MATH[1251] Core

The ideas Algebra and Calculus share, from linear structure and evolution to approximation and accumulation.

MATH1251 has two strands. Algebra develops linear structure: linear combinations, linear maps, coordinates and eigenvalues. Calculus studies change, approximation and accumulation. The two strands keep coming back to a few shared ideas, which this page gathers in one place.

The two strands

Algebra: complex numbers, vector spaces, linear transformations and eigenvalues.
Calculus: integration techniques, ordinary differential equations, Taylor series, functions of several variables and double integrals.

Shared ideas

Coordinates and linear structure

A basis gives every vector a unique list of coordinates, whether the vector is an arrow in space, a polynomial or an element of another finite-dimensional function space. Once bases are chosen, a linear transformation has a matrix. If an operator $V\to V$ has an eigenbasis, its matrix in that basis is diagonal, so it scales each basis direction independently.

Coordinates describe a vector; they are not the vector. Changing the basis changes the coordinate column and the matrix, but not the underlying vector or map.

The scalars are part of the structure. The notes work over $\mathbb R$ by default and over $\mathbb C$ when complex roots or eigenvectors are needed, so the same formula can describe different vector spaces.

TH / why coordinates are uniqueVector spacesRepresentation in a basis
If $B=\{u_1,\ldots,u_n\}$ is a basis for $V$ , then every vector $v\in V$ has a unique representation
$v=\lambda_1u_1+\cdots+\lambda_nu_n, \qquad \lambda_1,\ldots,\lambda_n\in F.$
DF / coordinates in a basisCoordinate vector
Let $B=(u_1,\ldots,u_n)$ be an ordered basis for a vector space $V$ . If
$u=x_1u_1+\cdots+x_nu_n,$
then the coordinate vector of $u$ with respect to $B$ is the column
$[u]_B= \begin{pmatrix} x_1\\ \vdots\\ x_n \end{pmatrix}.$
DF / coordinates of a polynomialPolynomial interpolation
Given $n$ distinct inputs $t_1,\ldots,t_n$ and prescribed outputs $y_1,\ldots,y_n$ , polynomial interpolation finds the polynomial $f\in P_{n-1}$ satisfying
$f(t_i)=y_i,\qquad i=1,\ldots,n.$
Using the standard basis
$B=(1,t,t^2,\ldots,t^{n-1}),$
write
$f(t)=a_0+a_1t+\cdots+a_{n-1}t^{n-1}.$
The coefficients form the coordinate vector
$[f]_B= \begin{pmatrix} a_0\\a_1\\\vdots\\a_{n-1} \end{pmatrix}.$
TH / a linear map in coordinatesLinear transformationsMatrix representation of a linear map
Let $T:V\to W$ be a linear map, where $\dim(V)=n$ and $\dim(W)=m$ . For ordered bases
$B_V=(v_1,\ldots,v_n),\qquad B_W=(w_1,\ldots,w_m),$
form the coordinate columns
$a_j=[T(v_j)]_{B_W}\qquad (j=1,\ldots,n).$
Then
$A=\begin{pmatrix}a_1&\cdots&a_n\end{pmatrix}$
is the unique $m\times n$ matrix satisfying
$[T(v)]_{B_W}=A[v]_{B_V}\qquad\text{for every }v\in V.$
We call $A$ the matrix representation of $T$ relative to $B_V$ and $B_W$ and may write it as $[T]_{B_V}^{B_W}$ .

With this notation, the subscript names the domain basis and the superscript names the codomain basis, matching the direction $[v]_{B_V}\mapsto[T(v)]_{B_W}$ in the displayed coordinate identity.
TH / a basis where the matrix is diagonalEigenvaluesWhen diagonalisation is possible
An $n\times n$ matrix over $F$ is diagonalisable over $F$ if and only if it has $n$ linearly independent eigenvectors in $F^n$ .

Kernels and solution sets

The kernel of a linear map $T$ contains the inputs that $T$ sends to zero. If $T(x)=b$ has one particular solution $x_p$ , then every solution has the form

x=x_p+x_h,\qquad x_h\in\ker T.

This one pattern describes the free directions in a matrix equation and the homogeneous part of a linear differential equation.

