Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2005 / Algebra

Expanding gives $(I-E{)}^2-(I-E)=E^2-E$, so either operator is idempotent exactly when the other is. For an idempotent $E$,If $w\in \ker E\cap \mathrm{Im}E$, write $w=Eu$; then $0=Ew=E^{2}u=Eu=w$. Hence $V=\ker E\oplus \mathrm{Im}E$.

Next,If the right side vanishes, left multiplication by $E_1$ gives $E_1{E}_2+E_1{E}_2{E}_1=0$, while right multiplication gives $E_1{E}_2{E}_1+E_2{E}_1=0$. Subtraction shows $E_1{E}_2=E_2{E}_1$, and their original sum now gives $2E_1{E}_2=0$. Over the real or complex numbers this implies $E_1{E}_2=E_2{E}_1=0$.

For three idempotents summing to $I$, each pair sum equals the complementary idempotent $I-E_k$. Thus every product of distinct operators vanishes. Every vector equals $E_{1}v+E_{2}v+E_{3}v$. If $v_1+v_2+v_3=0$ with $v_i\in \mathrm{Im}{E}_i$, applying $E_j$ gives $v_j=0$. ThereforeTake a basis of each summand and unite the three bases. A vector in the $j$th basis has eigenvalue one for $E_j$ and zero for the other two, so this is the requested common eigenbasis. Zero summands simply contribute no basis vectors.

A real inner product is a symmetric bilinear map $⟨\cdot ,\cdot ⟩:V\times V\to \mathbb{R}$ with $⟨v,v⟩>0$ for $v\ne 0$. It defines $\Vert v\Vert =\sqrt{⟨v,v⟩}$. Set $W=\mathrm{span}\{e_1,\dots ,e_k\}$ and $Pv={\sum }_i⟨v,e_i⟩e_i$. Orthonormality gives $v-Pv\perp W$, hence the Pythagorean identity givesThis proves Bessel's inequality. Equality holds exactly when $v\in W$.

If $v=0$, equality is immediate. If $v\ne 0$, the nonnegative squared normproves the Cauchy-Schwarz inequality. Equality holds precisely when the displayed vector vanishes. Thus equality is equivalent to linear dependence of $u,v$, including the cases with a zero vector.

The sum in question is $⟨Pu,Pv⟩$. Applying Cauchy-Schwarz and then Bessel's inequality givesIf either original vector is zero, equality holds. If both are nonzero, equality forces both projection norms to equal their original norms and also forces dependence of the projected vectors. Thus apart from the zero-vector cases, equality holds exactly when $u,v\in W$ and $u,v$ are linearly dependent.

For the final unnumbered request, apply Cauchy-Schwarz in ${\mathbb{R}}^k$ to $p_i=a^{i/2}$ and $q_i=a^{-i/2}$. Their inner product is $k$, soEquality requires the positive vectors $p,q$ to be proportional, equivalently $a^i$ constant for $i=1,\dots ,k$. For $k\ge 2$ this is $a=1$. If $k=1$, equality holds for every $a>0$.

An integral domain is a nonzero commutative ring with identity and no zero divisors. A principal ideal domain is an integral domain in which every ideal is generated by one element.

The leading coefficient of a product of nonzero real polynomials is nonzero, so $\mathbb{R}[x]$ is an integral domain. The zero ideal is principal. In any nonzero ideal $J$, choose a nonzero polynomial $g$ of least degree. Divide an arbitrary $f\in J$ by $g$: $f=qg+r$ with $r=0$ or $\mathrm{deg}r<\mathrm{deg}g$. Since $r\in J$, minimality forces $r=0$. Thus $J=(g)$ and $\mathbb{R}[x]$ is a principal ideal domain.

In $\mathbb{Z}[x]$ consider $J=(2,x)$. If $J=(h)$, then $h$ divides the constant polynomial 2, so $h$ is constant, and divisibility of $x$ forces $h=\pm 1$. That would give $J=\mathbb{Z}[x]$. But the constant coefficient of every element $2p+xq$ is even, so $1\notin J$. This contradiction proves $\mathbb{Z}[x]$ is not a principal ideal domain.

Let $J={\bigcup }_n{I}_n$. It contains zero. If $a,b\in J$, both belong to $I_N$ for some common $N$, so $a-b\in I_N\subseteq J$. If $r\in R$, then $ra\in I_N\subseteq J$. These are the ideal axioms, proving $J$ is an ideal.

In a principal ideal domain write $J=(g)$. Since $g\in J$, there is $n_0$ with $g\in I_{n_0}$. For $n\ge n_0$,All inclusions are equalities. Hence the chain stabilizes at $I_{n_0}$. This includes $J=(0)$ and is the ascending chain condition for ideals.