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Resolve the input along the three perpendicular eigendirections. The linear map removes the component along $(1,2,1)$, stretches the component along $(1,-1,1)$ by $12$, and reverses and stretches the component along $(1,0,-1)$ by $6$. Every output lies in the plane perpendicular to $(1,2,1)$. A nonzero input in the kernel of a linear map $A$ is sent to zero; other nonzero inputs remain in that output plane. Equivalently, take the orthogonal projection onto this plane, then scale by the two signed eigenvalues along its perpendicular eigendirections.

The inverse fails to exist exactly when at least one eigenvalue is zero. In that case a nonzero eigenvector lies in the kernel of a linear map, so $A$ is not one-to-one. Equivalently, its determinant ${\prod }_n{\lambda }_n$ is zero. Conversely, if every eigenvalue is nonzero, the preceding construction gives an inverse.

In $A^{\mathsf{T}}A$, every row sums to one, so $(1,1,1)$ is an eigenvector with eigenvalue $1$. The antisymmetric direction $(1,0,-1)$ gives eigenvalue $2$. A direction orthogonal to both is $(1,-2,1)$, and direct multiplication shows that it lies in the kernel of a linear map. Normalizing all three gives the complete systemThey form an orthonormal basis of ${\mathbb{R}}^3$. The extra zero eigenvalue reflects the one-dimensional kernel of a linear map for this rank-two rectangular matrix. The question's symbols ${\sigma }_m$ denote the eigenvalues of $A^{\mathsf{T}}A$; the nonzero singular values of $A$ are their square roots, namely $1$ and $\sqrt{2}$.

Translate the generalized coordinates so that the equilibrium is at zero. Its potential gradient vanishes, and its quadratic potential energy is $q^T\mathsf{V}q/2$, where $\mathsf{V}$ is the symmetric Hessian matrix. The Euler-Lagrange equations for small motion areA normal mode has fixed component ratios, $q(t)=v\cos (\omega t+\varphi )$, withThe angular frequency $\omega$ is its eigenfrequency. A zero mode instead admits a constant or linear time factor. For modes $v_{\alpha },v_{\beta }$, symmetry givesHence mass-matrix orthogonality of normal modes isWithin a degenerate eigenspace, a basis can be chosen orthogonal in this positive weighted inner product.

For three masses connected pairwise, the exact energies areA fixed common translation $x_i\mapsto x_i+a$ preserves all differences and velocities. A fixed rigid rotation $x_i\mapsto Q{x}_i$ preserves their Euclidean norms. Thus both energies have these symmetries. Here the transformation is fixed in time; a moving reference frame would change the kinetic energy. Every equilateral configuration of side $\sqrt{3}$ has zero spring extensions, hence zero force and minimum potential, proving that it is an equilibrium.

For the specified displacement coordinates, writeThe exact potential is $V=(k/2){\sum }_{ij=12,13,23}(d_{ij}-\sqrt{3}{)}^2$. For an equilibrium edge vector $d$ and small perturbation $\delta d$,Thus only each first-order extension is needed for the quadratic potential. These extensions, with an immaterial sign choice for the horizontal edge, areConsequently $V_2=(k/2)(s_1^2+s_2^2+s_3^2)$, with remainder $O(|q{|}^3)$. If $s=Bq$, then $\mathsf{V}=k{B}^{T}B$. Expanding the squares givesThe small-oscillation mass and stiffness matrices are $\mathsf{T}=m{I}_6$ and the displayed $\mathsf{V}$. Hence

The two common translations give zero-mode vectorsAn infinitesimal counterclockwise rotation sends each equilibrium vertex $r_i$ to $J{r}_i$, where $J(x,y)=(-y,x)$. Therefore a third zero-mode vector isEach vector makes all three $s_j$ vanish, so lies in the kernel of a linear map $B$ and therefore in the kernel of $\mathsf{V}$. The rows of $B$ are independent: the $q_4$ and $q_6$ entries first force the coefficients of its last two rows to vanish, then the first coefficient vanishes. Thus $\mathrm{rank}B=3$, and the rank-nullity theorem proves that these are exactly the three zero modes.

For radial motion takeso that $x_i(t)=(1+a(t))r_i$ to first order. The force on mass $i$ is $-ka{\sum }_{j\ne i}(r_i-r_j)=-3ka{r}_i$, since ${\sum }_i{r}_i=0$. Newton's equation therefore gives $m\ddot{a}=-3ka$. Equivalently $\mathsf{V}{v}_b=3k{v}_b$. Hence the breathing normal mode hasAll three masses move radially outward together, then inward together, preserving the triangular shape. The remaining two normal modes of an equilateral spring triangle are shear deformations with squared frequency $3k/(2m)$; together with the breathing mode and three rigid zero modes they account for all six degrees of freedom.

