Matrices

Attempt these from memory before reading.

• When is $AB$ defined, and what are its dimensions?
• Is $AB = BA$? Give a counterexample or say when it holds.
• What is $\det(AB)$ in terms of $\det(A)$ and $\det(B)$?
• What does the determinant measure geometrically?
• When is a matrix invertible, in three equivalent ways?

Check your answers against the text. Prompt 2 is the one that distinguishes matrix algebra from the algebra you learned at school, and the failure is not an anomaly — it is what makes order matter.

A matrix is presented as a rectangular array of numbers. That is the notation, not the object. A matrix is a linear map, written in a chosen basis, and every operation on matrices is the notation for an operation on maps.

Multiplication is the clearest example. If $A$ and $B$ are maps, then $(AB)v = A(Bv)$ means “apply $B$ first, then $A$” — multiplication is composition, which is why the order cannot be arbitrary and why the middle dimensions must agree. Once that is internalised, several rules stop being things to remember:

• $AB \neq BA$ because doing things in a different order is genuinely different — rotating then scaling is not scaling then rotating.
• $(AB)^T = B^TA^T$ because reversing a composition reverses the order.
• $\det(AB) = \det(A)\det(B)$ because the determinant measures how a map scales volume, and volumes scale multiplicatively under composition.

The determinant measures volume. For a $2\times2$ matrix it is the (oriented) area of the image of the unit square; for $3\times3$, the oriented volume of the image of the unit cube. A negative determinant means the map flips orientation. That single sentence explains the multiplicative rule, why a singular matrix has determinant zero (it squashes the cube flat), and why $\det(A^{-1}) = 1/\det A$.

Row reduction beats cofactor expansion. Cofactor expansion is $O(n!)$ and row reduction is $O(n^3)$; at $n = 20$ the difference is already $2.4 \times 10^{18}$ operations against $8\times10^3$. The efficient method uses the fact that row operations change the determinant in known ways — swap negates, scale multiplies, add leaves alone — so reduce to upper triangular and multiply the diagonal.

An $m \times n$ matrix $A$ over $F$ is a rectangular array of $mn$ elements from $F$ Arranged in $m$ rows and $n$ columns. The set of all such matrices is denoted $\mathcal{M}_{m \times n}(F)$.

Addition. For $A, B \in \mathcal{M}_{m \times n}(F)$, $(A + B)_{ij} = A_{ij} + B_{ij}$.

Scalar multiplication. For $\alpha \in F$ and $A \in \mathcal{M}_{m \times n}(F)$ $(\alpha A)_{ij} = \alpha A_{ij}$.

Matrix multiplication. For $A \in \mathcal{M}_{m \times n}(F)$ and $B \in \mathcal{M}_{n \times p}(F)$ The product $AB \in \mathcal{M}_{m \times p}(F)$ is defined by

Proposition 3.1. Matrix multiplication is associative but not commutative .

Proof. Associativity: $(AB)C$ has $(i,j)$-entry $\sum_l (\sum_k A_{ik} B_{kl}) C_{lj} = \sum_k A_{ik} (\sum_l B_{kl} C_{lj}) = (A(BC))_{ij}$ By interchanging the order of summation (both sums are finite). For non-commutativity, $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ give $AB = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \neq \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = BA$. $\blacksquare$

The transpose of $A \in \mathcal{M}_{m \times n}(F)$ Denoted $A^T$ Is the $n \times m$ matrix With $(A^T)_{ij} = A_{ji}$.

Properties of transpose:

1. $(A + B)^T = A^T + B^T$
2. $(\alpha A)^T = \alpha A^T$
3. $(AB)^T = B^T A^T$
4. $(A^T)^T = A$

A square matrix $A \in \mathcal{M}_{n \times n}(F)$ is invertible if there exists a matrix $A^{-1} \in \mathcal{M}_{n \times n}(F)$ such that

Theorem 3.1. The following are equivalent for $A \in \mathcal{M}_{n \times n}(F)$:

1. $A$ is invertible.
2. $\det(A) \neq 0$.
3. The columns of $A$ are linearly independent.
4. The rows of $A$ are linearly independent.
5. $\mathrm{rank}(A) = n$.
6. The equation $A\mathbf{x} = \mathbf{b}$ has a unique solution for every $\mathbf{b}$.
7. The only solution to $A\mathbf{x} = \mathbf{0}$ is $\mathbf{x} = \mathbf{0}$.

