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If $A$ is $m\times n$, then $Y$ is $n\times m$. The equation $AY=I_m$ makes every standard basis vector of ${\mathbb{R}}^m$ a linear combination of the columns of $A$: take the corresponding column of $Y$ as coefficients. Hence the column space is all of ${\mathbb{R}}^m$, giving $\mathrm{rank}(A)=m$. In particular, $n\ge m$ and $A$ has full row rank.

True for these square matrices. The determinant multiplication identity gives $\det (XY)=\det X\det Y$. Since $XY$ is invertible, its determinant is nonzero; both factors are therefore nonzero, so both matrices have matrix inverses.

Equivalently, $Y$ must be injective and $X$ must be surjective, and either property implies invertibility for an endomorphism of a finite-dimensional vector space. The equal square dimensions matter: a rectangular matrix and a right inverse can have identity product without being mutually invertible square matrices.

True. The rank-nullity theorem gives $\dim \mathcal{N}(A)=n-r=0$. The only vector in this zero-dimensional kernel of a linear map is the zero vector.

Choose $r$ independent columns of $A$ that form a basis of its column space, and put them in $B\in {\mathbb{R}}^{m\times r}$. For each column $a_j$ of $A$, let $c_j$ be its coordinate vector in that basis; put these coordinate vectors in $C\in {\mathbb{R}}^{r\times n}$. The column equations $a_j=B{c}_j$ give $A=BC$.

By construction $B$ has full column rank $r$. The columns of $C$ corresponding to the selected basis columns are the standard vectors of ${\mathbb{R}}^r$, so $C$ has full row rank $r$. This proves the requested rank factorization. When $r=0$, $A=0$ and the corresponding empty-column and empty-row factors have the required shapes and rank zero.

Real symmetry and positive definiteness are sufficient, and the preceding parts establish both properties for this matrix. If a residual becomes zero, stop: the solution has been reached and the displayed step quotient should not be evaluated as $0/0$.

For a convergence justification, let $x_{\ast }$ solve $A{x}_{\ast }=b$, let $e=x_{\ast }-x$, and let $r=Ae$. The quadratic energy error is $E=e^{T}Ae/2$. Moving by $\alpha r$, with $\alpha =(r^{T}r)/(r^{T}Ar)$, is exact line search for a positive-definite quadratic and givesWriting ${\lambda }_{\min },{\lambda }_{\max }>0$ for the extreme eigenvalues, one has $r^{T}Ar\le {\lambda }_{\max }{r}^{T}r$ and $r^{T}r=e^T{A}^{2}e\ge {\lambda }_{\min }{e}^{T}Ae$. ThereforeThe factor lies below one, so the energy tends to zero and positive definiteness forces $x\to x_{\ast }$. Here the bound is $E_{\mathrm{new}}\le (10/11)E$, since the eigenvalues are 1 and 11. This proves convergence for every initial vector, rather than merely checking one numerical iteration.

From the zero initial vector, the linear-system residual is $r_0=(1,1{)}^T$. Then $r_0^T{r}_0=2$, $A{r}_0=(5,13{)}^T$, and $r_0^{T}A{r}_0=18$. The step is ${\alpha }_0=1/9$, givingAs a check, the new residual is $(4/9,-4/9{)}^T$, orthogonal to the preceding residual as expected after this exact line search. It is nonzero, so one iteration has not already solved the system.

For the vector equation, the Jacobian matrix isWrite the updated coordinates as $(X,Y)$. The second equation of the multivariable Newton iteration gives $X-Y=0$. Substituting $X=Y$ into its first linearized equation givesHence $h(x,y)=6x^5-20y^3$, and, whenever $h\ne 0$,

Starting from $(0,1)$ gives $(x_1,y_1)=(21/20,21/20)$. Every subsequent iterate remains on the diagonal, where this is exactly the scalar Newton root-finding iteration from (b). The value $f(21/20)>0$, so $21/20<r$.