Fix $i$, write $\alpha =x_{i-1}$, $\beta =x_i$, and let $E=T-S$ on this interval. The Hermite interpolation conditions give bothTwo applications of integration by parts therefore giveOn this single cubic piece, the same calculation also givesHere $S^{(4)}=0$ inside the interval, and all four endpoint values of $E,E^{\prime }$ are zero. Since $T^{\prime \prime }=E^{\prime \prime }+S^{\prime \prime }$, we obtain the exact identityThis is an integral inequality, with no assertion that the integrand on the left has one sign pointwise. The stronger nodal derivative conditions make the boundary terms vanish on each interval separately; part (i) only needed their cancellation after summing.

The stated Hermite interpolating polynomial exists uniquely in ${\Pi }_{2n+1}$. Indeed, a polynomial in the kernel of a linear map of the value-and-derivative data map has a double root of a polynomial at each of $n+1$ distinct interpolation nodes. A nonzero such polynomial would have polynomial degree at least $2n+2$, which is impossible. The map is therefore an injective function and a linear map between vector spaces of the same finite dimension of a vector space, $2n+2$, and hence is a linear isomorphism. No higher smoothness of the original function $f$ is required to construct it.

An explicit illustration uses interpolation nodes $0,1,2$, values $0,1,0$, and zero derivatives at every interpolation node. The clamped cubic spline is $S(x)=3x^2-2x^3$ on $[0,1]$ and its reflection on $[1,2]$. Its Hermite interpolating polynomial is the quartic polynomial $T(x)=x^2(2-x{)}^2$. On $[0,1]$, $T-S=x^2(1-x{)}^2$, and the corresponding reflected formula holds on $[1,2]$. Direct integration gives $J(S)=24$, $J(T)=128/5$, and $J(T-S)=8/5$, exactly as the energy identity predicts.

Any nonzero vector in ${\mathbb{F}}_2^3$ extends to a basis. Sending one such basis to another gives an element of the general linear group, so its group action on the seven nonzero vectors is a transitive group action. Counting successive independent columns givesBy the orbit-stabilizer theorem, the stabilizer $H$ of a chosen nonzero vector $v$ has order $168/7=24$.

There are three two-dimensional linear subspaces containing $v$, corresponding to the three one-dimensional linear subspaces of ${\mathbb{F}}_2^3/⟨v⟩$. The stabilizer $H$ acts transitively on them: extend a basis of either plane beginning with $v$ to a basis of the whole vector space. Thus the stabilizer of a flag $⟨v⟩<P$ has order $24/3=8$.

If $Q\le G$ has order eight, each orbit of a group action of $Q$ on the seven nonzero vectors has size a power of two. Since seven is odd, at least one orbit of a group action has size one. Thus $Q$ fixes a nonzero vector $v$. Its group action on the three planes containing $v$ likewise has a fixed plane $P$. Therefore $Q$ is contained in, and by order equals, the stabilizer of that flag. The group action of $G$ is transitive on these flags, so all such subgroups are conjugate subgroups.

Distinct flags give distinct stabilizers. In a basis adapted to a flag, its order-eight stabilizer consists ofIts only common fixed nonzero vector is $e_1$. Its only invariant plane is $⟨e_1,e_2⟩$: an invariant plane is the kernel of a linear map of a nonzero linear functional, and over ${\mathbb{F}}_2$ that linear functional must be fixed; the displayed matrices have only the fixed nonzero linear functional $e_3^{\ast }$. Hence each subgroup specifies its flag uniquely. There are seven choices of $v$ and three choices of $P$, giving

The tangent space at the identity is the kernel of a linear map from part (i):This says precisely that $JH$ is skew-symmetric. Since $J^{-1}=J$, multiplication by $J$ identifies this tangent space with the three-dimensional space of skew-symmetric $3\times 3$ matrices. In entries it isSubstitution verifies the equation directly and displays its three independent parameters.

A map $g$ is differentiable at $a$ when a linear map $D{g}_a:{\mathbb{R}}^n\to {\mathbb{R}}^k$ approximates its increment to first order:This Fréchet derivative is unique, and its matrix in the standard bases is the Jacobian matrix. Differentiability on a domain means differentiability at each point there.

