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When $\alpha =3$, the third linear equation is $-3$ times the second, so it adds no constraint. Take $z=t$ freely. The remaining linear equations are $x-y=-1-3t$ and $x+6y=1-\beta t$. Solving gives the straight line of solutionsThis exists for every real $\beta$.

Call the three planes $P_1,P_2,P_3$ in their given order.

For $\alpha \ne 3,-1/2$, the nonzero determinant makes their normal vectors linearly independent. The three planes meet at the single point found in part (c)(i), for every $\beta$.

For $\alpha =3$, $P_2$ and $P_3$ coincide. Their common plane and $P_1$ have nonparallel normal vectors, since their first two coefficients are $(1,-1)$ and $(1,6)$. Thus they intersect in the straight line from part (c)(ii), for every $\beta$.

For $\alpha =-1/2$, write the three plane equations asThe last two meet on $z=0$, $x-y=-1$, which never lies on $P_1$. If $\beta =3$, $P_1$ and $P_2$ are distinct parallel planes, while $P_3$ cuts both. If $\beta =-1/2$, $P_1$ and $P_3$ are distinct parallel planes, while $P_2$ cuts both. For every other $\beta$, each pair of planes intersects in a straight line; these three pairwise intersection lines are distinct and parallel to $(1,1,0{)}^T$, with no common point. These exhaust the geometric cases.

The chord through the endpoints is the straight line $y=(x-1)\log 2$. At $\xi =1/\log 2$, the tangent line isThese two straight lines have the same slope, as required by the mean value theorem. Since $f^{\prime \prime }(x)=-1/x^2<0$, the natural logarithm is a concave function: its graph lies above the chord and below this tangent line.

Using the complex conjugate, $b\bar{z}+\bar{b}z=2\mathrm{Re}(\bar{b}z)$. The locus is therefore the straight lineIn the complex plane, the quantity $\mathrm{Re}(\bar{b}z)/|b|$ is the signed scalar projection of $z$ in the direction $b$. Thus the straight line is perpendicular to $b$ and passes through $-cb/(2|b{|}^2)$. This description uses no separation of $z$ into real coordinates.

For the reciprocal Möbius transformation, substitute $z=1/w$, $\bar{z}=1/\bar{w}$ and multiply by $w\bar{w}$, which is nonzero on the stipulated domain. This yieldsConsequentlyFor a nondegenerate original circle, $c=0$ says that it passes through the origin, and its image is a straight line. The restriction $w\ne 0$ must be retained: multiplying through does not authorize adding that excluded point. Degenerate choices can give a constant or an identically zero equation instead of a straight line.

Two intersecting straight lines are perpendicular if their angle is $90^{\circ }$. Their direction vectors have zero dot product.

Photocurrent is the electric current attributable to illumination. Here $I_{\mathrm{ph}}\ge 0$ denotes its magnitude after background subtraction. For a photodiode without internal multiplication, let $P$ be incident radiant flux concentrated at wavelength $\lambda$, and let $\eta$ be external quantum efficiency, including losses before collection. An incident photon has energy $E_{\gamma }=hc/\lambda$, with Planck constant $h$ and speed of light $c$. Dividing energy per second by energy per photon gives the incident photon rate ${\Phi }_{\gamma }$. Multiplying by $\eta$ gives the mean collected pair rate ${\dot{N}}_{\mathrm{pair}}$; multiplying by elementary charge $q$ givesThe factor is $q$: each collected pair replenishes an electron at one contact and removes an electron at the other. A cross-section of the external wire therefore counts one transferred electron per completed pair, consistent with charge conservation. The electron and electron hole are two internal charge carriers, not two independent transfers through that wire.

Choose an illustrative photodiode model with $\lambda =650\mathrm{nm}$, $\eta =0.75$ and $P=1.00\mu \mathrm{W}$. Using the exact SI values $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$ and $q=1.602176634\times 10^{-19}\mathrm{C}$ givesA mean fraction $0.75$ does not require each fourth photon to be lost in a repeating pattern. It specifies a mean conversion rate. Nor does $0.75$ mean $0.75\mathrm{A}{\mathrm{W}}^{-1}$: quantum efficiency is dimensionless, whereas spectral responsivity compares electric current with power.

At fixed radiant flux, longer-wavelength photons have less energy, so more arrive per second. This explains the rising straight line when quantum efficiency is held constant. A real photodiode's quantum efficiency varies with wavelength; the plotted assumption cannot be extended past its band gap absorption cutoff. At fixed incident photon rate instead, $I_{\mathrm{ph}}=\eta q{\Phi }_{\gamma }$ has no additional factor of $\lambda$.

