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Writing the columns as $a_1,a_2,a_3$, matrix multiplication givesThe column space is the linear span of these columns, so this particular linear combination belongs to it. The assertion holds for every vector of coefficients $y$.

With column-vector homogeneous coordinates, the translation matrix isIt maps $(x,y,1{)}^T$ to $(x+a,y+b,1{)}^T$. The final coordinate incorporates the constant displacement into matrix multiplication.

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.

By matrix multiplication, $D_{ij}={\sum }_{k=1}^N{B}_{ik}{C}_{kj}$. A potentially nonzero term requires $k=i$ because $B$ is a diagonal matrix, and $k=j$ because $C$ is a diagonal matrix. For $i\ne j$ these requirements cannot both hold. Thus every summand is zero and $D_{ij}=0$ whenever $i\ne j$.

For $D=\mathrm{diag}(d_1,d_2,d_3)$, the upper entry of $D^{-1}AD$ in positions $i,i+1$ is $b_i{d}_{i+1}/d_i$, and the corresponding lower entry is $c_i{d}_i/d_{i+1}$. They are equal when $(d_{i+1}/d_i{)}^2=c_i/b_i$. All coefficients used here are positive, so chooseDirect matrix multiplication gives the symmetric matrixIts pattern remains tridiagonal. This is positive diagonal symmetrization of a tridiagonal matrix.

The update must place the sum of the two current entries in the first coordinate and retain the old first coordinate in the second. Thus choose the matrixBy matrix multiplication, $A(L_n,L_{n-1}{)}^T=(L_n+L_{n-1},L_n{)}^T=(L_{n+1},L_n{)}^T$ for every $n\ge 1$, using the given recurrence relation. This is the companion-matrix representation of the Lucas numbers.

Multiply the original equation on the left by $A^{-1}$ to obtainEach term on the right commutes with $A$. ThusNo general commutativity of matrix multiplication is being assumed; it follows from this particular expression for $B$.

An integer-entry example isDirect matrix multiplication gives $A^2=\left(\begin{array}{cc}-1& 1\\ -1& 0\end{array}\right)$, so adding $A$ and $I_2$ gives the zero matrix. Equivalently, this is the companion matrix of $t^2+t+1$ and has the two nonreal cube roots of unity as eigenvalues. The direct multiplication is already a complete proof. It also shows why part (e) does not extend to all real square matrices: with this $A$, choosing $B=0$ satisfies the original product identity.

$\Theta (n^{{\log }_{2}7})$, approximately $\Theta (n^{2.8074})$, arithmetic operations. Strassen's algorithm replaces an $n\times n$ matrix multiplication by seven products of half-size blocks and a constant number of block additions and subtractions. ConsequentlyThe Master theorem applies with $a=7$, $b=2$, since $2<{\log }_{2}7$. Padding an arbitrary matrix dimension to the next power of two changes the dimension by less than a factor of two and preserves this asymptotic bound. The cost model counts scalar arithmetic as constant time; it is not a bit-complexity claim for arbitrarily large integer entries.

Apply Gaussian elimination to the supplied nonsymmetric matrix,The first pivot element is $-2$, with elimination multiplier $l_{21}=2/(-2)=-1$. Perform $R_2\leftarrow R_2+R_1$, givingThe second pivot element is $-1$, with elimination multiplier $l_{32}=-1$. Perform $R_3\leftarrow R_3+R_2$, then use the third pivot element $-2$ and elimination multiplier $l_{43}=-1$ to perform $R_4\leftarrow R_4+R_3$. The LU factorization isHere $L$ is a unit lower triangular matrix, hence an invertible matrix, and $U$ is an upper triangular matrix. No row interchange is needed: all three pivot elements used in elimination are nonzero. The final diagonal entry is zero, with an entirely zero last row; there is no further division by it. The three nonzero rows of $U$ are linearly independent, so the matrix rank is $3$ and $A$ is a singular matrix. Direct matrix multiplication gives $LU=A$.

