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The dual space isthe vector space of linear functionals on $V$. If $e_1,\dots ,e_n$ is a basis of a finite-dimensional $V$, define $e_i^{\ast }(e_j)={\delta }_{ij}$. Every $f\in V^{\ast }$ has the unique expansionso the dual basis $e_1^{\ast },\dots ,e_n^{\ast }$ proves

For $U\le V$, its annihilator of a vector subspace isIf $u_1,\dots ,u_k$ is a basis of $U$ and is extended to a basis $u_1,\dots ,u_n$ of $V$, thenConsequentlyIf $U\ne V$, this dimension is positive, giving a nonzero functional that vanishes on $U$.

For a linear map $\alpha :V\to W$, the dual map isNowsoEvery $g\circ \alpha$ vanishes on $\ker \alpha$, henceThe two spaces have the same dimension, sinceThus

Let $Q:V\to V/U$ be the quotient map. Since $Q$ is onto, $Q^{\ast }$ is injective, and the preceding identity givesThereforeFor the inclusion $J:U\hookrightarrow V$, the map $J^{\ast }:V^{\ast }\to U^{\ast }$ is restriction to $U$. It is onto and has kernelThe first isomorphism theorem now gives the other duals of a subspace and its quotient:

Fix $y\in V$. The map $x\mapsto \theta (x,y)$ is a linear functional, so the finite-dimensional Riesz representation theorem gives a unique vector $\beta (y)$ such thatFor scalars $\lambda ,\mu$,Uniqueness givesso $\beta$ is linear.

For $\alpha \in \mathrm{End}(V)$, apply this result toIt produces a unique linear map ${\alpha }^{\ast }$ satisfyingwhich proves existence and uniqueness of the adjoint.

Differentiability at $x\in {\mathbb{R}}^2$ means that there is a linear map $Df{|}_x:{\mathbb{R}}^2\to \mathbb{R}$ such thatThe partial derivatives are $D_{i}f(x)=Df{|}_x(e_i)$ when the derivative exists; independently, they may be defined by the corresponding one-variable difference quotients.

Suppose the partial derivatives exist near $x=(x_1,x_2)$ and are continuous at $x$. Apply the one-dimensional mean value theorem along the two coordinate segments from $x$ to $x+h$. It gives intermediate points ${\xi }_h,{\eta }_h\to x$ such thatContinuity of the partials makes the remainder after subtractingequal to $o(\Vert h\Vert )$. Hence $f$ is differentiable and this is its derivative.

For the given function, writing $r=\sqrt{x^2+y^2}$, at $r>0$ we haveAt the origin both partial derivatives are zero, since $f(h,0)/h$ and $f(0,h)/h$ tend to zero. Neither partial is continuous there: along its corresponding coordinate axis the cosine term oscillates without a limit. Neverthelessso $f$ is differentiable at the origin with derivative zero. This illustrates that existence of partial derivatives does not imply their continuity.

The final assertion is false. DefineAgain $|g|\le r^2$, so $g$ is differentiable at the origin and it is smooth elsewhere. But along the $x$-axis,is unbounded near zero, and similarly $D_{2}g(0,y)$ is unbounded. Thus both partial derivatives can be unbounded in every neighbourhood of a point even when the function is differentiable everywhere.

Two matrices are equivalent when $A^{\prime }=PAQ$ for invertible matrices $P$ and $Q$ of the appropriate sizes. A change of basis in the codomain multiplies a matrix on the left, while a change of basis in the domain multiplies it on the right. Hence two matrices of the same linear map in two pairs of bases are equivalent.

Conversely, start with the map $\alpha :F^n\to F^m$ having matrix $A$ in the standard bases. Given $A^{\prime }=PAQ$, choose domain and codomain bases whose change-of-coordinate matrices produce $Q$ and $P$; then $A^{\prime }$ represents the same map in those bases. This proves the equivalence.