A plane is a two-dimensional affine subspace: a point $\mathbf{a}$ together with every linear combination of two linearly independent direction vectors $\mathbf{u},\mathbf{v}$,In three-dimensional Euclidean space, a Cartesian equation of a plane fixes one linear equation $\mathbf{n}\cdot \mathbf{x}=d$, with a nonzero normal vector $\mathbf{n}$. The two free directions are orthogonal to $\mathbf{n}$. A plane extends indefinitely; a rectangular patch in a drawing marks only part of it, and does not define an edge of the plane.

For example, the plane $2x-y+2z=3$ contains $\mathbf{a}=(1,1,1)$. The vectors $\mathbf{u}=(1,2,0)$ and $\mathbf{v}=(0,2,1)$ are linearly independent and satisfy $(2,-1,2)\cdot \mathbf{u}=(2,-1,2)\cdot \mathbf{v}=0$. Hence its points are exactly $(1+s,1+2s+2t,1+t)$: substitution gives $2x-y+2z=3$ for every real $s,t$, and every solution has $s=x-1$, $t=z-1$.

The description separates position from direction. A translation of the plane changes $\mathbf{a}$ and possibly $d$, while retaining its parallel directions. A plane through the origin is a linear subspace; the displayed example is not, since the zero vector fails its linear equation. This distinction matters when applying orthogonal projection: subtract an anchor point before projecting onto the direction linear subspace.

Historically, Euclid's Euclid's Elements by Euclid, Book I Definition 7, describes a plane surface through the straight lines lying on it. David E. Joyce's accompanying guide notes that this older use can include a bounded square or circular figure. That usage differs from the whole unbounded affine subspace used here.^[ref]

A half-line beginning at $p$ with a nonzero direction vector $v$. Removing the endpoint gives an open ray; allowing every real parameter gives the whole straight line.

For $c\ne 0$ and real $r$, this equation defines a straight line. Writing $z=x+iy$ and $c=u+iv$ gives $2ux+2vy+r=0$, so $c$ supplies a normal direction. Conversely every real affine line $Ax+By+C=0$ has this representation with $c=(A+iB)/2$ and $r=C$. The condition $c\ne 0$ excludes an empty set or the entire complex plane.

Write a generalized circle equation as $a|z{|}^2+b\bar{z}+\bar{b}z+c=0$, where $a,c$ are real. Under the reciprocal Möbius transformation $z=1/w$, with $w\ne 0$, multiply by $|w{|}^2$ to obtainThus the coefficient triple transforms as $(a,b,c)\mapsto (c,\bar{b},a)$. A nondegenerate circle through zero has $c=0$ and becomes a straight line; a straight line missing zero becomes a circle through zero on the Riemann sphere. When working only with finite nonzero coordinates, retain the excluded point $w=0$ even if it satisfies the multiplied equation.

This reciprocal transformation is different from inversion in a circle, whose unit-circle formula is $z\mapsto 1/\bar{z}$. The latter reverses orientation; the reciprocal transformation is holomorphic away from its pole.

Consider the linear map $A:{\mathbb{R}}^3\to {\mathbb{R}}^2$ represented by the rectangular matrixTo compute its kernel, solve the homogeneous linear system $x+2y=0$, $y+z=0$. Taking $y=t$ gives $x=-2t$, $z=-t$, soThus $v$ represents a null direction of a linear map: decreasing the first input by two, increasing the second by one, and decreasing the third by one changes neither output. Solving the homogeneous linear system is the concrete computation behind a kernel; the column space instead decides which outputs are attainable.^[ref]

Now prescribe the nonzero output $b=(3,2{)}^T$. The linear system $Au=b$ becomes $x+2y=3$, $y+z=2$. Again take $y=t$:Direct substitution gives $A{u}_0=(3,2{)}^T$, and $A(u_0+tv)=A{u}_0+tAv=b$. Conversely, the two linear equations force exactly this form, so no solutions have been omitted. This fiber is an affine subspace of dimension one in the three-dimensional input vector space. It is parallel to the kernel, but does not pass through the origin. For instance, $u_0=(3,0,2{)}^T$ and $u_1=(1,1,1{)}^T$ both give $b$, while their difference $u_1-u_0=v$ gives zero. Adding $u_0$ and $u_1$ gives output $2b$, which is another way to see that this nonzero-output fiber is not a vector subspace.