The determinant is a function $\det : \mathcal{M}_{n \times n}(F) \to F$ defined recursively by Laplace expansion along the first row:

Where $M_{1j}$ is the $(1,j)$-minor (the determinant of the $(n-1) \times (n-1)$ matrix obtained by Deleting row 1 and column $j$).

The $(i,j)$-cofactor is $C_{ij} = (-1)^{i+j} M_{ij}$ So $\det(A) = \sum_{j=1}^n a_{ij} C_{ij}$ for any fixed row $i$.

Proposition 3.2 (Effect of Row Operations). Let $A \in \mathcal{M}_{n \times n}(F)$.

1. Swapping two rows of $A$ multiplies the determinant by $-1$.
2. Multiplying a row of $A$ by $\alpha \in F$ multiplies the determinant by $\alpha$.
3. Adding a multiple of one row to another leaves the determinant unchanged.

Proof. (1) This follows from the antisymmetry of the Leibniz formula $\det(A) = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma(i)}$. Swapping two rows Changes the sign of every permutation, hence the sign of the sum.

(2) Multiplying row $i$ by $\alpha$ multiplies every term in the Leibniz expansion by $\alpha$ Hence $\det$ is multiplied by $\alpha$.

(3) Adding $\alpha$ times row $j$ to row $i$ ($i \neq j$): by multilinearity in row $i$

A matrix with two equal rows has determinant 0 (by antisymmetry: swapping them leaves the matrix Unchanged but multiplies $\det$ by $-1$ So $\det = -\det$ Hence $\det = 0$). Therefore $\det(\mathrm{new}~A) = \det(A)$. $\blacksquare$

Theorem 3.3 (Multiplicativity). For $A, B \in \mathcal{M}_{n \times n}(F)$

Proof (via elementary matrices). Every matrix $B$ can be written as a product of elementary matrices Times an upper triangular matrix: $B = E_1 E_2 \cdots E_k U$. For an elementary matrix $E$:

• If $E$ swaps rows, $\det(E) = -1$ and $\det(AE) = -\det(A) = \det(A)\det(E)$.
• If $E$ multiplies a row by $\alpha$, $\det(E) = \alpha$ and $\det(AE) = \alpha\det(A) = \det(A)\det(E)$.
• If $E$ adds a multiple of one row to another, $\det(E) = 1$ and $\det(AE) = \det(A) = \det(A)\det(E)$.

Thus $\det(AE) = \det(A)\det(E)$ for every elementary matrix. By induction,

Since $\det(B) = \det(E_1)\cdots\det(E_k)\det(U)$. $\blacksquare$

Corollary 3.4. $\det(A^T) = \det(A)$ And for invertible $A$, $\det(A^{-1}) = 1/\det(A)$.

Proof. $AA^{-1} = I$ So $\det(A)\det(A^{-1}) = \det(I) = 1$ Giving $\det(A^{-1}) = 1/\det(A)$. For the transpose, use the Leibniz formula or observe that row Operations and column operations have the same effects on the determinant. $\blacksquare$

Definition. The adjugate (or adjoint) of $A \in \mathcal{M}_{n \times n}(F)$ is

Where $C_{ij}$ is the $(i,j)$-cofactor of $A$. That is, $\mathrm{adj}(A)$ is the transpose of the Cofactor matrix.

Theorem 3.5. For any $A \in \mathcal{M}_{n \times n}(F)$

In particular, if $\det(A) \neq 0$ Then $A^{-1} = \frac{1}{\det(A)} \mathrm{adj}(A)$.

Proof. The $(i,j)$-entry of $A \cdot \mathrm{adj}(A)$ is $\sum_{k=1}^n a_{ik} C_{jk}$. When $i = j$ This is $\sum_{k=1}^n a_{ik} C_{ik} = \det(A)$ (cofactor expansion along row $i$). When $i \neq j$ This is the cofactor expansion of a matrix obtained from $A$ by replacing row $j$ With row $i$ Which has two equal rows and hence determinant 0. $\blacksquare$

Problem. Compute $\det(A)$ where

Problem. Compute $\det(A)$ by row reduction where

Problem. Find $A^{-1}$ using the adjugate formula, where

• Quantum Mechanics
• Graph Theory

• Eigenvalues and Eigenvectors
• Inner Product Spaces
• Linear Independence, Span, Basis, and Dimension
• Problem Set