DF / the kernelLinear transformationsKernel
If $T:V\to W$ is a linear map, its kernel is
$\ker(T)=\{u\in V:T(u)=0\}.$
The same set is also called the null space. If $A\in\mathbb F^{m\times n}$ is regarded as the map $\mathbb F^n\to\mathbb F^m$ , then
$\ker(A)=\{x\in\mathbb F^n:Ax=0\}.$
TH / matrix equationsClasses of solutions of $Ax=b$
Let $A$ be a matrix over an infinite scalar field $\mathbb F$ (such as $\mathbb R$ or $\mathbb C$ ), let $b$ be a vector of the appropriate dimension and let $[A\mid b]$ be the augmented matrix formed by appending $b$ to $A$ . The equation $Ax=b$ has:
1. no solution if $\operatorname{rank}(A)\ne\operatorname{rank}([A\mid b]);$
2. a unique solution if $\operatorname{rank}(A)=\operatorname{rank}([A\mid b]) \quad\text{and}\quad \operatorname{nullity}(A)=0;$
3. infinitely many solutions if $\operatorname{rank}(A)=\operatorname{rank}([A\mid b]) \quad\text{and}\quad \operatorname{nullity}(A)>0.$
In the last case, if $\operatorname{nullity}(A)=k$ , then every solution has the form
$x=x_p+\omega_1u_1+\cdots+\omega_ku_k,$
where $x_p$ is any particular solution of $Ax=b$ , $\{u_1,\ldots,u_k\}$ is a basis for $\ker(A)$ and $\omega_1,\ldots,\omega_k\in\mathbb F$ .
TH / constant-coefficient ODEsOrdinary differential equationsHomogeneous part plus particular part
Write a linear equation as
$L[y]=f(x), \qquad L[y]=ay''+by'+cy.$
If $y_p$ is one particular solution and $y_h$ is any homogeneous solution, then the complete solution is
$y=y_h+y_p.$
MT / finding a particular solutionUndetermined coefficients
For suitable forcing functions:
- a polynomial of degree $n$ suggests a general polynomial of degree $n$ ;
- $e^{\beta x}$ suggests $Ce^{\beta x}$ ;
- $p_n(x)e^{\beta x}$ suggests $q_n(x)e^{\beta x}$ , where $q_n$ has degree $n$ ;
- $\sin(\beta x)$ or $\cos(\beta x)$ suggests $C_1\cos(\beta x)+C_2\sin(\beta x);$
- $e^{\alpha x}\sin(\beta x)$ or $e^{\alpha x}\cos(\beta x)$ suggests $e^{\alpha x} \bigl(C_1\cos(\beta x)+C_2\sin(\beta x)\bigr).$
Substitute the trial into the equation and match coefficients. If the trial overlaps with a homogeneous mode whose characteristic root has multiplicity $s$ , multiply the whole trial by $x^s$ (so a simple overlap requires one factor of $x$ , while a repeated root requires $x^2$ ).
DF / the kernel of a differential operatorKernel of an operator
The kernel of a linear differential operator $T$ is
$\ker(T)=\{y:T[y]=0\}.$
TH / linear differential equationsAll forced solutions are an affine translate
If $T[y]=f$ and $y_p$ is one particular solution, then
$\{\text{all solutions of }T[y]=f\} =y_p+\ker(T).$
Equivalently, every solution is $y_p+y_h$ for some homogeneous solution $y_h$ .

Roots, modes and evolution

A mode is a simple solution whose shape the evolution preserves: $r^n$ in discrete time and $e^{rt}$ in continuous time. The values of $r$ are characteristic roots and polar form reads each one as a scaling and a rotation.

In discrete time, $|r|$ controls growth or decay.
In continuous time, $\operatorname{Re}(r)$ controls growth or decay.
A nonzero imaginary part produces oscillation.

Scalar constant-coefficient ODEs build their solutions from these roots. In coupled linear ODEs, eigenvectors give the directions that evolve as single exponential modes. Matrix powers and Markov systems are the discrete-time versions of the same calculation.