The column rank of $A$ is the dimension of the span of its columns, and its row rank is the dimension of the span of its rows. If $A$ represents $\alpha :F^n\to F^m$, its column rank is $\mathrm{rank}\alpha$. The transpose represents the dual mapso the row rank is $\mathrm{rank}{\alpha }^{\ast }$.

If $\mathrm{rank}\alpha =r$, thenwhose dimension is $m-r$: extend a basis of $\mathrm{im}\alpha$ to one of $F^m$ and use the dual basis. Rank-nullity now gives $\mathrm{rank}{\alpha }^{\ast }=r$. This proves the equality of row rank and column rank.

Let $P$ be free with basis $(e_i{)}_{i\in I}$. Given a surjection $f:M\to N$ and a map $g:P\to N$, choose $m_i\in M$ satisfying $f(m_i)=g(e_i)$. There is a unique linear map $h:P\to M$ with $h(e_i)=m_i$, and on every basis vectorThus $f\circ h=g$, proving that free modules are projective.

Let $\Phi :M_d(\mathbb{C})\to M_D(\mathbb{C})$ be a linear map. It is positive when $X\ge 0$ implies $\Phi (X)\ge 0$, and completely positive whenis positive for every ancillary dimension $r$. In finite dimensions it is enough to check $r=d$.

A finite family of Kraus operators $A^{\alpha }:{\mathbb{C}}^d\to {\mathbb{C}}^D$ defines the Kraus representationThis map is completely positive because, for every positive semidefinite matrix $Y$ on the enlarged space,

Conversely, use the unnormalized maximally entangled vector $|\Omega ⟩={\sum }_{j=1}^d|j⟩|j⟩$. Complete positivity makes the Choi matrixpositive semidefinite. By the spectral theorem for normal operators, $C_{\Phi }={\sum }_{\alpha =1}^R|v_{\alpha }⟩⟨v_{\alpha }|$, where $R=\mathrm{rank}{C}_{\Phi }\le dD$. Reshape each $v_{\alpha }\in {\mathbb{C}}^D\otimes {\mathbb{C}}^d$ into a matrix $A^{\alpha }$ by $|v_{\alpha }⟩={\sum }_{\mu j}{A}_{\mu j}^{\alpha }|\mu ⟩|j⟩$. The Choi matrix inversion formulathen gives $\Phi (X)={\sum }_{\alpha }{A}^{\alpha }X(A^{\alpha }{)}^{†}$. Thus finite Kraus representations characterize finite-dimensional completely positive maps.

The additional normalizationmakes $\Phi$ trace preserving and hence a quantum channel; ${\sum }_{\alpha }{A}^{\alpha }(A^{\alpha }{)}^{†}=I_D$ instead makes it unital.

For a linear map $T:V\to W$ with $V$ finite-dimensional,Nowwhich is the required inequality after rank-nullity for $\alpha$.

If $X$ and $Y$ represent the same map, then $Y=C^{-1}XB$, where $B$ changes new domain coordinates to old ones and $C$ does the same in the codomain.

Invertible block row and column operations reduceto $\mathrm{diag}(P,S-R{P}^{-1}Q)$, proving the rank formula. Apply it to $\left(\begin{array}{cc}{I}_n& Q\\ R& I_m\end{array}\right)$ first with the upper-left block and then with the lower-right block. Equating the results gives

A map $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is differentiable at $x$ if there is a linear map $L$ such thatthen $Df{|}_x=L$. The inverse function theorem says that if $f$ is continuously differentiable near $x$ and $Df{|}_x$ is invertible, then $f$ restricts to a $C^1$ diffeomorphism between neighborhoods of $x$ and $f(x)$.

Since $F(A)=A^{T}A$ is polynomial,Thus $\ker DF{|}_I=T$, the space of skew-symmetric matrices.

