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The complex line integral along the smooth parametrized curve isFor the reversed curve, the chain rule gives ${\tilde{\gamma }}^{\prime }(t)=-{\gamma }^{\prime }(a+b-t)$. Substitution $s=a+b-t$ yieldsConsequently reversing orientation negates the complex line integral.

A set is compact if every open cover of it has a finite subcover. Let $K_1,\dots ,K_m$ be compact and let $\mathcal{U}$ cover their union. For each $j$, the same family covers $K_j$, so compactness supplies a finite subfamily ${\mathcal{U}}_j$ covering $K_j$. The union of these finitely many finite subfamilies is finite and covers ${\bigcup }_j{K}_j$. Hence a finite union of compact sets is compact.

An arbitrary union need not be compact. Each singleton $\{n\}$, $n\ge 1$, is compact, but their union $\mathbb{N}$ has the open cover $\{(n-1/3,n+1/3):n\ge 1\}$ with no finite subcover. Every member covers only one positive integer, so finitely many leave infinitely many integers uncovered.

Let $E_{ji}$ be the matrix unit having its only nonzero entry, one, in position $(j,i)$. Direct multiplication gives $\mathrm{tr}(A{E}_{ji})=a_{ij}$. If $A\in R$, every such pairing vanishes, forcing every entry $a_{ij}$ to be zero. Conversely, zero pairs to zero with every matrix. ThusThe matrix trace pairing is therefore a nondegenerate bilinear form, despite its failure to be positive definite when $n\ge 2$.

Interpret diagonalizability over the specified underlying field $\mathbb{R}$. Write $A=P\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)P^{-1}$ with real eigenvalues. Invariance of the trace under similarity givesIf $A\ne 0$ is diagonalizable, at least one eigenvalue is nonzero, proving $\phi (A,A)>0$. Real diagonalizability is essential: the rotation block from part (c) is diagonalizable over $\mathbb{C}$ but has eigenvalues $i,-i$ and negative trace of its square.

For $n\ge 2$, take the matrix whose leading $2\times 2$ block is $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ and whose other entries are zero. Its square has leading block $-I_2$, soThis contradicts the positive definiteness required of an inner product. Even the nonzero matrix unit $E_{12}$ has $\phi (E_{12},E_{12})=0$.

The printed assertion has an exception: when $n=1$, $\phi (a,b)=ab$ is an inner product on $\mathbb{R}$. Thus the requested failure requires $n\ge 2$, a restriction absent from the printed question.

Expand the trace by matrix entries:Hence $\phi (A,B)=\phi (B,A)$. Combined with part (a), this makes $\phi$ a symmetric bilinear form.

For real scalars $\alpha ,\beta$ and matrices $A,C,B$, distributivity of matrix multiplication and linearity of the trace giveThus $\phi$ is linear in its first argument.

The polynomial $X(X-1)$ annihilates $E$ by idempotence, so the minimal polynomial divides it. If $E=0$, its monic polynomial of smallest positive degree is $X$. If $E\ne 0$, both eigenvalues zero and one occur, so both factors must divide every annihilating polynomial. Therefore

If $Ev=\lambda v$ with $v\ne 0$, applying $E^2=E$ gives ${\lambda }^{2}v=\lambda v$, hence $\lambda \in \{0,1\}$. The kernel is nonzero since $E\ne I$, so zero always occurs. One occurs exactly when the image is nonzero. Thus the eigenvalue set is $\{0\}$ if $E=0$, and $\{0,1\}$ otherwise. Their algebraic multiplicities are $N-r$ and $r$ respectively.

In the adapted basis from part (a), the matrix has $r$ diagonal entries equal to one and $N-r$ equal to zero. With the characteristic polynomial convention ${\chi }_E(X)=\det (XI-E)$,The characteristic polynomial is independent of the choice of basis.

Every vector decomposes as $v=Ev+(v-Ev)$, where $Ev\in \mathrm{im}E$ and $E(v-Ev)=0$. If $w$ lies in both the image and kernel, write $w=Eu$ to obtain $w=E^{2}u=Ew=0$. Thusand $E$ is the projection onto its image along its kernel. It is the identity on its image and zero on its kernel. Let $N=\dim V$ and $r=\mathrm{rank}E$. A basis adapted to this direct sum gives matrix $\mathrm{diag}(I_r,0_{N-r})$. Since $E$ is not the identity, $N-r>0$, and therefore

The modulus map is continuous, the denominator $1+|z|$ is positive everywhere, and sums, products and quotients with nonzero denominator preserve continuity. Hence $f$ is continuous throughout $\mathbb{C}$.

For $g$, $|g(z)|=|z|$, so $g(z)\to 0=g(0)$ as $z\to 0$. The chosen argument is continuous locally away from the negative real axis, making $g$ continuous there as well. At a point $-r$, $r>0$, approach along $z=r{e}^{i\theta }$. From $\theta \uparrow \pi$, $g(z)\to ir$; from $\theta \downarrow -\pi$, $g(z)\to -ir$. The value assigned on the cut is $ir$. Thus $g$ is discontinuous at every strictly negative real point, and therefore is not a continuous map on the entire complex plane. Its radial factor is $r$, as printed, rather than the $\sqrt{r}$ of a square root.

Applying the derivative rule twice and using both zero initial data givesso $\bar{x}(p)=1/(p^2+1{)}^2$. By the convolution theorem for Laplace transforms,Use the product-to-sum identity $\sin s\sin (t-s)=\frac{1}{2}[\cos (2s-t)-\cos t]$ to integrate:It satisfies $x(0)=x^{\prime }(0)=0$, and differentiation gives $x^{\prime \prime }+x=\sin t$. The term proportional to $t\cos t$ is the resonant response of the forced harmonic oscillator.