View $X$ as the matrix of a linear map $T:F^m\to F^n$ of matrix rank $r$. Choose vectors $u_1,\dots ,u_r$ whose images form a basis of $\mathrm{im}T$, extend them by a basis of $\ker T$ to a basis of $F^m$, and extend $T{u}_1,\dots ,T{u}_r$ to a basis of $F^n$. In these two bases the matrix of $T$ is the rank normal formThe two changes of basis give invertible $P,Q$ with the displayed matrix equal to $Q^{-1}XP$.

For the block upper triangular matrix $A=\left(\begin{array}{cc}B& D\\ 0& C\end{array}\right)$, every nonzero term in the Leibniz formula for determinants sends the rows belonging to $C$ into the columns belonging to $C$; bijectivity then sends the remaining rows into the $B$ columns. The permutation sum consequently factors into the determinant sums for $B$ and $C$, proving

Finally suppose $L(X)=0$, so $AX=XB$: the map $X:{\mathbb{C}}^m\to {\mathbb{C}}^n$ is an intertwining operator. If $v$ lies in the generalized eigenspace of $B$ for $\lambda$, then for some $k$,Because $A$ and $B$ have no common eigenvalue, $A-\lambda I$ is invertible, hence $Xv=0$. The generalized eigenspaces of $B$ span ${\mathbb{C}}^m$, so $X=0$. Thus the Sylvester equation operator $L$ is injective.

A Jordan block $J_m(\lambda )$ has $\lambda$ on its diagonal, ones on its superdiagonal, and zeros elsewhere. A matrix is in Jordan normal form when it is block diagonal with Jordan blocks.

On a block $J_m(\lambda )$, $\ker (\alpha -\lambda I{)}^r$ has dimension $\min (r,m)$. Its increase from $r-1$ to $r$ is one exactly when $m\ge r$. Summing over blocks proves thatcounts the $\lambda$-blocks of size at least $r$.

If $\lambda \ne 0$, the polynomial $z^2$ has nonzero derivative at $\lambda$, soIf $\lambda =0$, squaring the nilpotent shift separates the odd and even basis chains:omitting a zero-size block.

The displayed invertible matrix preserves the subspaces spanned by $(e_1,e_4)$ and $(e_2,e_3)$. Its restrictions have characteristic polynomials $t^2-a_1{a}_4$ and $t^2-a_2{a}_3$. Since invertibility makes both products nonzero, each restriction has two distinct complex eigenvalues. Its Jordan form is thereforefor either choices of the square roots.

A transportation problem minimizes ${\sum }_{ij}{c}_{ij}{x}_{ij}$ subject to nonnegative shipments, prescribed row supplies, and prescribed column demands.

One optimal shipment table iswhich has the required row sums $(10,8,8)$ and column sums $(3,9,14)$ and costs $4(3)+6(1)+8(3)+3(3)+5(5)=\boxed{76}$.

For an optimality certificate, take row potentials $(1,3,3)$ and column potentials $(0,2,0)$. Their sums equal costs in occupied cells and do not exceed any cost:The dual value is $10+24+24+18=76$, so linear programming duality proves optimality.

The distribution function is $F(x)=x^2/{\theta }^2$ on $[0,\theta ]$, so its median is $\theta /\sqrt{2}$. The likelihood function iswhich decreases over its admissible range. Hence $\hat{\theta }=X_{(n)}={\max }_i{X}_i$. By the invariance property of maximum likelihood estimation,

Substituting the exact solution and expanding about $t_n$ gives the local truncation errorThus the method is consistent of order two; as a stable one-step method it has global convergence order $k=2$.

The order cannot be higher. For $y^{\prime }=t^2$, $y(0)=0$, the exact solution is $y=t^3/3$, and direct substitution makes the local truncation error exactly

The velocity is $\mathbf{u}=\nabla \varphi =q{\mathbf{e}}_r/(2\pi r)$. Its divergence in polar coordinates is zero for $r>0$, while a circle about the origin has fluxBy contour deformation the same holds for any enclosing contour. This is a two-dimensional point source of strength $q$.

For the second potential,which vanishes at $y=0$. In the full plane this is a pair of equal sources of strength $q$ at $(0,\pm a)$. In $y>0$, it is one source at $(0,a)$ beside an impermeable slip wall, with the lower source serving as its image.

For $f=x^2+y^2+z^2$ and $g=x^2+2y^2-z^2=1$, the Lagrange multiplier equations areThe four constrained stationary points areOn the constraint, $f=2x^2+3y^2-1$ with $x^2+2y^2\ge 1$. The two $y$-axis points are global minima, while the $x$-axis points are saddles: motion along the waist decreases $f$, whereas motion away from it increases $f$. There is no maximum.

If instead $g=x^2+2y^2-z^2$ is optimized on $x^2+y^2+z^2=1$, the stationary points are all six coordinate-axis points. The $y$-axis points are maxima with value $2$, the $z$-axis points are minima with value $-1$, and the $x$-axis points are saddles with value $1$.

Let the Markov chain state after each toss be $(S,I)$, where $S$ is player $A_1$'s net winnings and $I$ is the next player to toss. From $(S,i)$ it moves to $(S+1,1)$ with probability $p_i$ and to $(S-1,2)$ with probability $q_i=1-p_i$. The initial state is $(0,1)$, and a return means hitting any state with $S=0$ at positive time.

From $(1,1)$, the probability of hitting zero isIndeed, in the positive-drift case the bounded harmonic solution ofuses the decaying root $q_2/p_1$ and boundary value $h_2(0)=1$, yielding $h_1(1)=q_1/p_2$; with nonpositive drift the walk hits zero surely. By the reflected argument, from $(-1,2)$ the probability isConditioning on the first toss, the required return probability isBoth formulas equal one at the boundary $p_1+p_2=1$.

The rank-nullity theorem says that for a linear map $T:V\to W$ with finite-dimensional domain, $\dim V=\dim \ker T+\dim \mathrm{im}T$. Apply it to $\beta {|}_{\mathrm{im}\alpha }$. Its image is $\mathrm{im}(\beta \alpha )$ and its kernel is $\mathrm{im}\alpha \cap \ker \beta$, soSince $\mathrm{im}(\alpha \gamma )\subseteq \mathrm{im}\alpha$,Using the preceding identity for both restrictions and rearranging gives