Matrices

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

$(AB)_{ij} = \sum_{k=1}^n A_{ik} B_{kj}$

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

$AA^{-1} = A^{-1}A = I_n$

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:

$\det(A) = \sum_{j=1}^n (-1)^{1+j} a_{1j} M_{1j}$

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 $i$

\det(\mathrm{new}~A) = \det(A) + \alpha \cdot \det(\mathrm{matrix}~with~rows~i\mathrm{~and~j\mathrm}{~equal)}

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 $A, B \in \mathcal{M}_{n \times n}(F)$

$\det(AB) = \det(A)\det(B)$

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,

$\det(AB) = \det(A \cdot E_1 \cdots E_k U) = \det(A) \cdot \det(E_1) \cdots \det(E_k) \cdot \det(U) = \det(A) \cdot \det(B)$

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

$\mathrm{adj}(A) = (C_{ji})_{i,j=1}^n$

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 $A \in \mathcal{M}_{n \times n}(F)$

$A \cdot \mathrm{adj}(A) = \mathrm{adj}(A) \cdot A = \det(A) \cdot I_n$

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

$A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$

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

$A = \begin{pmatrix} 2 & 1 & 3 & 1 \\ 4 & 2 & 5 & 3 \\ 6 & 3 & 8 & 5 \\ 1 & 1 & 1 & 1 \end{pmatrix}$

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

$A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$

:::caution Common Pitfall The determinant is only defined for square matrices. There is no meaningful determinant for an $m \times n$ matrix with $m \neq n$. Do not confuse $\det(AB) = \det(A)\det(B)$ with a Non-existent formula for non-square matrices.

Problem. Compute $\det(A)$ where

$A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ 1 & 3 & 6 & 10 \\ 1 & 4 & 10 & 20 \end{pmatrix}$

Proposition 3.6 (Determinant of a Triangular Matrix). If $A$ is upper or lower triangular, Then $\det(A) = \prod_{i=1}^n a_{ii}$.

Proof. By repeated cofactor expansion along the first column (for upper triangular), or induction. At each step, all terms involving off-diagonal entries vanish due to the zero structure, leaving Only the product of diagonal entries. $\blacksquare$

• $\det(A + B) \neq \det(A) + \det(B)$ . For example, with $A = B = I_2$ $\det(A + B) = \det(2I_2) = 4$But $\det(A) + \det(B) = 2$.
• The adjugate formula is theoretically important but computationally inefficient. For large matrices, use Gaussian elimination or LU decomposition to compute inverses.
• A matrix with $\det(A) = 0$ has no inverse. Do not attempt to divide by zero.

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