For a polynomial $p(t)={\sum }_{j=0}^d{a}_j{t}^j$, substitute a fixed square matrix $A$ for $t$, taking $A^0=I$. In particular, $p(t)=t^2-3t+1$ becomes $p(A)=A^2-3A+I$, not $A^2-3A$ with the scalar $1$ added separately to each entry. Polynomials in the same matrix commute with each other, even though arbitrary matrix multiplication is noncommutative.

A prefix scan computes successive accumulated results for an associative operation $\oplus$ while keeping input order fixed. A scan based on addition produces prefix sums; one based on matrix multiplication produces successive ordered matrix products. Associativity allows regrouping, while commutativity is unnecessary because the order stays fixed. If an initial empty prefix is included, its value is the identity element for the chosen operation.

The ability to compute accumulated results does not by itself allow subtraction of two prefixes. For example, two accumulated minima cannot reveal a block minimum: the lists $(1,7,8)$ and $(1,2,3)$ both have successive minima $(1,1,1)$, but the minimum of positions two through three is seven for one and two for the other. Prefix sums support block recovery because addition has additive inverses; an arbitrary associative operation need not have that property.

A reduced QR decomposition of an $m\times k$ matrix over the real numbers with full column rank, $m\ge k$, writes $A=QR$, where $Q$ is $m\times k$ with orthonormal columns and $R$ is a $k\times k$ upper triangular matrix with positive diagonal. Applying the Gram-Schmidt process to the columns $v_j$ of $A$ constructs $Q=(q_1\cdots q_k)$ and records the coefficientsThis last column identity is exactly the matrix multiplication $A=QR$. The upper triangular matrix records that the $j$th input uses only the first $j$ new directions, while its positive diagonal records the norms of the nonzero orthogonal projection residuals. Since $Q^{T}Q=I$, one also has $R=Q^{T}A$; $Q^T$ denotes the transpose and $I$ the identity matrix. Over the complex numbers, use the conjugate transpose $Q^{\ast }$ in these identities.

For the three-dimensional Gram-Schmidt process example, the exact matrices areFor example, the last column reconstructs asThe diagonal entries are approximately $1.414214,1.224745,1.154701$. Each is a norm of the new direction before it was scaled to length one.

For a linear least-squares problem, the orthogonal projection of $b$ onto the column space of $A$ is $Q{Q}^{T}b$. Hence the minimizing coefficients solve the linear system $Rx=Q^{T}b$ by back substitution. Let $A_2=(v_1{v}_2)$ consist of the first two columns of the example; its reduced factors are $Q_2=(q_1{q}_2)$ and the leading $2\times 2$ block $R_2$ of $R$. For $b=(0,0,1{)}^T$, the linear system $R_{2}x=Q_2^{T}b$ isThus $A_{2}x=(1/3,-1/3,2/3{)}^T$, agreeing with the orthogonal projection computed above, and the minimum Euclidean norm of $A_{2}x-b$ is $1/\sqrt{3}$. The QR decomposition separates building an orthonormal basis from recovering coefficients in the original basis.^[ref]

Ramsey interferometry uses two coherent pulses with a free interval between them: the first creates a quantum superposition, the interval accumulates relative quantum phase, and the second converts that quantum phase into measurement probabilities. Its attraction is that an ordinary population readout can reveal the quantum coherence hidden from an energy measurement during the free interval.

Here is an ideal qubit model with an explicit pulse convention. Let $R_y(\theta )=\exp (-i\theta {\sigma }_y/2)$, prepare $|0⟩$, apply $R_y(\pi /2)$, evolve freely with $U_{\varphi }=\exp (-i\varphi {\sigma }_z/2)$, and apply the inverse pulse $R_y(-\pi /2)$ before a computational-basis measurement. The first pulse gives $|+⟩$. Direct matrix multiplication givesThus the Born rule gives $P(0)={\cos }^2(\varphi /2)$ and $P(1)={\sin }^2(\varphi /2)$. Before the second pulse, the same computational-basis measurement would give $1/2$ for both outcomes for every $\varphi$. At $\varphi =2\pi /3$, the second pulse changes these probabilities to $1/4$ and $3/4$. If the two pulses have the same sign instead, the output labels interchange: the pulse convention is part of the prediction.

