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For $z\ne0$, the polar form of $z$ is

where $\theta$ is an argument of $z$. The expressions $a+bi$ and $re^{i\theta}$ are two writings of the same complex number.

Write a characteristic root in polar form as

Then

Thus $\rho_j<1$ gives a decaying mode, $\rho_j>1$ gives an unbounded mode and $\rho_j=1$ gives a non-decaying mode.

For a characteristic root $r=\alpha+i\beta$,

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.

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, $\sqrt{a^2-a^2\sin^2\theta}=a|\cos\theta|$.

If $B=\{u_1,\ldots,u_n\}$ is a basis for $V$, then every vector $v\in V$ has a unique representation

Let $B=(u_1,\ldots,u_n)$ be an ordered basis for a vector space $V$. If

then the coordinate vector of $u$ with respect to $B$ is the column

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

Using the standard basis

write

The coefficients form the coordinate vector

For an equation of the form

set $y=xv(x)$. Then

so

The resulting equation is separable on an interval where $x\neq0$ and the required denominators are nonzero.

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 $$y_h=e^{\alpha x} \bigl(A\cos(\beta x)+B\sin(\beta x)\bigr).$$

Write a linear equation as

If $y_p$ is one particular solution and $y_h$ is any homogeneous solution, then the complete solution is

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 $x^2$).

The kernel of a linear differential operator $T$ is

If $T[y]=f$ and $y_p$ is one particular solution, then

Equivalently, every solution is $y_p+y_h$ for some homogeneous solution $y_h$.

If $T:V\to W$ is a linear map, its kernel is

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

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

where $x_p$ is any particular solution of $Ax=b$, $\{u_1,\ldots,u_k\}$ is a basis for $\ker(A)$ and $\omega_1,\ldots,\omega_k\in\mathbb F$.

Let $T:V\to W$ be a linear map, where $\dim(V)=n$ and $\dim(W)=m$. For ordered bases

form the coordinate columns

Then

is the unique $m\times n$ matrix satisfying

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.

If $f$ has derivatives through order $n$ at $a$, its Taylor polynomial of degree at most $n$ about $a$ is

Suppose $f$ has $n+1$ continuous derivatives on an interval containing $a$ and $x$. If

then

where the integral remainder is

For $x\ne a$, there is also a point $c$ between $a$ and $x$ for which the Lagrange remainder is

Suppose $k$ is an integer with $k\ge2$ and $f$ has $k+1$ continuous derivatives on a neighbourhood of $a$. If

then Taylor's theorem gives

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.

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

The columns of $P$ are the eigenvectors in the chosen eigenbasis and the corresponding eigenvalues appear on the diagonal of $D$.

An $n\times n$ matrix over $F$ is diagonalisable over $F$ if and only if it has $n$ linearly independent eigenvectors in $F^n$.

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).

For a constant matrix $A$ and a non-zero vector $\mathbf u$, the function

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 $\mathbf u=\mathbf0$.

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.

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

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

The degree-two part can be written with the Hessian as

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

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.

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

For $r\ge0$, the polar-coordinate change of variables is

and its Jacobian satisfies

Consequently,

and