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Orient the axis towards $n=(3/5,4/5,0)$ and measure positive angles by the right-hand rule about $n$. The vectors $n$, $m=(-4/5,3/5,0)$ and $e_3=(0,0,1)$ form a positively oriented orthonormal basis, since $n\times m=e_3$. Its change-of-basis matrix isRelative to this basis, the rotation fixes the first coordinate and rotates the last two. Writing $c=\cos \alpha$ and $s=\sin \alpha$, its matrix in the standard basis is thereforeIn particular $Rn=n$, $Rm=cm+s{e}_3$, and $R{e}_3=-sm+c{e}_3$, verifying the axis and the direction of rotation. Reversing the chosen orientation of the axis replaces $\alpha$ by $-\alpha$.

In an oriented orthonormal coordinate system, the dot product and cross product areThe scalar triple product is $a\cdot (b\times c)$; equivalently it is the determinant of the matrix whose columns are $a,b,c$. The three vectors have linear independence when

Let $U=(a_1{a}_2{a}_3)$ and $V=(b_1{b}_2{b}_3)$ be column matrices. Since $(U^{T}V{)}_{ij}=a_i\cdot b_j$, we have $S=U^{T}V$. Multiplicativity of the determinant and its invariance under matrix transpose yieldA square matrix has maximal matrix rank exactly when its determinant is nonzero, equivalently when its columns have linear independence. A product of two real numbers is nonzero exactly when both factors are nonzero. Hence $\boxed{\mathrm{rank}S=3⟺(a_i{)}_{i=1}^3\text{ and }(b_j{)}_{j=1}^3\text{ are both independent}}$.

The same argument works in every dimension. Set $U_n=(c_1\cdots c_n)$ and $V_n=(d_1\cdots d_n)$. Thenso $\boxed{\mathrm{rank}T=n⟺(c_i{)}_{i=1}^n\text{ and }(d_j{)}_{j=1}^n\text{ are both independent}}$.

For the last construction, choose two standard basis vectors $e_1,e_2$ in ${\mathbb{R}}^n$ and take $c_j=e_1+j{e}_2$ for $j=1,\dots ,n$. If $\alpha c_i+\beta c_j=0$ with $i\ne j$, its first two coordinates give $\alpha +\beta =0$ and $i\alpha +j\beta =0$, hence $\alpha =\beta =0$. Every pair therefore has linear independence. Every triple lies in the two-dimensional linear span of $e_1,e_2$ and has linear dependence. More explicitly, for distinct $i,j,k$,a nontrivial dependence. Thus $\boxed{c_j=e_1+j{e}_2(1\le j\le n)\text{ works for every }n>2}$.

If $S$ spans and has $n$ elements but is linearly dependent, a nontrivial linear combination equal to zero lets us remove one member without changing the span. This produces $n-1$ spanning vectors, whereas the standard basis contains $n$ linearly independent vectors, contradicting the Steinitz exchange lemma. Therefore $\boxed{S\text{ is linearly independent}}$. Together the three deductions establish that any two of the three properties imply the third. For $n=0$, the only independent or zero-element family is the empty basis, and the same conclusions hold.

Relative to the given basis $e_1,e_2$, the proposed vectors are the columns ofThe determinant is nonzero exactly when the columns form a basis. Thus $\boxed{\lambda \ne \pm 1}$. At $\lambda =1$ the vectors coincide, and at $\lambda =-1$ they are negatives of one another.

If $S$ is linearly independent and has $n$ elements, apply the Steinitz exchange lemma with $S$ as the independent family and the standard basis as the spanning family. All $n$ members of the standard basis are replaced, leaving $S$ itself spanning. Thus $\boxed{\mathrm{span}S={\mathbb{R}}^n}$, so $S$ is a basis.

The Steinitz exchange lemma says that if $w_1,\dots ,w_m$ span a vector space and $v_1,\dots ,v_r$ are linearly independent, then $r\le m$, and after reordering the $w_j$, the family $v_1,\dots ,v_r,w_{r+1},\dots ,w_m$ still spans. Here is a proof. Suppose the first $k-1$ replacements have been made. Express $v_k$ in that spanning family. Some coefficient of a remaining $w_j$ must be nonzero, since otherwise $v_k$ belongs to the span of $v_1,\dots ,v_{k-1}$, contrary to linear independence. Solve for that $w_j$ and replace it by $v_k$; this preserves spanning. If $r>m$, the first $m$ replacements leave a spanning family $v_1,\dots ,v_m$, contradicting linear independence of the first $m+1$ vectors. This proves both assertions.