The first and third columns of $A$ are $(1,0{)}^T$ and $(0,1{)}^T$, a basis of the output vector space. Hence $A$ has matrix rank two and reaches every output. The rank-nullity theorem predicts nullity $3-2=1$ before we solve any linear system. It provides a useful completeness check: one nonzero kernel vector already forms a basis here, so a second independent null direction of a linear map would signal a calculation error. In general, matrix rank $r$ for an $m\times n$ matrix leaves $n-r$ independent kernel directions, rather than $m-r$.^[ref]

Choose a fixed positive reference $A_{\mathrm{ref}}$ with the same units as $A$. Taking the natural logarithm of a dimensionless ratio givesThus the Richardson plot is a straight line when $A$ and $W$ are constant, with slope $-W/k_B$ and intercept $\ln (A/A_{\mathrm{ref}})$. The familiar notation $\ln (J/T^2)$ silently chooses units; the reference makes that choice explicit. Changing $A_{\mathrm{ref}}$ shifts the intercept and leaves the inferred work function unchanged. If the horizontal axis instead uses $1000\mathrm{K}/T$, its slope is $-W/(1000\mathrm{K}{k}_B)$.

For the chosen $2.0\mathrm{eV}$ example, choose $A_{\mathrm{ref}}=120\mathrm{A}\mathrm{c}{\mathrm{m}}^{-2}{\mathrm{K}}^{-2}$. Then $y(1000\mathrm{K})=-23.2090$ and $y(1200\mathrm{K})=-19.3409$. Their slope is approximately $-23209\mathrm{K}$, giving $W=2.0\mathrm{eV}$. Two temperatures determine a line algebraically; several temperatures allow curvature and drift to be detected. These values are calculated, rather than experimental observations.

A straight trace alone does not establish a unique surface work function. Adsorption, changing surface coverage, or a mixture of regions with different barriers can change the inferred slope. For independent emitting patches with area fractions $f_i$, constant $A_i$ and barriers $W_i$,Lower barriers contribute disproportionately. Even with fixed patches, the slope of the Richardson plot gives a temperature-dependent weighted mean of the $W_i$, rather than their simple area average. If $A$ or $W$ themselves vary with temperature, the apparent slope-derived barrier becomesCheck collection, temperature calibration, and surface stability before assigning curvature to a new emission mechanism.

A rotational spectral line assignment associates a measured spectral line with a molecule, an internal quantum state, and an initial and final set of quantum numbers. For the spin-free linear rigid rotor, absorption is labelled $J\to J+1$. A regularly spaced group suggests a candidate rotational constant, but the lowest line inside a scanned band need not be $0\to 1$. Several molecules can contribute to the same spectrum, and an apparent missing member can lie outside the band or below the detection threshold.

First determine line centers and their uncertainties from the instrument's readout, accounting for resolved Doppler shifts or hyperfine structure rather than treating every peak as a separate rotational transition. Then propose integer assignments, fit a rotational Hamiltonian, and predict additional spectral lines. For a weakly distorted linear molecule within one vibrational quantum state, the centrifugal distortion model gives the useful diagnosticHere $B,D,{\nu }_m$ are all frequencies in $\mathrm{Hz}$. Correct assignments make the second expression a straight line against $m^2$, with intercept $B$ and slope $-2D$. This follows by division of the transition formula; no additional model of molecular geometry is required. A direct weighted least squares fit of ${\nu }_m$ to $2Bm-4D{m}^3$ uses the measured uncertainties. If instead the transformed values are fitted, their uncertainties must also be divided by $2m$.^[ref]

For an original example, stipulate $B=5.000000\mathrm{GHz}$ and $D=2.000\mathrm{kHz}$, and suppose a scan covers $38$–$72\mathrm{GHz}$. The model places four consecutive transitions at $39.999488$, $49.999000$, $59.998272$, and $69.997256\mathrm{GHz}$. Their assignments are $m=4,5,6,7$, hence $J=3,4,5,6$. An extra stipulated line at $53.2\mathrm{GHz}$ is unrelated to this series. All these frequencies are calculated example inputs, not measured OCS data.

At $m=4$ and $m=7$, the transformed values are $4999.936$ and $4999.804\mathrm{MHz}$. Their straight line has slope $(4999.804-4999.936)/(49-16)=-0.004\mathrm{MHz}$, recovering $D=0.002\mathrm{MHz}$ and $B=5000\mathrm{MHz}$. Both intermediate transitions satisfy the same line exactly. Merely relabelling the first visible transition $0\to 1$ fails this diagnostic. The model predicts a withheld $7\to 8$ transition at $79.995904\mathrm{GHz}$: an additional measurement there would test the assignment rather than merely refit the original four lines.

