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The rank-nullity theorem for a linear map $T:V\to W$, with $V$ a finite-dimensional vector space over a field $F$, isHere the nullity is the dimension of the kernel of a linear map, and the rank of a linear map is the dimension of its image of a linear map.

The transpose is a linear map, so $T(A)=A+A^T$ is a linear map on the $n^2$-dimensional vector space $M_{n\times n}(F)$. Let $E_{ij}$ denote a matrix unit. The answer depends on the characteristic of a field.

If $\mathrm{char}F\ne 2$, then $T(A)=0$ means $A^T=-A$. Such skew-symmetric matrices have zero diagonal, and their entries above the diagonal can be chosen freely. A basis of the kernel of a linear map is $E_{ij}-E_{ji}$ for $i<j$, giving nullity $n(n-1)/2$. The image of a linear map consists of all symmetric matrices: $T(A)$ is symmetric, and every symmetric matrix $S$ is $T(S/2)$. Its basis comprises $E_{ii}$ and $E_{ij}+E_{ji}$ for $i<j$, giving rank of a linear map $n(n+1)/2$.

If $\mathrm{char}F=2$, then $-A=A$, so $\ker T$ consists of all symmetric matrices, with dimension $n(n+1)/2$. Every $T(A)$ is symmetric with zero diagonal, and every symmetric zero-diagonal matrix $S$ arises by taking $A_{ij}=S_{ij}$ for $i<j$ and all other entries zero. The image of a linear map therefore has basis $E_{ij}+E_{ji}$ for $i<j$, and dimension $n(n-1)/2$. This is the symmetrization map in characteristic two.

In both cases the rank-nullity theorem is verified explicitly:

Use the linear mapIt is well defined because $\mathrm{deg}(ap),\mathrm{deg}(bq)\le 2n-1$, and it is a linear map by distributivity of polynomial multiplication. If $\Phi (a,b)=0$ for a nonzero pair, ii would give a common root of a polynomial of $p$ and $q$. The hypothesis therefore implies $\ker \Phi =\{(0,0)\}$, so $\Phi$ is injective.

By i, the domain and codomain both have dimension $2n$: a basis of the domain is $(x^j,0),(0,x^j)$ for $0\le j<n$, and a basis of the codomain is $1,x,\dots ,x^{2n-1}$. The stated finite-dimensional result makes $\Phi$ surjective. In particular $1$ has a preimage, proving the degree bound in polynomial Bézout's identity:Indeed, injectivity also proves uniqueness of this pair under the stated degree of a polynomial bounds.

Every $v\in V$ has a unique basis expansion $v={\sum }_{i=1}^n{a}_i{v}_i$. Define the linear functionalUniqueness of the coordinates makes this definition well defined. The coordinates of a sum or a scalar multiple are the corresponding sums or scalar multiples of the coordinates, so this formula defines a linear map. It has $T(v_i)={\mu }_i$. Conversely, any linear map with those values must satisfy the same formula by linearity, proving uniqueness.

If the finite field $\mathbb{F}$ has $q$ elements, each of the $n$ values ${\mu }_i$ can be chosen independently in $q$ ways. The number of linear maps $V\to \mathbb{F}$ is therefore $\boxed{q^n}$, including the zero linear map.

A map $T:V\to W$ between vector spaces over the same field $\mathbb{F}$ is a linear map precisely when it preserves vector addition and scalar multiplication:Equivalently, $T(au+bv)=aT(u)+bT(v)$ for all $u,v\in V$ and $a,b\in \mathbb{F}$.

Use basis extension to choose a basis $(v,v_2,\dots ,v_n)$ of the vector space $V$ over the finite field ${\mathbb{F}}_p={\mathbb{Z}}_p$. The condition $T(v)=0$ fixes the first value of a linear functional; its remaining $n-1$ values are arbitrary. There are $p^{n-1}$ such linear maps, exactly one of which is zero. Hence the required number isFor $n=1$ this is zero, as it should be: vanishing at a basis vector then forces the entire linear functional to vanish.

For the linear map $T_A:x\mapsto Ax$ on the finite-dimensional vector space ${\mathbb{R}}^n$, the rank-nullity theorem implies that injectivity and surjectivity are equivalent. Consequently $T_A$ is invertible if and only if $\ker T_A=\{0\}$.

Using the permitted equivalence between invertibility and a nonzero determinant, we obtain

Write $B=[b_1,\dots ,b_n]$. Since $AB=[A{b}_1,\dots ,A{b}_n]$ and $b\mapsto Ab$ is a linear map, the multilinearity of the determinant gives, for any column position $j$,If $b_i=b_j$ for $i\ne j$, the corresponding columns $A{b}_i,A{b}_j$ also agree, so $D(B)=0$. Thus $D$ is an alternating multilinear map of the columns of $B$, even when $A$ is not invertible.