DefineThen $D\Phi {|}_I(H)=2H$, so the inverse function theorem supplies open neighborhoods $I\in U$, $0\in V$ on which $\Phi :U\to V$ is a $C^1$ diffeomorphism. Its symmetric part is $A^{T}A-I$ and its skew part is $A-A^T$. Consequentlyand therefore $\Phi (U\cap \mathcal{O})=V\cap T$.

For any $R\in \mathcal{O}$, left multiplication $A\mapsto R^{T}A$ is a homeomorphism preserving $\mathcal{O}$ and carrying $R$ to $I$. Transporting the preceding chart gives a neighborhood of $R$ in $\mathcal{O}$ homeomorphic to an open subset of $T$.

The dual space $V^{\ast }$ is the vector space of linear maps $V\to \mathbb{F}$. If $B=(v_1,\dots ,v_n)$, its dual family satisfies $v_i^{\ast }(v_j)={\delta }_{ij}$. Every functional $f$ obeysso the family spans; evaluation on each $v_j$ proves linear independence. Thus it is a basis without any prior dimension argument.

For $p=a+bt+c{t}^2$,The basis dual to $(f_0,f_1,f_2)$ isas direct substitution gives $f_i(p_j)={\delta }_{ij}$. Hence the three functionals form a basis of $P_2^{\ast }$.

The dual $X^{\ast }$ is the vector space of bounded linear maps $f:X\to \mathbb{F}$, with normIf $(f_n)$ is Cauchy in this norm, then $(f_n(x))$ is Cauchy in the complete scalar field for every $x$. Define $f(x)={\lim }_n{f}_n(x)$. Pointwise passage to the limit preserves linearity. Given $\epsilon >0$, choose $N$ such that $\Vert f_n-f_m\Vert <\epsilon$ for $m,n\ge N$; sending $m\to \infty$ givesThus $f$ is bounded and $\Vert f_n-f\Vert \le \epsilon$. Hence $X^{\ast }$ is Banach, even if $X$ was only normed.

Yes. For $x=(x_n)\in c$, let $L={\lim }_n{x}_n$ and defineThe tail tends to zero, so $T(x)\in c_0$, and $\Vert T(x){\Vert }_{\infty }\le 2\Vert x{\Vert }_{\infty }$. Conversely, for $y=(y_n)\in c_0$, defineThis sequence converges to $y_1$ and has norm at most $2\Vert y{\Vert }_{\infty }$. The formulas are inverse bounded linear maps, so $c$ and $c_0$ are isomorphic Banach spaces.

Differentiability at $p$ means $f(p+h)=f(p)+Lh+o(|h|)$ for a linear map $L=Df(p)$; partial derivatives are its values on coordinate vectors. Integrating the partial derivatives along the two coordinate segments and using continuity at $p$ proves the stated sufficient condition. Here $f(x,y)=g(x)+g(y)$ with $g(t)=t^2\sin (1/t)$ and $g(0)=0$. Since $g\prime (0)=0$, $f$ is differentiable everywhere. But $g\prime (t)=2t\sin (1/t)-\cos (1/t)$ does not approach $0$ at zero, so $f$ is $C^1$ exactly where $xy\ne 0$.

The open mapping lemma says: if a bounded linear map between Banach spaces satisfies $\overline{T(B_X)}\supset r{B}_Y$, then $T(B_X)$ contains a (possibly smaller) ball. Given $y$, repeatedly choose corrections $x_n$ whose residuals shrink geometrically; completeness makes $\sum x_n$ converge and maps it to $y$. Baire category applied to $Y={\bigcup }_{n}T(n{B}_X)$ supplies the closure hypothesis for a surjective map. This proves the Open mapping theorem.

For a linear map $T:V\to V$, its kernel of a linear map and image of a linear map areIts nullity of a linear map is $n(T)=\dim \ker T$, and its rank of a linear map is $r(T)=\dim \mathrm{im}T$.

