For a linear map $\alpha :{\mathbb{R}}^m\to {\mathbb{R}}^n$, its kernel and image of a linear map areThe kernel is a subspace of the domain and the image a subspace of the codomain. Given the stated bases, define the matrix of a linear map byIts $j$th column is the coordinate vector of $\alpha (e_j)$; if $v={\sum }_j{v}_j{e}_j$, the output coordinates are $(\alpha v{)}_i={\sum }_j{A}_{ij}{v}_j$.

For the change of basis, write $e_j^{\prime }={\sum }_k{P}_{kj}{e}_k$ and $f_i^{\prime }={\sum }_{\ell }{Q}_{\ell i}{f}_{\ell }$. The matrices $P,Q$ are invertible. Input coordinates change by $v_{\mathrm{old}}=P{v}_{\mathrm{new}}$ and output coordinates by $w_{\mathrm{old}}=Q{w}_{\mathrm{new}}$, soThe domain and codomain basis changes need not be the same matrix.

For $\beta$, the two input vectors form a basis of ${\mathbb{R}}^2$, so its image is the span of $v=(1,2,3{)}^T$ and $w=(6,4,2{)}^T$. The vector $n=(1,-2,1{)}^T$ is orthogonal to both, since $n\cdot v=n\cdot w=0$. Every possible output therefore satisfies $n\cdot \beta x=0$, whereas the target itself is $n$ and $n\cdot n=6$. Hence there is no $x$ with the required output. This uses an image obstruction by an annihilating functional, rather than an inconsistent guessed input.

For $\gamma$, use the input basis $v_1=(1,2,0{)}^T$, $v_2=(0,1,1{)}^T$, $v_3=(0,1,0{)}^T$. Its determinant is $-1$, so it is genuinely a basis. Write $x=s{v}_1+t{v}_2+w{v}_3$. ThenSubstitution back into physical coordinates givesFor a direct check, its standard matrix is $\left(\begin{array}{ccc}1& 0& -2\\ 1& 1& 0\end{array}\right)$, which annihilates exactly that line.

For $x\ge 0$, the function $h(x)=e^x-1-x$ has $h(0)=0$ and $h^{\prime }(x)=e^x-1\ge 0$, because the exponential function is increasing and $e^0=1$. Therefore $\boxed{e^x\ge 1+x}$.

Write $S_n={\sum }_{j=1}^n{a}_j$ and $P_n={\prod }_{j=1}^n(1+a_j)$. Expanding the product produces $1+S_n$ and additional nonnegative terms, giving $P_n\ge 1+S_n\ge S_n$. Applying the exponential inequality to every factor givesBoth sequences are increasing, since the terms $a_j$ are positive. If $S_n$ converges to a finite value $S$, then $P_n\le e^S$, so the convergence theorem for monotone bounded sequences makes $P_n$ converge. Conversely, if $P_n$ has a finite limit, then $S_n\le P_n$ is bounded and increasing, hence converges by the same theorem. Thus the positive series and the product both converge to finite limits, or both diverge to $+\infty$. This proves the positive sum-product convergence criterion. The word limit here means a finite real limit; allowing extended infinite limits would make the conclusion vacuous for these monotone sequences.

For a partition $D=\{a=x_0<x_1<\cdots <x_n=b\}$, put $\Delta x_k=x_k-x_{k-1}$ andBoundedness makes these finite. The lower Darboux sum and upper Darboux sum areSince $m_k\le M_k$ and the widths are positive, $s(f,D)\le S(f,D)$.

Splitting a partition interval into smaller intervals can only increase each infimum and decrease each supremum. The smaller widths sum to the original width. Its new lower contribution is therefore at least its old lower contribution, and its new upper contribution at most its old upper contribution. Repeating this for every inserted point proves Darboux sum refinement monotonicity:For arbitrary partitions, take their common partition refinement $D_{\ast }=D_1\cup D_2$. ThenNo compatibility of the original partitions is required.

Finally suppose $f\ge 0$ is Riemann integrable. Then $t_k=m_k\Delta x_k\ge 0$, so every product factor is nonnegative and the exponential inequality can be multiplied safely:A lower Darboux sum is no greater than the Riemann integral, since the integral equals the supremum of all lower sums. Monotonicity of the exponential therefore gives the exponential bound for a lower-sum productFor a degenerate interval $a=b$, the empty product and the exponential are both one.