Course map and matrix algebra

Size and entrywise operations. A matrix $A$ of size $m\times n$ is a rectangular array of scalars (real or complex) with $m$ rows and $n$ columns. Addition and scalar multiplication act entrywise: $(A+B)_{ij}=A_{ij}+B_{ij}$ and $(cA)_{ij}=cA_{ij}$, and they require $A,B$ to be the same size.

Matrix multiplication is not entrywise. For $A$ of size $m\times n$ and $B$ of size $n\times p$, the product $C=AB$ has size $m\times p$, with each entry a dot product of a row of $A$ with a column of $B$:

where: $A_{ik}$ runs along row $i$ of $A$, $B_{kj}$ runs down column $j$ of $B$, and the inner dimension $n$ must match. Crucially, matrix multiplication is not commutative: $AB\neq BA$ in general, and often only one of the two products is even defined.

Transpose and determinant. The transpose $A^T$ swaps rows and columns, $(A^T)_{ij}=A_{ji}$. For a square matrix, the determinant $\det A$ is a scalar; for $2\times2$,

For an $n\times n$ matrix, cofactor expansion along row $i$ gives $\det A=\sum_{j=1}^n(-1)^{i+j}A_{ij}M_{ij}$, where $M_{ij}$ is the minor formed by deleting row $i$ and column $j$; expanding along any row or column gives the same value, which is a useful hand-check.

Adjoint and inverse. The adjoint (adjugate) $\operatorname{adj}A$ is the transpose of the cofactor matrix. When $\det A\neq0$,

where: $\operatorname{adj}A$ is built entirely from signed minors of $A$, and $\det A$ scales the result. A matrix with $\det A=0$ has no inverse and is called singular.

Special families. Over $\mathbb{R}$: $A$ is symmetric if $A=A^T$, skew-symmetric if $A=-A^T$, orthogonal if $A^TA=I$. Over $\mathbb{C}$, replace transpose by conjugate transpose $A^{*}=\bar A^{T}$: Hermitian if $A=A^{*}$, skew-Hermitian if $A=-A^{*}$, unitary if $A^{*}A=I$. Real symmetric/skew-symmetric/orthogonal matrices are the real-entry special cases of Hermitian/skew-Hermitian/unitary.

Bilinear and quadratic forms. A bilinear form is $\mathbf{x}^TA\mathbf{y}=\sum_{i,j}A_{ij}x_iy_j$, linear in $\mathbf x$ and in $\mathbf y$ separately. Setting $\mathbf y=\mathbf x$ gives the quadratic form $\mathbf{x}^TA\mathbf{x}$; for symmetric $A$ it expands cleanly as $\sum_i A_{ii}x_i^2+2\sum_{i<j}A_{ij}x_ix_j$.

Standing identities. Two reversal rules recur throughout the chapter: transpose of a product reverses order, $(AB)^T=B^TA^T$, and so does inverse of a product, $(AB)^{-1}=B^{-1}A^{-1}$ (both $A,B$ invertible, same order). Determinants are multiplicative, $\det(AB)=\det A\,\det B$, which immediately gives $\det(A^{-1})=1/\det A$. It is worth pausing on why the reversal happens, since it is the single most-forgotten sign or order slip in this chapter. Entrywise, $((AB)^T)_{ij}=(AB)_{ji}=\sum_kA_{jk}B_{ki}=\sum_k(B^T)_{ik}(A^T)_{kj}=(B^TA^T)_{ij}$, so the two sides agree term by term; the same bookkeeping, run through the definition of an inverse, forces $(AB)^{-1}=B^{-1}A^{-1}$ rather than $A^{-1}B^{-1}$. A closely related pair of matrices, the identity $I$ (ones on the diagonal, zeros elsewhere) and the zero matrix $O$, act the way $1$ and $0$ do for numbers: $AI=IA=A$ for any square $A$ of matching size, and $A+O=A$. Unlike ordinary numbers, though, $AB=O$ does not force $A=O$ or $B=O$ — two nonzero matrices can multiply to give the zero matrix, another reminder that matrix algebra only mimics ordinary arithmetic up to a point, and every identity above needs to be verified from the definitions rather than assumed by analogy with scalars.

Let $A=\begin{pmatrix}2&-1\\3&4\end{pmatrix}$, $B=\begin{pmatrix}1&0\\2&5\end{pmatrix}$. Find $AB$, $\det A$, $A^{-1}$, and verify $(AB)^{-1}=B^{-1}A^{-1}$.

Step 1. Compute the product using $C_{ij}=\sum_kA_{ik}B_{kj}$:

Step 2. Compute $\det A=2(4)-(-1)(3)=8+3=11$, nonzero, so $A$ is invertible.

Step 3. For a $2\times2$ matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ the adjoint is $\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$, so

Check: $AA^{-1}=\frac{1}{11}\begin{pmatrix}8+3&2-2\\12-12&3+8\end{pmatrix}=\frac{1}{11}\begin{pmatrix}11&0\\0&11\end{pmatrix}=I.$

Competitive stretch

Step 4. Verify the reversal identity numerically. $\det B=1(5)-0(2)=5$, so $B^{-1}=\frac{1}{5}\begin{pmatrix}5&0\\-2&1\end{pmatrix}$. Then $B^{-1}A^{-1}=\frac{1}{5}\begin{pmatrix}5&0\\-2&1\end{pmatrix}\cdot\frac{1}{11}\begin{pmatrix}4&1\\-3&2\end{pmatrix}=\frac{1}{55}\begin{pmatrix}20&5\\-11&0\end{pmatrix}$. Independently, $\det(AB)=0(20)-(-5)(11)=55$, and for $\begin{pmatrix}0&-5\\11&20\end{pmatrix}$ the adjoint is $\begin{pmatrix}20&5\\-11&0\end{pmatrix}$, so $(AB)^{-1}=\frac{1}{55}\begin{pmatrix}20&5\\-11&0\end{pmatrix}$. The two computations agree, confirming $(AB)^{-1}=B^{-1}A^{-1}=\dfrac{1}{55}\begin{pmatrix}20&5\\-11&0\end{pmatrix}$.

• Assuming $AB=BA$: matrix multiplication is generally non-commutative, and even when both products exist they can differ.
• Forgetting the adjoint uses the transpose of the cofactor matrix, not the cofactor matrix itself — a very common $2\times2$/$3\times3$ slip.
• Confusing "symmetric" (real, $A=A^T$) with "Hermitian" (complex, $A=\bar A^T$); for a real matrix the two coincide, but a complex matrix must be checked with the conjugate.
• Getting the order wrong in reversal identities: writing $A^TB^T$ instead of $(AB)^T=B^TA^T$, or $A^{-1}B^{-1}$ instead of $(AB)^{-1}=B^{-1}A^{-1}$.
• Trying to invert a singular matrix ($\det A=0$), or dividing by $\det A$ without first checking it is nonzero.

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