Now suppose $S$ is linearly independent and spans ${\mathbb{R}}^n$. Express each member of the standard basis using finitely many members of $S$, and let $S_0$ be the finite union of those members. Then $S_0$ spans. No element of $S\setminus S_0$ can exist, since it would be a linear combination of $S_0$ and violate linear independence. Thus $S$ is finite. Applying the Steinitz exchange lemma in both directions to $S$ and the standard basis gives $|S|\le n$ and $n\le |S|$. Hence $\boxed{|S|=n}$.

Suppose $y$ in the capped simplex has a coordinate strictly between zero and one. Since ${\sum }_i{y}_i=k$ is an integer, it cannot have exactly one such coordinate: the other coordinates would contribute an integer sum. Hence there are distinct $i,j$ with $0<y_i,y_j<1$.

Choose $0<\epsilon <\min \{y_i,1-y_i,y_j,1-y_j\}$ and set $y^{\pm }=y\pm \epsilon (e_i-e_j)$, where $e_i,e_j$ are standard basis vectors. Both perturbed vectors satisfy the coordinate bounds and have the same coordinate sum. They are distinct and $y=(y^{+}+y^{-})/2$, so $y$ is not an extreme point.

Conversely, every zero-one vector in this convex polytope is an extreme point. If it were a nontrivial convex combination of two feasible vectors, each coordinate equal to zero would force both corresponding coordinates to be zero, and each coordinate equal to one would force both to be one. Thus the two vectors would equal the original one. The extreme points are therefore exactly the indicators of $k$-element subsets.

For algebraic stability, the weights must be nonnegative and the real symmetric matrix $M=BA+A^{T}B-b{b}^T$, where $B=\mathrm{diag}(b)$, must be a positive semidefinite matrix.

The weights are positive, but $a_{11}=0$ and $b_1=1/6$, soA positive semidefinite matrix cannot have a negative diagonal entry, as its quadratic form at the first standard basis vector would be negative. Thus this Lobatto IIIA method is not algebraically stable, despite its A-stability.

Choose the rotation taking $e_1$ to $e_2$, $e_2$ to $-e_1$ and $e_3$ to itself. The columns of the matrix of a linear map are these images of the standard basis, soThis fixes the orientation used throughout this question.

The straight line $2x=y=z$ has direction vector $(1,2,2)$, so take the unit vector $\mathbf{n}=(1,2,2{)}^T/3$. A half-turn has $\cos \pi =-1$ and $\sin \pi =0$. The preceding rotation formula becomesThus the rotation matrix in the standard basis isIt fixes $(1,2,2{)}^T$ and negates every vector in its orthogonal complement, which also checks the half-turn geometrically.

For the specified standard basis vectors, $\mathbf{a}\times \mathbf{b}=\hat{\mathbf{k}}$, soOn the other hand, $\mathbf{b}\times \mathbf{c}=\hat{\mathbf{i}}-\hat{\mathbf{j}}$ and henceThese results are unequal, proving that the cross product does not satisfy the associative property.

The integrand is positive, so the minimum-variance importance distribution is its target-weighted probability density function, normalized by $\theta =\sqrt{e}$:By completing the square,Thus $q^{\star }$ is the multivariate normal distribution $N(-e_1,I_d)$, where $e_1=(1,0,\dots ,0)$ is the first standard basis vector. Its importance weight and weighted integrand areA variance cannot be negative, and this proposal makes every summand constant, proving optimality directly. It is admissible under the arbitrary-proposal condition even though its mean vector is not zero. Although optimal proposals usually involve an unknown normalizing integral, here both that integral and the resulting normal distribution are explicit.

Associativity of matrix multiplication identifies $(q_i^{T}x)q_i=q_i(q_i^{T}x)=(q_i{q}_i^T)x$. Thus each matrix $P_i=q_i{q}_i^T$ has the required action and $x={\sum }_i{P}_{i}x$.

Consequently $({\sum }_i{P}_i-I_n)x=0$ for every $x$. Taking $x$ successively as the standard basis vectors shows that every column of this difference is zero. Hence ${\sum }_{i=1}^n{P}_i=I_n$. Equality on one selected vector would not suffice; here it holds on all vectors.