A standard sufficient condition for a zero locus to be a submanifold is the regular level set theorem: if $g$ is $C^1$ on an open domain and $D{g}_a$ is surjective at every $a\in g^{-1}(0)$, thenFor smooth $g$, the submanifold is smooth. The full-rank condition is the relevant constraint qualification.

Under this regularity hypothesis, its tangent space isIndeed, if a differentiable curve $\gamma$ lies in $M$ and passes through $a$, the chain rule gives $D{g}_a({\gamma }^{\prime }(0))=0$. Thus all tangent velocities lie in the kernel of a linear map. The rank-nullity theorem makes this kernel $(n-k)$-dimensional, the same dimension as the tangent space, proving equality. In coordinates, solve the homogeneous linear equations given by the Jacobian matrix.

The regularity qualification matters: $g(x)=x^2$ has zero set $\{0\}$, a zero-dimensional submanifold, but $D{g}_0=0$. Its kernel is all of $\mathbb{R}$, whereas its tangent space is zero. This is the phenomenon that singular defining maps can overestimate tangent spaces.

The real projective plane is the set of one-dimensional linear subspaces of ${\mathbb{R}}^3$. Equivalently,A projective point is written in homogeneous coordinates $[x:y:z]$. A projective line is the set of projective points whose representatives belong to one fixed two-dimensional linear subspace of ${\mathbb{R}}^3$. Such a linear subspace is the kernel of a linear map of a nonzero linear functional, so its equation has the form $\alpha x+\beta y+\gamma z=0$ with $(\alpha ,\beta ,\gamma )\ne 0$.

Two distinct projective points have linearly independent representatives $p,q$. Their linear span has dimension two and hence defines a projective line containing both. Conversely, any two-dimensional linear subspace containing both representatives contains their entire linear span and therefore equals it. Thus exactly one projective line passes through two distinct projective points.

The Lorenz gauge condition is $\nabla \cdot \mathbf{A}=-c^{-2}{\partial }_t\varphi$. Substitute the electric potential and magnetic vector potential into Gauss's law:Next the Ampère-Maxwell equation, the curl of the curl identity, and the Lorenz gauge condition giveThe gradient terms cancel. Since ${\mu }_0{ϵ}_0{c}^2=1$, the time component obeys $\square (\varphi /c)=-{\mu }_{0}c\rho$, so togetherThe four-potential can be chosen to transform as a four-vector. To make the argument rigorous, choose the retarded potential with no freely added homogeneous solution. If $\square G_{\mathrm{ret}}(x)={\delta }^{(4)}(x)$, this choice isThe retarded Green function is a Lorentz scalar under proper orthochronous Lorentz transformations, and $d^{4}y$ is invariant. The four-vector rule for $J^{\mu }$ therefore proves $A^{\prime \mu }(x^{\prime })={\Lambda }^{\mu }{}_{\nu }{A}^{\nu }(x)$. More generally, transform any homogeneous initial data by this same rule. Both the wave equation and the Lorenz gauge condition are then preserved.

Qualification. The wave equation by itself cannot prove the transformation rule for every independently chosen gauge potential: the d'Alembert operator has a nontrivial kernel of a linear map. For a concrete counterexample, zero sources and zero fields admit $A^{\mu }=0$ in one inertial frame and $A^{\prime \mu }=(0,C,0,0)$ in another. Both satisfy the wave equation and the Lorenz gauge condition, but the latter is not the Lorentz transformation of zero. It is a residual Lorenz gauge symmetry, generated by $\chi =C{x}^{\prime }$. Thus the usual assertion is a covariant choice of four-potential, with a consistent gauge transformation and homogeneous data, rather than a uniqueness result from the displayed equation.

The rank-nullity theorem states that a linear map $T:V\to W$ with finite-dimensional $V$ satisfiesChoose a basis $k_1,\dots ,k_r$ of $\ker T$ and use basis extension to obtain a basis $k_1,\dots ,k_r,u_1,\dots ,u_s$ of $V$. We prove that $T{u}_1,\dots ,T{u}_s$ is a basis of $\mathrm{im}T$.

Every $v\in V$ is a linear combination of the extended basis. Applying $T$ kills the $k_i$ terms, so $Tv$ is a linear combination of $T{u}_1,\dots ,T{u}_s$. This proves the linear span assertion. For linear independence, if ${\sum }_j{b}_{j}T{u}_j=0$, then ${\sum }_j{b}_j{u}_j$ belongs to the kernel of a linear map. Hence ${\sum }_j{b}_j{u}_j={\sum }_i{a}_i{k}_i$ for some scalars $a_i$. The linear independence of the extended basis forces every $a_i$ and $b_j$ to vanish. Consequently $\mathrm{rank}T=s$ and $\mathrm{nullity}T=r$, while $\dim V=r+s$, proving the theorem.