An assignment earns confidence when several independently observed spectral lines, including lines outside the fitted interval, agree within uncertainty and expected strong accessible lines are present. Inspect the residual of an inverse problem for each line against $J$: systematic curvature can signal inadequate centrifugal distortion, whereas isolated discrepancies suggest a blend or wrong assignment. Comparing a known sample, a blank, and an isotope shift of a linear rotor supplies different checks. These are general inference controls; the actual reported controls of the NIST trace-gas FTMW validation experiment are described separately. One isolated match or one fitted rotational constant alone cannot certify molecular identity.

A secant line is a straight line passing through two distinct points of a curve. For a graph $y=f(x)$ at $x=a$ and $x=a+h$, its slope is the difference quotient $(f(a+h)-f(a))/h$. As the second point approaches the first, a limiting secant line can give the tangent line.

Let $f$ have continuous second partial derivatives near an interior critical point $x_{\ast }$ of an open subset of ${\mathbb{R}}^n$, with $\nabla f(x_{\ast })=0$. The Hessian matrix gives the following sufficient conditions:

A positive semidefinite or negative semidefinite Hessian matrix with a zero eigenvalue does not settle the classification. The higher-order test for separated extrema gives two quartic examples with exactly the same zero Hessian. These conditions apply to an interior critical point; a boundary optimum or a constrained optimization problem requires attention to the allowed directions. ^[ref]

The proof outline is the Taylor expansionFor a positive-definite matrix, the smallest eigenvalue supplies a positive bound proportional to $\Vert h{\Vert }^2$, which dominates the remainder for small enough $h$. Negate the function for the local maximum case. For opposite-sign eigenvalues, choose the two corresponding eigenvectors: the leading terms have opposite signs along their lines.

There are also necessary conditions. At an interior local minimum, the Hessian matrix is positive semidefinite; at an interior local maximum, it is negative semidefinite. Restriction to each straight line reduces this to the one-variable second derivative condition. Since the Laplacian is the trace of the Hessian, it is nonnegative at such a local minimum and nonpositive at such a local maximum. The sign of the Laplacian alone is insufficient: $x^2-2y^2$ has negative Laplacian but a saddle point of a scalar function at the origin.

A semi-log plot uses a logarithmic scale on one axis and a linear scale on the other. Which axis is logarithmic matters. Plotting $x$ against $\log y$ makes positive exponential functions appear as straight lines; plotting $\log x$ against $y$ instead makes an affine function of a logarithm appear straight. The plotting operations change positions while retaining the original tick values.^[ref]

Take the original dimensionless mathematical model $y(t)=3e^{0.4t}$ for $0\le t\le 8$. It has exponential growth, with $y(0)=3$, $y(4)\approx 14.8591$, and $y(8)\approx 73.5976$. Taking a base-ten logarithm givesThe slope in these transformed positions is $0.4/\ln 10\approx 0.173718$, although the original relative growth rate $y^{\prime }/y$ is $0.4$. Equal steps of two in $t$ multiply $y$ by $e^{0.8}\approx 2.22554$. The corresponding ordinary gains increase, so the left plot bends while the right plot is straight.

One practical use is checking a proposed exponential distribution for failure times. For rate $\lambda >0$ and $t\ge 0$, its cumulative distribution function is $F(t)=1-e^{-\lambda t}$, so the reciprocal survival function $y=1/[1-F(t)]$ satisfies ${\log }_{10}y=\lambda t/\ln 10$. Estimated probabilities can therefore be placed on a logarithmic vertical axis and compared with a straight line. Agreement supports compatibility with the chosen statistical model; a roughly straight finite sample does not establish a unique physical mechanism.^[ref]

For a p-n junction at fixed temperature, the Shockley diode equation models its electric current asHere $I_S>0$ is the diode saturation current, $n>0$ the diode ideality factor, $V_T$ the thermal voltage, $k_B$ the Boltzmann constant, and $q>0$ the elementary charge.^[ref]

For an original model example, choose $I_S=1\mathrm{nA}$, $n=2$, and $T=300\mathrm{K}$. Using $k_B=1.380649\times 10^{-23}J/K$ and $q=1.602176634\times 10^{-19}\mathrm{C}$ gives $V_T=25.8520\mathrm{mV}$. Then $V_D=0.600\mathrm{V}$ gives $I_D=0.109591\mathrm{mA}$, whereas $V_D=0.800\mathrm{V}$ gives $I_D=5.24450\mathrm{mA}$. These are predictions of the declared model. There is no universal turn-on electric potential difference: solving the Shockley diode equation for the value needed at a chosen electric current givesAt $I_D\gg I_S$, increasing $V_D$ by $0.100\mathrm{V}$ multiplies the electric current by approximately $e^{0.100/(n{V}_T)}=6.91772$. At $V_D=-0.300\mathrm{V}$ the same model gives $I_D=-0.996979\mathrm{nA}$. Its reverse branch approaches $-I_S$; it cannot describe diode reverse breakdown.