In a pulse experiment the phase is measured relative to the oscillator supplying the coherent pulses. In a rotating frame, write the free Hamiltonian operator as $H_{\mathrm{rot}}=\hslash \delta {\sigma }_z/2$ with signed angular frequency difference $\delta$ in the chosen labeling convention. A free interval $T$ then accumulates $\varphi =\delta T$; changing the second pulse's phase changes the analysis direction. This comparison makes Ramsey interferometry useful for measuring frequency differences and diagnosing loss of quantum coherence.

For example, suppose the first pulse has prepared a density matrix with equal populations and transverse Bloch vector length $0\le V\le 1$:The same inverse analysis pulse yields $P(0)=(1+V\cos \varphi )/2$. The pure state case $V=1$ gives the full ideal fringe; the mixed state $I/2$ gives the flat result $1/2$. This is a controlled mathematical comparison, not a fit to an experiment. Experimental contrast alone does not identify quantum decoherence: imperfect preparation, pulse errors and readout errors can also reduce it.

For an invertible lower triangular matrix, the scalar pivot $a$ and the remaining triangular block $C$ are both invertible. The displayed formula follows by matrix multiplication: the lower-left block in the product is $b/a-C(C^{-1}b/a)=0$, and the diagonal blocks are identities. Recursively invert $C$, compute $-C^{-1}b/a$, and assemble the inverse matrix; the one-dimensional base case is $a^{-1}$. This recursion reduces the size by one at each step.

When a fixed matrix $A$ is used with several forcing or measurement vectors, compute its Pivoted LU decomposition once. Collect $r$ right-hand sides into the columns of $B=(b^{(1)}\cdots b^{(r)})$, and the corresponding solutions into $X$. ThenEach column undergoes the same row permutation, forward substitution and back substitution. Neither $L$, $U$ nor $P$ depends on $B$. This is the useful separation between preparing a linear system and applying its solution procedure repeatedly. LAPACK implements it by taking previously computed factors, permuting all right-hand sides, and performing the two triangular solves.^[ref]

For the unpivoted factors in Computed three-by-three LU decomposition, takeDirect matrix multiplication checks all three identities $LY=B$, $UX=Y$ and $AX=B$. The first column reproduces the previous solve; the other two use exactly the same factors. New right-hand sides can also arrive one at a time. If $A$ changes, these stored factors generally need to be recomputed.

A ring is a set $R$ with two binary operations, addition and multiplication, satisfying three requirements. Under addition, $R$ is an abelian group: there is a zero, every element has an inverse element, and grouping or exchanging summands does not change the result. Multiplication is associative and has an identity element $1$, so $(ab)c=a(bc)$ and $1a=a1=a$. Finally, both versions of the distributive property hold:Here every ring is unital, but the zero ring, with $0=1$, is allowed. Commutativity of multiplication is an additional condition defining a commutative ring. These conventions agree with ^[ref].

The integers form a ring with their usual operations. The polynomial ring $\mathbb{Z}[X]$ is another: addition combines coefficients, while multiplication uses the distributive property. For example,The same rules justify symbolic calculations in both rings; this is why one definition can organize apparently different kinds of arithmetic. In every ring, $a0=0a=0$, and $a(-b)=-(ab)$. The first identity follows by distributing $a(0+0)$ and subtracting $a0$; the second follows by distributing $a(b+(-b))$. Thus familiar sign rules are consequences of the structure.

Order still matters in some rings. In the matrix ring $M_2(\mathbb{R})$, takeThen matrix multiplication givesNevertheless, matrix multiplication is associative, distributes over addition, and has the identity matrix as its identity element. This is a valid ring. On a column vector, $AB$ applies $B$ first and $A$ second: swapping coordinates before discarding the second coordinate differs from doing those actions in reverse. The failure of commutativity records a meaningful difference in function composition.