For $1\le x<r$, concavity again puts the tangent's zero at or above $r$. Alsoso $r\le N(21/20)\le N(1)=8/7<6/5$. After one scalar step the sequence therefore enters the invariant interval proved in (b). It converges to $(r,r)$ at least quadratically. If the vector error is measured in the Euclidean norm, then $\Vert (x_k,y_k)-(r,r)\Vert =\sqrt{2}|x_k-r|$, so the scalar quadratic estimate also gives a quadratic vector estimate. The initial off-diagonal step causes no difficulty because its Jacobian matrix is invertible and lands exactly on the diagonal.

Use a column gradient. By the multivariable chain rule,The scalar coefficients multiply the constant vectors ${\mathbf{a}}_i$. In particular $f^{\prime }(t)=-t{e}^{-t^2/2}$, although explicit differentiation of $f$ is not required.

Put $\mathbf{z}=(x,y{)}^T$, $r^2={\mathbf{z}}^T\mathbf{z}$ and $f=\sqrt{1+r^2}$. The chain rule yieldsFor every nonzero vector $\mathbf{v}$, the Cauchy-Schwarz inequality givesThus the Hessian is a positive-definite matrix everywhere. The Hessian criterion for convexity implies that $f$ is strictly convex on ${\mathbb{R}}^2$. Its unique global minimum is at $\mathbf{z}=0$, where $f=1$.

The characteristic polynomial is $\det (\lambda I-A)={\lambda }^2-\lambda -1$. Its distinct eigenvalues are $\alpha =(1+\sqrt{5})/2$ and $\beta =(1-\sqrt{5})/2$. For either eigenvalue $\lambda$, the vector $(\lambda ,1{)}^{\top }$ is an eigenvector, since $A(\lambda ,1{)}^{\top }=(\lambda +1,\lambda {)}^{\top }=\lambda (\lambda ,1{)}^{\top }$.

ChooseThe determinant of $S$ is $\alpha -\beta =\sqrt{5}\ne 0$, soIts columns being eigenvectors gives $AS=SD$, and left multiplication by $S^{-1}$ proves $S^{-1}AS=D$.

The initial vector records both initial Fibonacci numbers: $V_1=(U_2,U_1{)}^{\top }=(1,1{)}^{\top }$. For $n\ge 2$,Conversely this vector recurrence forces the second component to be the preceding first component and forces their sum to be the next first component, recovering the scalar recurrence and its initial conditions. Iterating, or using mathematical induction, yields $V_n=A^{n-1}(1,1{)}^{\top }$ for every $n\ge 1$.

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. $E^2=I$.

Replacing row one by $R_1-R_2-R_3-R_4$ does not change the determinant: adding a multiple of another row preserves a determinant. It produces first row $(-2,0,0,0)$, whose cofactor minor is the three-dimensional identity matrix. Thus the determinant is $-2$.

The columns are homogeneous coordinate vectors $(1,v_i{)}^{\top }$ of the four vertices. Subtracting the first column from each other column preserves the determinant and reduces its magnitude to the absolute determinant of the three edge vectors based at $v_1$. That absolute determinant is the parallelepiped volume. A tetrahedron occupies one sixth of that parallelepiped, so $V=|-2|/3!=1/3$ in cubic coordinate units.

No. The nonzero vector $v=(2,-9,-3,21{)}^{\top }$ is in the kernel of the linear map, so $T(v)=T(0)=0$ although $v\ne 0$. A linear map is injective exactly when its kernel consists only of zero.

The reduced row echelon form gives $u_1=2u_4/21$, $u_2=-3u_4/7$, and $u_3=-u_4/7$. Taking $u_4=21t$ gives every vector in the null space as $t(2,-9,-3,21{)}^{\top }$. Thus a basis for the kernel of the linear map isDirect multiplication gives $A(2,-9,-3,21{)}^{\top }=0$. The rank-nullity theorem also checks the count: matrix rank three plus nullity one equals the domain dimension four.

