The eigenvalues in $K$ are precisely the roots in $K$ of the companion polynomial $d$; over an algebraic closure take all its roots. Thus the form determines their root set, but an explicit list generally requires factoring or solving $d$ and cannot be read off by inspection. This is the relevant distinction in the qualification “with no further calculations”. Over a general field the polynomial may have no roots in $K$ at all, and its diagonal entries are not its eigenvalues. If “determinable” means only uniquely specified rather than immediately readable, all the information is of course fixed by the matrix, including this root set.

An $R$-module isomorphism $T:V_{\alpha }\to V_{\beta }$ is a $K$-linear isomorphism satisfying $T\alpha =\beta T$, by applying it to multiplication by $X$. Conversely that identity implies $Tf(\alpha )=f(\beta )T$ for every polynomial, so a similarity is an $R$-module isomorphism.

The cyclic decomposition theorem over the principal ideal domain $R=K[X]$ states that any finitely generated module iswith nonunit monic invariant factors, uniquely determined, and free rank $r$ uniquely determined. A finite-dimensional $K$-space has $r=0$. In each cyclic summand the basis $1,X,\dots ,X^{\mathrm{deg}{d}_i-1}$ makes multiplication by $X$ the companion matrix of $d_i$. Their block diagonal sum is the rational canonical form.

Here the two polynomials are already in divisibility order. With the companion convention having ones below the diagonal,Integer coefficients are interpreted in $K$, so this remains valid in characteristics two and three.

In the cyclic case there is one companion block $C(d)$. Its defining polynomial is both characteristic and minimal: the vector represented by $1$ is cyclic and a polynomial annihilates it exactly when divisible by $d$. Thus the minimal polynomial is read directly from that block.

Coordinatewise multiplication gives $V=A(1,1,0)$, so the third module is cyclic.

A submodule of a cyclic module need not be cyclic. In the second example the first-row space is a left ideal and hence a submodule of the regular module $A$. The regular module is cyclic, generated by $1_A$, whereas its first-row submodule has just been shown not to be cyclic. Thus the proposed inheritance property fails even for a finite-dimensional algebra.

For a matrix $v$ supported in its first row and any upper triangular matrix $a$, multiplication gives $av=a_{11}v$. Thus $Av=Kv$ has dimension at most one for every $v$, whereas the entire module has dimension three. Hence this module is not cyclic, over every field $K$.

A left module over the unital $K$-algebra $A$ is a $K$-vector space $V$ with a $K$-bilinear action $A\times V\to V$ satisfying $(ab)v=a(bv)$ and $1v=v$. For a vector $v$, the map $A\to V$, $a\mapsto av$, is an $A$-module homomorphism. Its kernel is $\mathrm{Ann}(v)$; explicitly this is a linear subspace and $b(av)=(ba)v=0$ for $a\in \mathrm{Ann}(v)$, so it is a left ideal.

If $V=Av$, this homomorphism is surjective, and the module isomorphism theorem gives $V\cong A/\mathrm{Ann}(v)$. Conversely the quotient by a left ideal is generated by $1+I$. Thus cyclic modules are exactly the quotients $A/I$ by left ideals.

For the first example set $e=\mathrm{diag}(1,1,0)$. Right multiplication by $e$ keeps the first two columns and zeroes the third. Every matrix with third column zero equals its product with $e$. Hence $V=Ae$ is cyclic, generated by $e$.

In ${\mathbb{R}}^{2n}$, take the two unit vectorsTheir supports are disjoint, so they are orthonormal. Apply part (a) to $a=(a_1,\dots ,a_{2n})$:Multiplying by $n$ gives

In part (i), set every $b_i=1$. Its squared norm is $n$, givingThe equality condition is $a_1=\cdots =a_n$, because $a$ must be a scalar multiple of $(1,\dots ,1)$.

Use the Euclidean inner product on ${\mathbb{R}}^n$, with $a=(a_1,\dots ,a_n)$ and $b=(b_1,\dots ,b_n)$. If $b=0$, both sides of the desired inequality are zero. Otherwise take the single unit vector $v=b/\Vert b\Vert$ in part (a). ThensoThis is the real Cauchy-Schwarz inequality. Equality holds when the two vectors are linearly dependent.

The complex sine has simple zeros exactly at $z=k\pi$, $k\in \mathbb{Z}$, because $\sin (k\pi )=0$ and $\cos (k\pi )=(-1{)}^k\ne 0$. Hence the denominator has a zero of order three at each of these points and nowhere else.

At $0$, the numerator has order two, so the quotient has a simple pole. Indeed its leading term is $-{\pi }^4/z$. At $\pi$, the numerator has order one, so there is a pole of order two. At $-\pi$, the numerator has order three, cancelling the denominator's order-three zero, so the singularity is removable. Its extension value isAt every remaining $k\pi$, the numerator is nonzero, giving a pole of order three. Thus the complete finite-plane classification isAway from this discrete set the denominator is nonzero, so the quotient is holomorphic.

Use the branch from part (b) and close the real interval $[-R,R]$ with the upper semicircle, oriented counterclockwise. For $\mathrm{Im}z\ge 0$, the number $z+i\alpha$ has positive imaginary part, so the numerator is a holomorphic logarithm throughout a neighbourhood of the contour and its interior. The branch point $-i\alpha$ lies strictly below the contour. The only enclosed pole ofis the simple pole at $i\beta$, with residueTo justify discarding the arc, take $R>\max \{\beta ,\alpha +1\}$. On it, $|z+i\alpha |$ lies between $R-\alpha$ and $R+\alpha$, and the argument has bounded absolute value. HenceThe real-axis integral is absolutely convergent, since its integrand is $O(\log |x|/x^2)$ at infinity and has no real singularities. The residue theorem therefore givesFinally, $\mathrm{Re}\mathrm{Log}(x+i\alpha )=\frac{1}{2}\log (x^2+{\alpha }^2)$. Taking twice the real part yieldsThe branch choice affects the first integral's imaginary part; the second integral is real and independent of that choice.

Choose the continuous argument with values in $(-\pi /2,3\pi /2)$ and defineIts branch cut is the nonpositive imaginary axis, including zero, exactly the removed ray. Locally away from that ray, the Cauchy-Riemann equations or the local inverse of the exponential show that it is a holomorphic logarithm, with derivative $1/z$ and $e^{\mathrm{Log}z}=z$. The chosen interval fixes the additive ambiguity by integer multiples of $2\pi i$.

For a piecewise continuously differentiable contour $\gamma :[a,b]\to U$, the contour integral is defined byIf the contour has corners, integrate on its finitely many differentiable pieces and sum the results. The integral on the right is taken componentwise as an ordinary real-parameter integral of a complex-valued function. The factor ${\gamma }^{\prime }(t)$ encodes both the speed and orientation of the contour.

Use the reciprocal Möbius transformationIt is a holomorphic function on $U_1$ by part (b). For $z\ne 0$,Part (c) now gives $z\in U_1$ if and only if $|F(z)+i|>1$, so $F(U_1)\subseteq U_2$. Conversely, $0\notin U_2$ because $|0+i|=1$. For any $w\in U_2$, put $z=1/w$; the same identity gives $|z-i|>|z|$, so $z\in U_1$ and $F(z)=w$. Finally, reciprocation is injective on nonzero complex numbers. This proves bijectivity, with inverseThe inverse is also holomorphic, so this is a biholomorphism.

Write $z=x+iy$. Both moduli are nonnegative, so squaring preserves their strict inequality:ThusThe boundary is the perpendicular bisector of $0$ and $i$ in the complex plane; the strict inequality selects the half-plane containing $0$.