A constructed extension adds internal series electrical resistance $R_s$ to a junction whose electric potential difference is $V_j$. Applying the Kirchhoff cycle law givesFor $R_s=20\Omega$ and the same $I_S,n,T$, sustaining $I_D=5.24450\mathrm{mA}$ needs $V_D=0.904890\mathrm{V}$: $0.800\mathrm{V}$ across the junction plus $0.104890\mathrm{V}$ across its internal electrical resistance. The derivativeshows why the current-voltage characteristic becomes nearly a straight line when the internal series electrical resistance dominates. This is still a fixed-temperature model; self-heating can change $I_S$, $n$ and $V_T$. Real diodes also have leakage and charge-storage effects omitted here.^[ref]

Unroll a circular cylinder of radius $R>0$ using the coordinate $u=R\varphi$. Its line element becomes $d{s}^2=d{u}^2+d{z}^2$, so a geodesic lifts to a straight line in the plane. An endpoint has infinitely many lifts differing by $2\pi Rk$. Minimize the ordinary Euclidean distance over those lifts to obtain the unrestricted shortest path. A fixed lift instead specifies a winding class and need not give the global shortest path.

For angular difference $0\le \delta \le 2\pi$, put $m=\min (\delta ,2\pi -\delta )$. Then $d=\sqrt{R^2{m}^2+(z_1-z_0{)}^2}$. At $\delta =\pi$, two different geodesics have the same shortest length. This is the source of the corner in the distance-versus-angle graph at the opposite generator of the circular cylinder.

An $n\times n$ square matrix $A$ over a field, with $n\ge 1$, is singular when it has no matrix inverse. Geometrically, its linear map loses at least one input direction: some nonzero vector $v$ satisfies $Av=0$, so $A(x+v)=Ax$. Over the real numbers, the following family makes that loss visible:

For this diagonal matrix, $\det A(t)=t$. Whenever $t\ne 0$, the input can be recovered as $(x,y)=(b_1,b_2/t)$ and the kernel is $\{0\}$. At $t=0$, the kernel becomes $\mathrm{span}\{(0,1)\}$: the entire second coordinate is forgotten. Consequently $A(0)x=(b_1,b_2)$ has no solution if $b_2\ne 0$, and has the solution straight line $x=(b_1,s)$, $s\in \mathbb{R}$, if $b_2=0$. A singular matrix therefore need not make a particular linear system inconsistent; it prevents unique recovery when that linear system is consistent.

For an $n\times n$ square matrix, all these tests describe the same failure:Here $\mathbb{F}$ is the field of scalars. Equivalently, the columns are linearly dependent, the linear map is neither injective nor surjective, and some outputs cannot be reached. See the invertible matrix equivalences and Strang's MIT 18.06SC lecture summary, Multiplication and inverse matrices, p. 2. Over an infinite field, every consistent linear system for a singular matrix has infinitely many solutions; over a finite field with $q$ elements and nullity $d$, it has exactly $q^d$ solutions. This follows by adding each kernel vector to one solution.

The square hypothesis is essential. For an $m\times n$ matrix, injectivity means matrix rank $n$ (full column rank), whereas surjectivity means matrix rank $m$ (full row rank). A rectangular matrix can satisfy one condition without the other; its kernel and image of a linear map give the useful tests without calling it a singular matrix.

Exact singularity also differs from sensitivity. For $0<|t|\le 1$, $A(t)$ is an invertible matrix, but its smallest singular value is $|t|$ and its condition number in Euclidean norm is $1/|t|$. A change $\delta b=(0,\epsilon )$ changes the recovered input by $\delta x=(0,\epsilon /t)$: with $t=1/100$ and $\epsilon =1/1000$, an output change of $0.001$ causes an input change of $0.1$. These are exact toy values, not measurements. The condition number quantifies this sensitivity before a direction is lost; see Steve Marschner, Condition numbers, §2. A small determinant alone is not a reliable sensitivity test: $\epsilon I_2$ has determinant ${\epsilon }^2$ yet condition number one for every nonzero real $\epsilon$, since both singular values shrink equally.