sigma \mathbf{w}$ first (matrix-vector product), then take the dot product with $\mathbf{w}$ .

$\Sigma = \begin{pmatrix}9&6\\6&16\end{pmatrix}$ , $\mathbf{w} = \begin{pmatrix}0.6\\0.4\end{pmatrix}$ .

Step 1 — Compute $\Sigma\mathbf{w}$ : row 1: $9(0.6)+6(0.4) = 5.4+2.4 = 7.8$ . Row 2: $6(0.6)+16(0.4) = 3.6+6.4 = 10.0$ . So $\Sigma\mathbf{w} = \begin{pmatrix}7.8\\10.0\end{pmatrix}$ .

Step 2 — Compute $\mathbf{w}^T(\Sigma\mathbf{w}) = 0.6(7.8) + 0.4(10.0) = 4.68 + 4.0 = 8.68$ .

Portfolio variance $= 8.68$ . Portfolio standard deviation $= \sqrt{8.68} \approx 2.946\%$ . The total variance of $9 + 16 = 25$ is reduced to $8.68$ because the assets are positively correlated ( $\sigma_{12} = 6 > 0$ ) — but the weights concentrate less than $50\%$ in the high-variance asset.

A two-asset portfolio has covariance matrix $\Sigma = \begin{pmatrix}9&6\\6&16\end{pmatrix}$ (variances in $\%^2$ ) and weight vector $\mathbf{w} = \begin{pmatrix}0.6\\0.4\end{pmatrix}$ . Compute the portfolio variance $\sigma_p^2 = \mathbf{w}^T \Sigma \mathbf{w}$ . Show the matrix-vector product $\Sigma\mathbf{w}$ first, then the scalar product with $\mathbf{w}^T$ .

08 · Summary

Term	Definition
Matrix	Rectangular array $A \in \mathbb{R}^{m \times n}$ with $m$ rows, $n$ columns. Entry $a_{ij}$ : row $i$ , column $j$ .
Addition	$(A+B)_{ij} = a_{ij}+b_{ij}$ . Requires identical dimensions.
Scalar mult.	$(cA)_{ij} = c\,a_{ij}$ . Every entry scaled.
Matrix product	$(AB)_{ij} = \sum_k a_{ik}b_{kj}$ . Row-by-column dot products. Inner dimensions must match.
Non-commutativity	$AB \neq BA$ in general. Order matters in every matrix product.
Transpose	$(A^T)_{ij} = a_{ji}$ . Dimensions flip. $(AB)^T = B^TA^T$ — order reverses.
Identity $I_n$	Ones on diagonal, zeros elsewhere. $AI_n = I_mA = A$ .
Symmetric	$A = A^T$ . Off-diagonal entries satisfy $a_{ij} = a_{ji}$ .
Trace	$\text{tr}(A) = \sum_i a_{ii}$ . Square matrices only. Equals total variance for covariance matrices.

Next: Systems of Linear Equations — how the equation $A\mathbf{x} = \mathbf{b}$ organises multiple unknowns, and what the solution set looks like geometrically.