DF / a root as a scaling and a rotationComplex numbersPolar form
For $z\ne0$ , the polar form of $z$ is
$z=r(\cos\theta+i\sin\theta)=re^{i\theta}, \qquad r=|z|>0,$
where $\theta$ is an argument of $z$ . The expressions $a+bi$ and $re^{i\theta}$ are two writings of the same complex number.
TH / discrete time: $|r|$ decidesDiscrete stability from root moduli
Write a characteristic root in polar form as
$r_j=\rho_j e^{i\theta_j}.$
Then
$|r_j^n|=\rho_j^n.$
Thus $\rho_j<1$ gives a decaying mode, $\rho_j>1$ gives an unbounded mode and $\rho_j=1$ gives a non-decaying mode.
TH / continuous time: $\operatorname{Re} r$ decidesContinuous stability from real parts
For a characteristic root $r=\alpha+i\beta$ ,
$|e^{rt}|=e^{\alpha t}.$
Therefore $\alpha<0$ gives decay, $\alpha>0$ gives an unbounded mode and, for a simple root, $\alpha=0$ gives a non-decaying oscillatory or constant mode. A repeated root on the imaginary axis introduces a term $t e^{rt}$ , whose magnitude is unbounded.
TH / matrix powersEigenvaluesLong-term behaviour of the modes
Suppose $A=PDP^{-1}$ is diagonalisable, so it has an eigenbasis. A mode contribution $c\lambda^kv$ with $c\ne0$ decays to zero when $|\lambda|<1$ , has constant magnitude when $|\lambda|=1$ and grows when $|\lambda|>1$ . Without an eigenbasis, a Jordan block can produce additional factors such as $k\lambda^{k-1}$ , so the diagonalisation hypothesis is what makes this scalar classification complete (especially at the unit circle).
TH / coupled ODEs: eigenvectors as modesExponential modes
For a constant matrix $A$ and a non-zero vector $\mathbf u$ , the function
$\mathbf{x}(t)=\mathbf{u}e^{\lambda t}$
is a solution of $\mathbf{x}'=A\mathbf{x}$ if and only if $\lambda$ is an eigenvalue of $A$ and $\mathbf{u}$ is a corresponding eigenvector. The zero solution is handled separately by taking $\mathbf u=\mathbf0$ .
TH / Markov systemsA sufficient condition for long-term Markov behaviour
Suppose that a transition matrix $A$ satisfies:
1. $1$ is an eigenvalue with algebraic multiplicity $1$ ;
2. every other eigenvalue $\lambda_j$ has $|\lambda_j|<1$ ;
3. there is a basis of eigenvectors for $\mathbb R^n$ .
Then $\mathbf{x}(k+1)=A\mathbf{x}(k)$ has a limit $\mathbf{p}$ in the eigenspace for $\lambda=1$ . It is the unique vector in that eigenspace whose component sum equals the component sum of $\mathbf{x}(0)$ ; it may be zero when that sum is zero. If $\mathbf{x}(0)$ is a probability vector, then $\mathbf{p}$ is the stationary distribution and does not depend on the initial probabilities.
MT / scalar ODEsOrdinary differential equationsRead the characteristic roots
Solve the characteristic equation and choose the real homogeneous modes according to the roots:
- If $r_1\neq r_2$ are distinct real roots, use $y_h=Ae^{r_1x}+Be^{r_2x}.$
- If $r$ is a repeated real root, use $y_h=(A+Bx)e^{rx}.$
- If the roots are $r=\alpha\pm i\beta$ with $\beta\neq0$ , use $y_h=e^{\alpha x} \bigl(A\cos(\beta x)+B\sin(\beta x)\bigr).$

Local approximation

An approximation replaces a hard object with a simpler one and keeps track of the difference:

\text{original object}=\text{simple model}+\text{error}.

A Taylor polynomial matches derivatives at a centre and the remainder measures the error away from that centre. An infinite Taylor series needs a separate convergence argument: matching every derivative does not by itself show that the series equals the function.

In several variables, the tangent plane is the first-order model. The second-order model adds curvature through the Hessian, which can classify interior stationary points. These models live in the polynomial spaces from Algebra, where interpolation instead fixes a polynomial by its values at distinct inputs.