Left multiplication by $E$ interchanges rows three and four; applying it twice restores every row. Equivalently, it sends the standard basis vectors $e_3,e_4$ to each other and fixes $e_1,e_2$, so its square fixes a basis. $E^2=I$.

Write $\hat{A}=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$ and $v_i=P^{-1}{e}_i$, where $e_i$ are the standard basis vectors. Since $A{P}^{-1}=P^{-1}\hat{A}$,Thus the columns of $P^{-1}$ form an eigenbasis, and the requested ordered basis is $D=(P^{-1}{e}_1,\dots ,P^{-1}{e}_n)$. Its basis matrix is $P^{-1}$, giving $[t{]}_D=PA{P}^{-1}=\hat{A}$. More completely, for each distinct eigenvalue $\lambda$, its eigenspace isAll nonzero vectors in these eigenspaces are eigenvectors; when an eigenvalue repeats, they need not be individual columns of $P^{-1}$.

If $A^{\top }=-A$, then the dot product satisfiesConversely, the assumed dot product identity gives $x^{\top }(A^{\top }+A)y=0$ for every $x,y$. Take $x=e_i$ and $y=e_j$ from the standard basis: every entry of $A^{\top }+A$ is zero. Hence $A^{\top }=-A$, proving the equivalence.

Proof. For a skew-symmetric matrix, the scalar $x^{\top }Ax$ equals its transpose $x^{\top }{A}^{\top }x=-x^{\top }Ax$, so it vanishes. Conversely, substituting standard basis vectors $e_i$ into the vanishing quadratic form gives $a_{ii}=0$. Substituting $e_i+e_j$ gives $a_{ij}+a_{ji}=0$. Thus $A^{\top }=-A$. In particular, a real quadratic form depends only on a matrix's symmetric part.

The row rank of a matrix is the dimension of the vector space spanned by its rows. Elementary row operations preserve this row space; multiplication on the right by an invertible matrix preserves all linear relations among the rows.

To see why row rank equals column rank, write a reduced row echelon form as $R=EA$, where $E$ is a product of elementary matrices and is an invertible matrix. Reversibility of the row operations gives equal row spaces for $R$ and $A$. Also $Rz=0$ if and only if $Az=0$, so the columns have exactly the same linear relations. If $R$ has $r$ nonzero rows, its pivot columns are the first $r$ standard basis vectors of ${\mathbb{F}}^m$ and span its column space. Its nonzero rows are linearly independent, since their matrix entries in the pivot positions form an identity matrix. Both dimensions therefore equal $r$, and both counts transfer back to $A$.

The Sherman–Morrison formula computes the matrix inverse after a rank-one matrix update. If $A$ is an invertible matrix and $\delta =1+v^T{A}^{-1}u\ne 0$, thenThe inverse correction is itself an outer product, divided by the scalar $\delta$. The matrix determinant lemma shows that the nonzero-denominator condition is exactly the condition that the updated matrix remains an invertible matrix.

For $A=I$, $u=(1,2{)}^T$, and $v=(3,-1{)}^T$, we have $\delta =2$. Thus the computed inverse isMultiplication in either order gives the identity matrix. The result reuses the original inverse $I$ and the same correction vectors, rather than solving independently for all four inverse entries.

Often only the solution of an updated linear system is needed. If $A{x}_0=b$ and $Az=u$, thenprovided $1+v^{T}z\ne 0$. The two solves use the same LU decomposition of $A$. This expresses the computational benefit without requiring an explicitly stored matrix inverse. Strang's Lecture 14, transcript pp. 9–10 explains this reuse of the original factorization.

To see why the inverse correction works, first consider $A=I$ and put $s=v^{T}u$. The rank-one matrix $R=u{v}^T$ satisfies $R^2=sR$. Expandingalso works in the reverse order. The general case follows by factoring out $A$ and replacing $u$ by $A^{-1}u$. A matrix identity has reduced to cancellation of scalar coefficients.

Historically, Jack Sherman and Winifred J. Morrison studied updating an inverse when a single matrix entry changes in their ^[ref]. Such an entry change is the special rank-one matrix update $\alpha e_i{e}_j^T$, with standard basis vectors $e_i,e_j$. The reference and the earlier 1949 abstract on row or column changes are recorded in Henderson and Searle, On Deriving the Inverse of a Sum of Matrices (1981), references 40–41. This distinguishes the specific original setting from the more general $u{v}^T$ formula used here.