Write $r=(r_1,\dots ,r_n{)}^T$. The linear map represented by $A$ has the formThe image of a linear map here is contained in the one-dimensional linear span of $e_i$, so its matrix rank is at most one. The rank-nullity theorem yieldsThus the solution space of $Ax=0$, namely the zero eigenspace when zero is an eigenvalue, contains at least $n-1$ linearly independent vectors. If $r=0$ its dimension is $n$; otherwise it is $n-1$. When $n=1$ the asserted lower bound is zero, even if zero is not an eigenvalue.

The two-by-two matrix inverse formula givesMultiplication in the reversed order givesThe order reversal is needed because applying the inverse undoes the original two linear maps in reverse sequence.

The eigenvector equations give $AB=B\mathrm{diag}(4,-2,-2)$. Multiplying on the left by $B^{-1}$ therefore givesThis is an orthogonal diagonalization: the linear map acts independently by multiplication by its eigenvalues in the new orthonormal basis. The repeated eigenvalue explains the freedom to rotate the last two basis directions.

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.

An eigenvalue equation has the form $A\mathbf{a}=\lambda \mathbf{a}$, where $\mathbf{a}\ne 0$ is an eigenvector and $\lambda$ its eigenvalue. It means that the linear map preserves the vector's line and acts there by multiplication by $\lambda$. Nonzero solutions occur precisely when $\det (A-\lambda I)=0$; the zero vector is excluded because it would satisfy the equation for every $\lambda$.

The symmetric group $S_n$ consists of all bijections of $\{1,\dots ,n\}$, with composition as its operation; its group order is $n!$.

An $m$-dimensional complex group representation of $G$ is a group homomorphism $\rho :G\to GL(V)$, where $V$ is an $m$-dimensional complex vector space. After a choice of basis, each $\rho (g)$ is an invertible $m\times m$ complex matrix, with $\rho (gh)=\rho (g)\rho (h)$.

An irreducible representation has $V\ne 0$ and no invariant vector subspace other than $\{0\}$ and $V$. Two group representations $\rho$ on $V$ and $\sigma$ on $W$ are isomorphic representations if there exists an invertible linear map $S:V\to W$ such that $S\rho (g)=\sigma (g)S$ for every $g$. They are inequivalent if no such map exists; in equal-dimensional matrix form this says no fixed change of basis conjugates every $\rho (g)$ to $\sigma (g)$.

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.

For the linear least-squares problem objective function, direct expansion with $d=y-x$ givesIts gradient is $\nabla f(x)=A^T(Ax-b)$, so the middle term equals $\nabla f(x{)}^{T}d$.

Use the supplied thin singular value decomposition $A=US{V}^T$. Since $U^{T}U=I_n$, the linear map $w\mapsto Uw$ preserves the Euclidean norm. The square matrix $V$ is an orthogonal matrix, so $\Vert V^{T}d\Vert =\Vert d\Vert$. ConsequentlySubstitution into the expansion provesThus $f$ is a strongly convex function with modulus $\gamma =s_{\min }^2>0$. This modulus is sharp: choose $d=V{e}_i$ for an index attaining the smallest singular value. The Hessian matrix is $A^{T}A=V{S}^2{V}^T$, so the same conclusion says that its smallest eigenvalue is $s_{\min }^2$.

The total derivative is the unique linear map $D{f}_a:{\mathbb{R}}^m\to {\mathbb{R}}^n$ satisfyingThis describes the best first-order linear approximation without choosing coordinates. The chain rule isThe evaluation point $f(a)$ in the outer derivative matters, particularly when composing a quadratic matrix map with the exponential.

The total derivative is a linear map $Df(a):{\mathbb{R}}^n\to {\mathbb{R}}^n$ such thatIt is unique: the difference of two such linear maps, evaluated on $h=tv$, must vanish on every fixed vector $v$ after division by $t$ and passage to the limit.

The three nonzero vectors in ${\mathbb{F}}_2^2$ are $e_1$, $e_2$ and $e_1+e_2$. An invertible linear map permutes this three-element set, so part (b) gives a group homomorphismIf a matrix acts trivially on all three vectors, it fixes both basis vectors $e_1,e_2$, so it is the identity matrix. Thus the kernel of a group homomorphism is trivial and $\rho$ is injective.

To count the domain, its first column may be any of the three nonzero vectors. The second column must be outside the two-element span of the first, leaving two choices. HenceAn injection between these finite sets of equal size is surjective, proving

Let $L(x,y,z)=(x-y,y-z,z-x)$. This is a linear map, hence smooth; composing it with the given $C^1$ function proves that $g=f\circ L$ is continuously differentiable by the chain rule. Write $f_1,f_2,f_3$ for the first partial derivatives, evaluated at $L(p)$. ThenAdding these identities givesGeometrically, $L(p+t(1,1,1))=L(p)$, so $g$ is constant along every line in the diagonal direction; its derivative in that direction must vanish.