DF / match values instead of derivativesVector spacesPolynomial interpolation
Given $n$ distinct inputs $t_1,\ldots,t_n$ and prescribed outputs $y_1,\ldots,y_n$ , polynomial interpolation finds the polynomial $f\in P_{n-1}$ satisfying
$f(t_i)=y_i,\qquad i=1,\ldots,n.$
Using the standard basis
$B=(1,t,t^2,\ldots,t^{n-1}),$
write
$f(t)=a_0+a_1t+\cdots+a_{n-1}t^{n-1}.$
The coefficients form the coordinate vector
$[f]_B= \begin{pmatrix} a_0\\a_1\\\vdots\\a_{n-1} \end{pmatrix}.$
DF / one variable: match derivatives at a centreTaylor series, sequences and infinite seriesTaylor polynomial
If $f$ has derivatives through order $n$ at $a$ , its Taylor polynomial of degree at most $n$ about $a$ is
$P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k.$
TH / the error termTaylor's theorem with remainder
Suppose $f$ has $n+1$ continuous derivatives on an interval containing $a$ and $x$ . If
$P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k,$
then
$f(x)=P_n(x)+R_{n+1}(x),$
where the integral remainder is
$R_{n+1}(x) =\frac1{n!}\int_a^x f^{(n+1)}(t)(x-t)^n\,dt.$
For $x\ne a$ , there is also a point $c$ between $a$ and $x$ for which the Lagrange remainder is
$R_{n+1}(x) =\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}.$
TH / classifying stationary points in one variableClassify a stationary point with the first nonzero derivative
Suppose $k$ is an integer with $k\ge2$ and $f$ has $k+1$ continuous derivatives on a neighbourhood of $a$ . If
$f^{(j)}(a)=0\quad(1\le j<k), \qquad f^{(k)}(a)\ne0.$
then Taylor's theorem gives
$f(x)-f(a) =\frac{f^{(k)}(a)}{k!}(x-a)^k+R_{k+1}(x), \qquad R_{k+1}(x)=O\bigl(|x-a|^{k+1}\bigr),\qquad R_{k+1}(x)=o\bigl(|x-a|^k\bigr).$
Thus, locally:
- if $k$ is even and $f^{(k)}(a)>0$ , $a$ is a local minimum;
- if $k$ is even and $f^{(k)}(a)<0$ , $a$ is a local maximum;
- if $k$ is odd, the leading term changes sign across $a$ , giving a horizontal point of inflection.
TH / first-order model in several variablesFunctions of several variablesTangent plane to a graph
If $f$ is differentiable at $\mathbf{a}=(x_0,y_0)$ and $z_0=f(x_0,y_0)$ , then the tangent plane to $z=f(x,y)$ at $(x_0,y_0,z_0)$ is
$z =f(x_0,y_0) +f_x(x_0,y_0)(x-x_0) +f_y(x_0,y_0)(y-y_0).$
DF / second-order model: the HessianTaylor polynomial through degree three
Let $\mathbf{a}=(x_0,y_0)$ , $h_x=x-x_0$ and $h_y=y-y_0$ . If $f$ has continuous partial derivatives through order three near $\mathbf{a}$ , its Taylor polynomial through total degree three is
$\begin{aligned} T_3(x,y) ={}&f(\mathbf{a}) +f_x(\mathbf{a})h_x +f_y(\mathbf{a})h_y\\ &+\frac{1}{2!} \left( f_{xx}(\mathbf{a})h_x^2 +2f_{xy}(\mathbf{a})h_xh_y +f_{yy}(\mathbf{a})h_y^2 \right)\\ &+\frac{1}{3!} \left( f_{xxx}(\mathbf{a})h_x^3 +3f_{xxy}(\mathbf{a})h_x^2h_y +3f_{xyy}(\mathbf{a})h_xh_y^2 +f_{yyy}(\mathbf{a})h_y^3 \right). \end{aligned}$
The degree-two part can be written with the Hessian as
$T_2(\mathbf{x}) = f(\mathbf{a}) +\nabla f(\mathbf{a})\cdot(\mathbf{x}-\mathbf{a}) +\frac12 (\mathbf{x}-\mathbf{a})^{\mathsf T} H_f(\mathbf{a}) (\mathbf{x}-\mathbf{a}).$
TH / classifying stationary points in several variablesThe second-order partial derivative test
Let $\mathbf{x}_0=(x_0,y_0)$ be a stationary point. Assume that $f$ has continuous second-order partial derivatives on a neighbourhood of $\mathbf{x}_0$ (so $f$ is $C^2$ there). Define
$D =\det H_f(\mathbf{x}_0) =f_{xx}(\mathbf{x}_0)f_{yy}(\mathbf{x}_0) -f_{xy}(\mathbf{x}_0)^2.$
Then:
- if $D>0$ and $f_{xx}(\mathbf{x}_0)>0$ , $\mathbf{x}_0$ is a local minimum;
- if $D>0$ and $f_{xx}(\mathbf{x}_0)<0$ , $\mathbf{x}_0$ is a local maximum;
- if $D<0$ , $\mathbf{x}_0$ is a saddle point;
- if $D=0$ , the test is inconclusive.