Let $\mathbb{N}=\{1,2,\dots \}$ and give each factor of ${\mathbb{R}}^{\mathbb{N}}$ the usual topology of the real line. The familiar standard basis vectors extend to unit sequences $e_n$ with $(e_n{)}_j=1$ when $j=n$ and zero otherwise. Their finite linear combinations are exactly the finitely supported sequences, rather than all of ${\mathbb{R}}^{\mathbb{N}}$.

For every fixed coordinate $j$, $(e_n{)}_j=0$ for all $n>j$. Therefore $e_n\to 0$ in the product topology. More directly, any basic cylinder set $B(F,U)$ containing zero contains every $e_n$ with $n>\max (F\cup \{0\})$: all of its tested coordinates are then zero. The moving unit entry escapes every fixed finite collection of observations.

In the l-infinity sequence space of bounded real sequences, however,Thus the same sequence does not converge to zero in the supremum norm. Here the comparison is between the subspace topology inherited from the product topology and the norm topology on ${\ell }^{\infty }$. The supremum norm is not a finite-valued norm on all of ${\mathbb{R}}^{\mathbb{N}}$, which also contains unbounded sequences. In the box topology on ${\mathbb{R}}^{\mathbb{N}}$, the neighborhood ${\prod }_{j\ge 1}(-1/2,1/2)$ of zero excludes every $e_n$, so $e_n$ is not a convergent sequence with limit zero there either.

For any fixed real $n$-by-$n$ matrix $A$, as the real parameter $t$ tends to zero,The big O notation means the remainder is bounded in absolute value by $C{t}^2$ for some constant $C$ near zero. The determinant stays positive for sufficiently small $t$, so its oriented volume factor is also its ordinary volume factor. Thus the trace is the first-order fractional volume change, measured from the identity matrix.

A proof follows directly from multilinearity of the determinant. Each column of $I+tA$ is $e_j+tA{e}_j$. The constant term chooses all the standard basis columns and equals one. A term containing exactly one factor $t$ replaces just column $j$ by $A{e}_j$; expanding that vector in the standard basis leaves only its $e_j$ component, $a_{jj}$, because every other choice repeats a column. Summing gives $\mathrm{tr}A$. Terms with two or more replacements have order $t^2$ or higher. This also proves the derivative of the determinant at the identity matrix.

This coefficient has a direct visual meaning. At the point $e_j$, the linear map $x\mapsto Ax$ has component $a_{jj}$ along direction $e_j$. In a small step $x\mapsto x+tAx$, these are the first-order changes of the square's or box's sides in their own directions. Summing them gives its fractional area or volume change; the off-diagonal entries initially change shape rather than contributing to this first-order total.

For the constant-coefficient linear system of differential equations $x^{\prime }=Ax$, the exact linear flow is $x(t)=e^{tA}x(0)$, using the matrix exponential. Its volume factor isThis is the matrix exponential determinant identity. One can derive it without diagonalization: set $E(t)=e^{tA}$, whose matrix inverse is $E(-t)$, and use $E^{\prime }=AE$ in Jacobi's formula to get $V^{\prime }=\det E\mathrm{tr}(E^{-1}AE)=(\mathrm{tr}A)V$. The scalar differential equation with initial value one gives the displayed exponential function. Thus positive trace causes volume expansion, negative trace causes contraction, and zero trace preserves volume, even when individual directions behave differently.

In two dimensions, compare four generators at $t=0.4$:The first and third diagonal matrices expand or contract both directions. The second expands horizontally and contracts vertically by reciprocal factors, making a rectangle of unchanged area. For the last nilpotent matrix, $A^2=0$, so its matrix exponential is exactly $I+tA$; it sends $(x,y)$ to $(x+ty,y)$, a shear mapping. Zero trace allows substantial shape change and does not mean the flow map is stationary.

The first-order statement and the exact flow map should be distinguished. For $A=\mathrm{diag}(1,-1)$, $\det (I+tA)=1-t^2$, whereas $\det (e^{tA})=1$ for all real $t$. Zero trace eliminates the first-order change in the approximate step, and it preserves volume exactly in the matrix exponential flow. The examples use an abstract parameter $t$ and coordinate vectors; no physical units are assumed.