Choose a basis $u_1,\dots ,u_k$ of $\ker T$ and extend it to a basis $u_1,\dots ,u_k,v_1,\dots ,v_m$ of $V$. The vectors $T(v_1),\dots ,T(v_m)$ span $\mathrm{im}T$, because applying $T$ to an arbitrary linear combination of the full basis eliminates the $u_i$ terms. They are also linearly independent: if ${\sum }_j{b}_{j}T(v_j)=0$, then ${\sum }_j{b}_j{v}_j\in \ker T$, so ${\sum }_j{b}_j{v}_j={\sum }_i{a}_i{u}_i$ for some coefficients. Independence of the extended basis forces all $a_i,b_j$ to vanish.

Consequently $r(T)=m$, $n(T)=k$ and $\dim V=k+m$. This proves the rank-nullity theorem:

For the linear map $T:V\to W$, its image of a linear map and kernel of a linear map areBoth contain the zero vector, since $T(0)=0$. If $y_1=T(v_1)$ and $y_2=T(v_2)$ belong to the image of a linear map, then for any real numbers $a,b$,Thus $\mathrm{im}T$ is a vector subspace of $W$. If $u_1,u_2\in \ker T$, thenso $\ker T$ is a vector subspace of $V$.

The necessary and sufficient condition isFor necessity, if $ST={\mathrm{id}}_V$ and $v\in \ker T$, then $v=S(Tv)=S(0)=0$.

For sufficiency, let $e_1,e_2$ be a basis of $V$ and suppose the kernel of a linear map is trivial. The vectors $T{e}_1,T{e}_2$ are linearly independent, since $aT{e}_1+bT{e}_2=0$ implies $a{e}_1+b{e}_2\in \ker T$. Extend them to a basis $T{e}_1,T{e}_2,w_3$ of $W$. Define a linear map $S$ byThen $ST$ agrees with the identity operator on a basis, hence on all of $V$. Thus a left inverse exists precisely when $T$ is injective; this is the linear left inverse criterion.

The prescribed form is a sum of products of real polynomial evaluations. It is a symmetric bilinear form, since each evaluation is a linear map. AlsoEquality forces $f(-1)=f(0)=f(1)=0$. A nonzero polynomial of degree of a polynomial at most two cannot have three distinct roots of a polynomial, by the root bound for a polynomial. Thus equality holds only for the zero polynomial, establishing positive definiteness. Therefore the form is a real inner product, and $V$ is an inner product space. This is a discrete sampling inner product on polynomial spaces.

The definition of an eigenvalue and the kernel criterion for a singular square matrix giveFor the second equivalence, a linear map from a finite-dimensional vector space to itself has zero kernel exactly when its rank of a linear map is the full dimension, by the rank-nullity theorem, and therefore exactly when it is invertible. This proves both directions.

For any matrix $Y$ and compatible invertible matrix $P$, the linear map $r\mapsto rP$ maps $\mathrm{rowspace}(Y)$ onto $\mathrm{rowspace}(YP)$: a linear combination of rows of $Y$ is sent to the same linear combination of rows of $YP$. Its inverse function is $s\mapsto s{P}^{-1}$, so these row spaces have equal dimensions. Hence right matrix multiplication by $P$ also preserves row rank.

If $X=P^{-1}YP$, ii now givesThis proves the required matrix similarity invariant without needing to assume equality of row rank and column rank.

Define the linear functional $L:F^n\to F$ by $L(x)={\sum }_{i=1}^n{a}_i{x}_i$. The given hyperplane is $H=\ker L$. Choose $j$ with $a_j\ne 0$. For any $t\in F$, the vector whose only nonzero coordinate is $x_j=t/a_j$ satisfies $L(x)=t$. Hence $L$ is surjective and has rank of a linear map equal to ${\dim }_{F}F=1$.

Applying the rank-nullity theorem to this particular linear map givesThe nonzero-coefficient hypothesis is essential: if all $a_i$ were zero, the kernel of a linear map would instead be all of $F^n$.