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Eigenvectors belonging to distinct eigenvalues are linearly independent. Under the hypothesis they therefore form a basis $v_1,\dots ,v_n$. For every $j$,because ${\chi }_A({\lambda }_j)=0$. A linear map that vanishes on a basis is zero, henceThis proves the Cayley-Hamilton theorem in the distinct-eigenvalue case.

A basis of a vector space $V$ is a family of vectors that is both linearly independent and a spanning set of $V$.

Write the given finite basis as $\mathcal{B}=\{v_1,\dots ,v_n\}$. First, any other basis ${\mathcal{B}}^{\prime }$ must also be finite. Each $v_i$ is a finite linear combination of elements of ${\mathcal{B}}^{\prime }$, so the union $S$ of the finitely many elements of ${\mathcal{B}}^{\prime }$ occurring in these $n$ expressions is finite. Since $S$ spans every $v_i$, it spans $V$. If some $w\in {\mathcal{B}}^{\prime }\setminus S$ existed, then $w$ would lie in the span of $S$, contradicting the linear independence of ${\mathcal{B}}^{\prime }$. Hence ${\mathcal{B}}^{\prime }=S$; write $|{\mathcal{B}}^{\prime }|=m$.

It remains to prove the elementary Steinitz exchange lemma directly. If independent vectors $w_1,\dots ,w_r$ lie in the span of $v_1,\dots ,v_n$, express $w_1$ in terms of the $v_i$. Some coefficient is nonzero, so the corresponding $v_i$ can be solved for in terms of $w_1$ and the other $v_i$; replacing it by $w_1$ preserves the span. Inductively, after replacing $k$ of the $v_i$ by $w_1,\dots ,w_k$, the expression for $w_{k+1}$ must have a nonzero coefficient on one of the unreplaced $v_i$, since otherwise $w_{k+1}$ would be a linear combination of $w_1,\dots ,w_k$. That $v_i$ can again be replaced. There are only $n$ original vectors to replace, so $r\le n$.

Apply this argument first to the independent family ${\mathcal{B}}^{\prime }$ and the spanning family $\mathcal{B}$ to obtain $m\le n$, and then with the two bases interchanged to obtain $n\le m$. Therefore

Let $r=dim(U\cap W)$. Choose a basis $e_1,\dots ,e_r$ of the intersection of vector subspaces, extend it to a basis $e_1,\dots ,e_r,u_{r+1},\dots ,u_m$ of $U$, and independently extend it to a basis $e_1,\dots ,e_r,w_{r+1},\dots ,w_n$ of $W$. The combined listspans the sum of vector subspaces $U+W$. It is also linearly independent: a vanishing linear combination says that a combination of the $u_i$ belongs to $U\cap W$, and its expression in the chosen basis of $U$ forces all its coefficients to vanish; the coefficients of the $w_i$ then vanish in the same way. This proves the dimension formula for a sum of subspaces

Parametrize the two linear subspaces asPutting a vector of $U$ in $W$ first forces its fourth coordinate to vanish; the remaining defining equation is then automatic. Thusand the dimension formula gives $\dim (U+W)=2+2-1=3$.

The linear functionalvanishes on each displayed generator of $U$ and $W$. Its kernel is three-dimensional because $\ell \ne 0$, so the inclusion $U+W\subseteq \ker \ell$ between two three-dimensional subspaces is equality.

The rank of a matrix is the dimension of its column space, equivalently the dimension of the image of a linear map represented by the matrix. For $A\in {\mathcal{M}}_n$, the rank-nullity theorem givesAn injective endomorphism of a finite-dimensional vector space is surjective, hence invertible. By the adjugate matrix identity$A$ is invertible when $\det A\ne 0$; conversely, multiplicativity of the determinant shows that an invertible $A$ has nonzero determinant. Therefore

Let $E_{ij}$ be the matrix unit with its only nonzero entry at $(i,j)$. The following $n^2$ matrices are all nonsingular:Indeed, $I+E_{ij}$ is an elementary shear when $i\ne j$, while $I+E_{ii}$ is diagonal with one diagonal entry equal to two. Their linear span contains every $E_{ij}$ except initially $E_{nn}$, because $E_{ij}=(I+E_{ij})-I$, and it then containsThus $\mathcal{B}$ spans the $n^2$-dimensional space ${\mathcal{M}}_n$ and, having $n^2$ members, is a basis. This also covers $n=1$, when $\mathcal{B}=\{I\}$.

Now let $A$ be a nonsingular zero-one matrix. If it had fewer than $n-1$ zero entries, at least two rows would contain no zero at all. Those two rows would both be the all-one row, contradicting linear independence. Hence every such matrix has at mostones. The bound is attained. Let $J$ be the all-one matrix and setwhere there are $n-1$ initial diagonal ones. If $A\mathbf{x}=0$ and $S={\sum }_j{x}_j$, the first $n-1$ row equations give $x_i=S$, while the last gives $S=0$. Hence every $x_i=0$, so $A$ is nonsingular and has exactly $n^2-n+1$ ones. Therefore

Supposethroughout the punctured neighbourhood, and let $k$ be the least index with $c_k\ne 0$. Part (i) implies ${\varphi }_j=o({\varphi }_k)$ for every $j>k$. Dividing by ${\varphi }_k$ and taking $x\to x_0$ therefore givesa contradiction. Hence every $c_j$ vanishes and ${\varphi }_0,\dots ,{\varphi }_N$ are linearly independent.

We use induction on $d$. SupposeApplying $f-{\alpha }_{d}I$ givesBy induction, every $c_i({\alpha }_i-{\alpha }_d)=0$ for $i<d$. The eigenvalues are distinct, so $c_i=0$ for $i<d$, and the original relation then gives $c_d=0$. Thus eigenvectors with distinct eigenvalues are linearly independent.

A vector $x$ is a basic feasible solution whenand the columns $A_i$ for which $x_i>0$ are linearly independent. Equivalently, after extending those columns to a basis of the column space of $A$, one sets all nonbasic variables to zero and solves for the basic variables, obtaining a nonnegative vector.

Substitution ofand integration by parts giveThus takeand require the endpoint term to vanish. Infinite contour ends must lie in sectors where $\mathrm{Re}(t^4)>0$, namely sectors centred on the positive and negative real and imaginary axes, each of angular width $\pi /4$.

Let the common final segment be the positive real ray from $0$ to $+\infty$, and choose the initial rays from $-\infty$, $+i\infty$, and $-i\infty$ to $0$. Call the resulting contours ${\gamma }_1,{\gamma }_2,{\gamma }_3$. PutSince $w_i^{(r)}(0)={\int }_{{\gamma }_i}{t}^r{e}^{-t^4/4}dt$, for $r=0,1,2$ the three rows areConsequentlyUsing the Gamma function reflection identity,The initial-data vectors are therefore linearly independent, so the three contour integrals are linearly independent solutions.

Part (b) gives paths with the same endpoint $z$ but integral valuesApplying the local inverse and then continuing it shows thatSince $K,L>0$, the first period is real and the second is purely imaginary, so they are linearly independent over $\mathbb{R}$. Hence $\mathcal{H}$ is an elliptic function with two periodsIndeed, it is the Jacobi elliptic sine, and these generators give its period lattice.