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The minimal polynomial divides the annihilating polynomial $t^2(t-1)$, since $A^3-A^2=0$. Every nonconstant monic divisor can occur in dimension four: use Jordan blocks of sizes at most two at eigenvalue $0$ and of size one at eigenvalue $1$, adding scalar blocks as needed. The complete list isFor example the first four occur for $0$, $J_2(0)\oplus 0\oplus 0$, $I$, and $\mathrm{diag}(0,0,1,1)$, respectively; the final one occurs for $J_2(0)\oplus 0\oplus 1$.

Failure to be diagonalizable requires a size-two Jordan block at zero. The condition $A^2\ne 0$ requires at least one block at one, since every allowed zero block squares to zero. A size-two zero block and one scalar block at one leave only one dimension. It can carry zero or one. Thus, up to permutation of blocks, the only possible Jordan normal forms are

Let $a_1,a_2,a_3$ denote the columns. The Gram-Schmidt process begins with $\Vert a_1\Vert =2$, soThe second column has $r_{12}=q_1^{\mathsf{T}}{a}_2=4$ and residualThis residual has norm $2$, giving $q_2=\frac{1}{2}(-1,1,-1,1{)}^{\mathsf{T}}$ and $r_{22}=2$.

For the third column the orthogonal projections are $r_{13}=6$ and $r_{23}=4$. Subtracting them givesIts norm is $2$, so $q_3=\frac{1}{2}(1,1,-1,-1{)}^{\mathsf{T}}$ and $r_{33}=2$. Thus the requested thin QR decomposition isThe three displayed columns have pairwise zero inner products and unit norms; direct multiplication gives $QR=A$.

Use the one-sided product bound for an entire function by consideringThis is an entire function, and $f^2=(u^2-v^2)+2iuv$ gives $|F|=e^{2uv}$. If $uv\le M$, then $|F|\le e^{2M}$ everywhere. The Liouville theorem therefore makes $F$ constant.

Because an exponential never vanishes, differentiating yields $0=F^{\prime }=-2if{f}^{\prime }F$, and hence $(f^2{)}^{\prime }=2f{f}^{\prime }=0$. Thus $f^2$ is constant. If that constant is zero, $f=0$; otherwise $f$ takes values in the two square roots of one nonzero constant. Continuity on the connected space $\mathbb{C}$ prevents switching between those two values. Therefore $f$ is constant.

Represent a possibly randomized statistical test by a function $\phi (Y)$ taking values in $[0,1]$, its conditional probability of rejection. Its power function is $\pi (\rho )={\mathbb{E}}_{\rho }\phi (Y)$, and its size of a statistical test is the supremum of this probability over the null hypothesis. At a prescribed significance level $\alpha$, a uniformly most powerful test has at least as much power as every other level-$\alpha$ test at every parameter in the alternative.

Put $S={\sum }_{j=1}^n{Y}_j$. For any fixed alternative $0<{\rho }_1<{\rho }_0$, the likelihood ratio iswhich is strictly increasing in $S$. The Neyman-Pearson lemma therefore selects the upper tail $S>c_{\alpha }$. Its cutoff is determined solely by the null distribution, not by ${\rho }_1$, so the same statistical test is most powerful against every allowed alternative and hence uniformly most powerful.

Under rate $\rho$, $S$ has a gamma distribution with shape $n$ and rate $\rho$. For completeness, convolution starts with $\rho e^{-\rho s}$ and, if the $n$-term density is ${\rho }^n{s}^{n-1}{e}^{-\rho s}/(n-1)!$, convolving once more gives ${\rho }^{n+1}{s}^n{e}^{-\rho s}/n!$. Integrating this density by parts repeatedly gives its upper-tail probability. Consequently reject precisely when $S>c_{\alpha }$, whereand the power function isFor $0<\alpha <1$ the continuous strictly decreasing null tail gives a unique positive cutoff, with exact size $\alpha$ and no boundary randomization.

Let $R_{ij}$ be the components of an orthogonal matrix relating two Cartesian frames by $v_i^{\prime }=R_{ij}{v}_j$. Repeated indices are summed from $1$ to $3$. The tensor component transformation law for a Cartesian second-rank tensor isFor a proper rotation, $\det R=1$; the transformation law itself also holds for orthogonal changes of Cartesian frame with determinant $-1$.

To find a cubically invariant second-rank tensor, first square each of the prescribed quarter-turns. For example the half-turn about the first axis is $S_x=\mathrm{diag}(1,-1,-1)$. Invariance under $S_{x}A{S}_x^{\mathsf{T}}$ forces $A_{12},A_{13},A_{21},A_{31}$ to vanish. The corresponding half-turns about the other axes eliminate the remaining off-diagonal components, even though symmetry of $A$ was never assumed.

Now write $A=\mathrm{diag}(a,b,c)$. A quarter-turn about the third axis interchanges $a,b$, so $a=b$; a quarter-turn about the first axis gives $b=c$. Conversely a scalar multiple of the identity matrix is invariant under every rotation. Thus the most general answer is

A metric space is a complete metric space when every Cauchy sequence converges to a point of that space. It is a totally bounded space when, for every $\epsilon >0$, finitely many open metric balls of radius $\epsilon$ cover the space.

With their usual distances, $\mathbb{R}$ is complete but not totally bounded, and $(0,1)$ is totally bounded but not complete. The Cauchy sequence $1/n$ in $(0,1)$ has its limit outside the space; no finite collection of balls of a fixed radius covers $\mathbb{R}$.

For the requested continuous functions, takeandBoth are continuous and onto. The domain of $f$ is a complete metric space, but its image is not complete because $1/n$ tends to the missing point $0$. The domain of $g$ is a totally bounded space, but its unbounded image is not totally bounded. Uniform continuity, unlike mere continuity, would preserve total boundedness.