For a linear map $T:V\to W$, its image of a linear map and kernel of a linear map areTheir dimensions define the rank of a linear map and nullity:The image of a linear map measures which output vectors can occur, while the kernel of a linear map measures which input vectors are sent to zero.

Take the linear mapIts kernel of a linear map consists of all $(x,0)$, and its image of a linear map is the same line:Their set intersection is nonzero and their sum is only one-dimensional. In particular, $(0,1)$ cannot be written as a sum of one vector from each of these vector subspaces. The nonzero set intersection also prevents an internal direct sum of vector spaces. ThusThe matrix is a nonzero nilpotent matrix; its failure of diagonalizability explains why the preceding result does not apply.

The image of a linear map and kernel of a linear map areA linear map sends zero to zero, since $T(0)=T(0+0)=T(0)+T(0)$. Thus both displayed sets contain zero. If $w_1=T(v_1)$ and $w_2=T(v_2)$ belong to the image of a linear map, then for any real scalars $a,b$,If $v_1,v_2$ belong to the kernel of a linear map, thenso $a{v}_1+b{v}_2\in \ker T$. Containing zero and being closed under linear combinations is the vector subspace criterion. Hence the image of a linear map is a vector subspace of $W$, and the kernel of a linear map is a vector subspace of $V$.

The rank-nullity theorem says that for a linear map $T:V\to W$ with finite-dimensional domain of a function $V$,Take a basis $k_1,\dots ,k_r$ of the kernel of a linear map and use basis extension to obtain a basis $k_1,\dots ,k_r,u_1,\dots ,u_s$ of $V$. We show that $T(u_1),\dots ,T(u_s)$ is a basis of the image of a linear map.

Every $v\in V$ is a linear combination of the extended basis, and its $k_i$ terms vanish under $T$. Therefore $T(v)$ is a linear combination of the $T(u_j)$, proving the required linear span property. For linear independence, suppose ${\sum }_j{b}_{j}T(u_j)=0$. Then ${\sum }_j{b}_j{u}_j\in \ker T$, so ${\sum }_j{b}_j{u}_j={\sum }_i{a}_i{k}_i$ for some real scalars $a_i$. The extended basis is linearly independent, forcing every $a_i$ and $b_j$ to be zero. Thus the image of a linear map has dimension $s$, the kernel of a linear map has dimension $r$, and $\dim V=r+s$ proves the result.

Here $v_3=2v_1-v_2$, so the image of a linear map is the linear span of $v_1,v_2$. These two vectors are linearly independent: if $a{v}_1+b{v}_2=0$, the $e_1$ and $e_2$ coordinates give $a+b=0$ and $a-b=0$, hence $a=b=0$. Consequently the rank of a linear map is two. By the rank-nullity theorem, the nullity is $3-2=1$:More explicitly, the relation $-2v_1+v_2+v_3=0$ gives the kernel of a linear map

For an $m\times n$ matrix $A$ over the real numbers, write its rows as $r_1,\dots ,r_m$ and its columns as $c_1,\dots ,c_n$. ThenThus the row space and column space consist of all linear combinations of the corresponding rows and columns; the null space is the kernel of a linear map represented by $A$. The matrix rank counts independent row or column directions, with row rank equal to column rank.

The independent-column representation gives $T(x)=c_1(x_1+3x_3)+c_2(x_2+2x_4)$. Thus a vector lies in the kernel of a linear map represented by this matrix precisely when $x_1=-3x_3$ and $x_2=-2x_4$. Writing the two free coordinates separately givesThe two displayed vectors are linearly independent, so they form a basis of the kernel. Its dimension two also agrees with the rank-nullity theorem, $4=2+2$.

The system is consistent for every $b\in {\mathbb{R}}^m$: take $x=Yb$, since $A(Yb)=(AY)b=b$. Thus $Y$ is a right inverse giving an explicit solution for every right-hand side.

All solutions are $Yb+u$ with $u$ in the kernel of a linear map represented by $A$. The rank-nullity theorem gives kernel dimension $n-m$; the solution is unique when $n=m$, and is nonunique when $n>m$.

True. The rank-nullity theorem gives $\dim \mathcal{N}(A)=n-r=0$. The only vector in this zero-dimensional kernel of a linear map is the zero vector.