Accumulation and changes of variables

Integration techniques use the form of an integrand to find an antiderivative and they return as tools for solving differential equations. Double integrals accumulate a quantity over a planar region. Slicing reduces them to iterated single integrals and a change of variables can simplify the region or the integrand, with the Jacobian determinant accounting for local area scaling. It plays the same role as choosing a good basis: a better description makes the calculation simpler.

DF / linear maps: a basis that decouplesEigenvaluesDiagonalisation
An $n\times n$ matrix $A$ over a field $F$ is diagonalisable over $F$ if there are an invertible matrix $P\in M_n(F)$ and a diagonal matrix $D\in M_n(F)$ such that
$P^{-1}AP=D, \qquad\text{equivalently}\qquad A=PDP^{-1}.$
The columns of $P$ are the eigenvectors in the chosen eigenbasis and the corresponding eigenvalues appear on the diagonal of $D$ .

MT / single integralsIntegration techniquesSubstitutions for

\sqrt{a^2\pm x^2}

and

\sqrt{x^2-a^2}

For $a>0$ , choose the indicated trigonometric or hyperbolic branch so that the substitution is one-to-one on the interval being integrated. Keep the absolute value when simplifying a square root until the chosen range fixes its sign; for example, $\sqrt{a^2-a^2\sin^2\theta}=a|\cos\theta|$ .

Integrand factor	Substitutions	Identities
$\sqrt{a^2-x^2}$	$x=a\sin\theta$ , $dx=a\cos\theta\,d\theta$ or $x=a\tanh\theta$ , $dx=a\operatorname{sech}^2\theta\,d\theta$	$\cos^2\theta+\sin^2\theta=1$ , $\operatorname{sech}^2\theta+\tanh^2\theta=1$
$\sqrt{a^2+x^2}$	$x=a\tan\theta$ , $dx=a\sec^2\theta\,d\theta$ or $x=a\sinh\theta$ , $dx=a\cosh\theta\,d\theta$	$\sec^2\theta=\tan^2\theta+1$ , $\cosh^2\theta=1+\sinh^2\theta$
$\sqrt{x^2-a^2}$	$x=a\sec\theta$ , $dx=a\sec\theta\tan\theta\,d\theta$ or, for $x\ge a$ , $x=a\cosh\theta$ , $dx=a\sinh\theta\,d\theta$	$\sec^2\theta=1+\tan^2\theta$ , $\cosh^2\theta=1+\sinh^2\theta$

MT / ODEs: a substitution makes it separableOrdinary differential equationsThe substitution $y=xv$
For an equation of the form
$\frac{dy}{dx}=f\left(\frac{y}{x}\right),$
set $y=xv(x)$ . Then
$y'=v+xv',$
so
$xv'=f(v)-v, \qquad \frac{dv}{dx}=\frac{f(v)-v}{x}.$
The resulting equation is separable on an interval where $x\neq0$ and the required denominators are nonzero.
TH / double integrals: the Jacobian scales areaDouble integralsChange of variables
Let $\Omega$ be a region in $(x,y)$ -coordinates and let $\Phi$ be a region in $(u,v)$ -coordinates. Assume that $f$ is integrable on $\Omega$ , that $g$ is $C^1$ on $\Phi$ , that $\Phi$ can be split into finitely many pieces on which $g$ is one-to-one with non-overlapping interiors and that the Jacobian is nonzero away from boundary or other exceptional sets of area zero. The images of those pieces must cover $\Omega$ once, apart from boundaries of area zero; otherwise the transformed integral can omit or count part of the region more than once. If a map covers the target with multiplicity, split the parameter region into one-to-one pieces (or account for that multiplicity) before applying the formula. Then
$\iint_\Omega f(x,y)\,dx\,dy = \iint_\Phi f\bigl(x(u,v),y(u,v)\bigr) \left|\det J_g(u,v)\right|\,du\,dv.$
TH / polar coordinatesPolar area element
For $r\ge0$ , the polar-coordinate change of variables is
$x=r\cos\theta,\qquad y=r\sin\theta,$
and its Jacobian satisfies
$\left|\frac{\partial(x,y)}{\partial(r,\theta)}\right| =r.$
Consequently,
$dA=r\,dr\,d\theta,$
and
$\iint_\Omega f(x,y)\,dA =\iint_\Phi f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta.$

Where next

The concept map shows how the course's ideas depend on one another and gives a reading path through each strand. Mathematical language covers the general vocabulary and notation the chapters assume. Each chapter defines its own terms where they first appear.

Core

On this page