Work over the rational numbers, allowing unknowns in the real numbers when interpreting the geometry. Consider four equations in three unknowns:The third row of $A$ is the sum of its first two rows, and the fourth is their difference, second minus first. The same relations must hold on the right: consistency requires $\tau =1+4=5$ and $3=4-1$. The latter condition already holds. Gaussian elimination makes this compatibility test explicit while exposing the remaining freedom.

Apply the elementary row operations in this order, using each row's current value:then $R_3\leftarrow R_3-R_2$ and $R_4\leftarrow R_4-R_2$. The resulting augmented matrix is in row echelon form for every $\tau$:When $\tau \ne 5$, the third row has its pivot element in the augmented column. It says $0=\tau -5$, so the linear system has no solution. When $\tau =5$, that entire row vanishes and only two pivot elements remain. More equations than unknowns therefore do not force a unique solution: some equations may repeat information.

The final operation $R_1\leftarrow R_1-2R_2$ reduces the coefficient block toThe product of all six elementary matrices isDirect matrix multiplication verifies both displayed transformations. Its determinant is one, so this transformation is reversible. In particular, the third row of $E$ extracts the discrepancy between the third equation and the sum of the first two.

For $\tau =5$, $[R\mid Eb]$ is the reduced row echelon form of the full augmented matrix. For $\tau \ne 5$, only its coefficient block is reduced: scale the third row by $1/(\tau -5)$ and clear the entries above its augmented pivot element. The full reduced row echelon form is thenThe last column is part of the matrix when testing the echelon conditions. Calling $[R\mid Eb]$ fully reduced before clearing that column would miss this distinction.

Use the cross-product Lie algebra convention $X=\widehat{e_1}$, $Y=\widehat{e_2}$, $Z=\widehat{e_3}$, so $[X,Y]=Z$. Put $R_x(t)=\exp (tX)$ and $R_y(t)=\exp (tY)$, using the matrix exponential. The finite group commutator isHere $\epsilon \to 0$ and big O notation refers to any fixed matrix norm. Active rotation matrices act on column vectors from right to left: first rotate about $y$ by $-\epsilon$, then about $x$ by $-\epsilon$, then about $y$ by $+\epsilon$, and finally about $x$ by $+\epsilon$. Swapping $x$ and $y$ replaces the product by its inverse and changes the leading Lie bracket to $-Z$.

Writing $c=\cos \epsilon$ and $s=\sin \epsilon$, direct matrix multiplication givesFor $\epsilon =0.1$ radians, $C_{\epsilon }{e}_1\approx (0.999950208,0.009966462,0.000501230)$, whereas $e_1+{\epsilon }^{2}Z{e}_1=(1,0.01,0)$. Thus the residual is a positive rotation about $z$ to leading order; at finite $\epsilon$ it also has smaller components. The Lie bracket captures the leading effect of the four moves rather than their entire finite motion. See Etingof, Section 7.2, Proposition 7.11.

The left panel projects the actual four arcs onto the $yz$-plane for $\epsilon =0.25$; the start and endpoint are close but distinct. The right panel divides the residual coordinates by ${\epsilon }^2$, making the limit $(\Delta y/{\epsilon }^2,\Delta z/{\epsilon }^2)\to (1,0)$ visible. This is a numerical picture of $[X,Y]=Z$ in the SO(3) Lie algebra.

Every finite Markov chain has a stationary distribution, without requiring Markov chain irreducibility or aperiodicity. For a row-stochastic matrix $P$, put $B=P^T$ and choose a column probability vector $q$. Every $B^{j}q$ lies in the probability simplex. Their arithmetic averages $x_N=N^{-1}{\sum }_{j=0}^{N-1}{B}^{j}q$ therefore lie in this compact set, while telescoping givesA convergent subsequence has a limit $x$ in the probability simplex. Continuity of matrix multiplication gives $Bx=x$, so $\pi =x^T$ is a stationary distribution. The individual iterates need not converge: a deterministic two-state swap alternates its point masses forever.