Complete the square to write the objective asLet $z=(x,y)$ and $a=(1,2)$, so $\Vert a\Vert =\sqrt{5}$ while feasibility requires $\Vert z\Vert \ge \sqrt{20}=2\sqrt{5}$. The reverse triangle inequality givesThe objective is therefore at least zero. Equality is attained by $z=2a=(2,4)$, which lies on the boundary. Equality requires the vectors to have the same direction and $\Vert z\Vert =2\sqrt{5}$, so this minimizer is unique. HenceIt also satisfies the Karush-Kuhn-Tucker conditions with $\lambda =1/2$. The direct distance argument proves global optimality even though the exterior of a disk is not a convex set; this is nearest-point minimization outside a Euclidean ball.

Differentiate the column gradient $2e^{r^{T}r}r$. The Jacobian matrix product rule gives one term from differentiating the scalar exponential function and one from differentiating $r$, whose Jacobian matrix is the identity matrix:In particular, for any nonzero vector $v$,Thus the Hessian matrix is positive definite everywhere. This calculation proves strict convexity and supplies the derivatives needed for sums of these terms.

Here a dense $100000$-by-$100000$ matrix has $10^{10}$ entries, about $80$ GB even with eight-byte storage per entry. Use limited-memory BFGS on the negative of the objective, or equivalently its positive-definite approximate-metric form of gradient ascent. It stores only a fixed small number $\ell$ of vector pairs, costing $O(\ell d)$ memory rather than $O(d^2)$.

Let $G(x)={\sum }_{i=1}^{100}{e}^{e^{-\Vert x-a_i{\Vert }^2}}$ and put $c_i(x)=e^{e^{-\Vert x-a_i{\Vert }^2}}{e}^{-\Vert x-a_i{\Vert }^2}$. Direct use of the chain rule givesIf $H_n$ denotes the positive-definite inverse-Hessian matrix approximation for $-G$, the iteration iswith a line search increasing $G$ and the usual safeguarded limited-memory updates. No exact Hessian matrix is required. With fixed memory and the usual bounded-metric and line search hypotheses, expect local linear convergence near a nondegenerate strict maximum; fixed-memory limited-memory BFGS does not generally inherit the superlinear theorem for full-memory BFGS.

Unlike the preceding objective, $G$ is not generally a concave function. Thus this is a method for finding a local maximum; comparison of different converged candidates or a global search is needed to certify a global maximum for arbitrary centres. The symbolic problem supplies no assumption that would justify an unconditional global-convergence claim.^[ref]

A general-purpose high-dimensional choice is Monte Carlo integration. Draw independent vectors $W_r$ uniformly from the unit cube and evaluate $H(W_r)={\sum }_{i=1}^{100}{e}^{\Vert W_r-a_i{\Vert }^2}$. The cube has volume $1$, so its integral is the expected value of $H(W_1)$. The iterative estimator isThe integrand is continuous on a compact set, hence bounded with finite variance. The strong law of large numbers gives almost sure convergence, and the square root of the mean squared error is exactly $\sigma /\sqrt{N}$, where ${\sigma }^2=\mathrm{Var}(H(W_1))$. Thus the standard stochastic accuracy rate is $O(N^{-1/2})$; the dimension affects the variance and evaluation cost, not this exponent.

There is also an important structure-specific improvement here: $e^{\Vert x-a_i{\Vert }^2}={\prod }_{j=1}^{100}{e}^{(x_j-a_{ij}{)}^2}$ is a separable integrand. The Fubini theorem reduces the requested value exactly toOne can therefore replace high-dimensional sampling by $10000$ one-dimensional composite Simpson's rule calculations. For each factor use even $n$, $h=1/n$, andthen iterate $n\leftarrow 2n$ and form the sum of the products. These smooth factors give $O(n^{-4})$ deterministic discretization error for fixed centres. Thus Monte Carlo integration is a valid generic answer, while exploiting separability is the more efficient specialized alternative; a large dimension alone does not force a tensor-product grid.

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 $P^{-1}$.