Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2009 / Algebra

If $c(\lambda )=\det (\lambda I-T)=0$, then $\lambda I-T$ has nonzero kernel of a linear map. A nonzero vector $v$ in that kernel of a linear map satisfies $Tv=\lambda v$, proving that $\lambda$ is an eigenvalue. Extend $v$ to a basis. Its first column is $(\lambda ,0,\dots ,0{)}^T$, so the matrix has block formHere $B$ represents the induced operator on $V/⟨v⟩$.

If the algebraic multiplicity is $r$, repeat this argument in that quotient: its characteristic polynomial still has $\lambda$ with multiplicity $r-1$. Induction and lifting a quotient basis gives a basis whose first $r$ columns have the prescribed upper-triangular form. With $U$ the span of those $r$ vectors, the full matrix isThus $U$ is invariant and is contained in $W_{\lambda }$. Conversely, if $(T-\lambda I{)}^{s}v=0$, project to $V/U$. The induced map $C-\lambda I$ is invertible, so its $s$th power cannot kill a nonzero class. Hence $v\in U$. Consequently $\boxed{W_{\lambda }=U,\dim W_{\lambda }=r}$; this proves simultaneously that it is a subspace and $T$-invariant. It also proves $(T-\lambda I{)}^r=0$ on this space, without using Cayley-Hamilton theorem in advance.

To prove independence of the different spaces, take a relation ${\sum }_{\lambda }{v}_{\lambda }=0$ with $v_{\lambda }\in W_{\lambda }$. Fix $\lambda$ and apply $P_{\lambda }(T)={\prod }_{\mu \ne \lambda }(T-\mu I{)}^{r_{\mu }}$. It kills all terms except $v_{\lambda }$. On $W_{\lambda }$, each factor $T-\mu I=(\lambda -\mu )I+N$ is invertible: its inverse is a finite geometric series in the nilpotent $N$. Hence $P_{\lambda }(T)v_{\lambda }=0$ implies $v_{\lambda }=0$. The sum is direct, and its dimension is $\sum r_{\lambda }=\dim V$, soFinally $c(T)={\prod }_{\lambda }(T-\lambda I{)}^{r_{\lambda }}$ kills each summand, hence every vector. This proves $\boxed{c(T)=0}$ without circularly assuming the theorem during the decomposition.

The orthogonal complement is $W^{\perp }=\{v\in V:⟨v,w⟩=0\text{ for every }w\in W\}$. Put $Pv={\sum }_i⟨v,e_i⟩e_i$, so $T=I-P$. Orthonormality gives $P^2=P$, and therefore $T^2=T$. Also $⟨Tv,e_j⟩=0$ for every $j$, so $Tv\in W^{\perp }$. Every $v=Pv+Tv$ is thus a sum from $W$ and $W^{\perp }$, and their intersection is zero, since a vector orthogonal to itself has squared norm zero. This proves $V=W\oplus W^{\perp }$. The decomposition is unique, so its two projection maps depend only on $W$, not on the chosen orthonormal basis.

Finite dimensionality of $W$ cannot simply be dropped. Take the real Hilbert space $V={\ell }^2(\mathbb{N};\mathbb{R})$ and $W=c_{00}$, the finitely supported sequences. This is an infinite-dimensional proper subspace. A vector orthogonal to every coordinate unit vector is zero, so $W^{\perp }=\{0\}$. Yet, for example, $(2^{-j}{)}_{j\ge 1}$ belongs to $V\setminus W$. Hence $W\oplus W^{\perp }=W\ne V$. The failure occurs even though $V$ itself is complete; the proper dense subspace $W$ is not closed.

For any $w\in W$, the decomposition $v-w=Tv+(Pv-w)$ is orthogonal. The Pythagorean identity givesEquality is attained at $w=Pv$. Thus $\boxed{\underset{w\in W}{\inf }\Vert v-w\Vert =\Vert Tv\Vert }$, and the minimizer in $W$ is uniquely $Pv$.

The assumption $v\notin W^{\perp }$ means $p=Pv\ne 0$. For every unit vector $w\in W$, orthogonality and Cauchy–Schwarz give $⟨v,w⟩=⟨p,w⟩\le \Vert p\Vert$. Equality is attained atEquality in the real Cauchy–Schwarz inequality requires $w$ to be a scalar multiple of $p$, and attaining the positive supremum fixes its sign; the unit norm fixes its magnitude. This proves uniqueness.

Use ordinary matrix addition and multiplication on $R=K[A]$. Sums and products of polynomials in $A$ remain such polynomials, and they commute because powers of the same matrix commute. The identity is $A^0=I$. Evaluation $\varphi (\sum c_i{X}^i)=\sum c_i{A}^i$ is a surjective unital algebra homomorphism over $K$. It is unique among maps fixing the scalar matrices and sending $X$ to $A$.

The scalar condition is needed for the word “unique”: as an unrestricted ring-homomorphism assertion over an arbitrary field, prescribing only the image of $X$ does not suffice. For example, over $\mathbb{C}$ with a real scalar matrix $A$, conjugating every coefficient gives another unital ring homomorphism with the same image of $X$. The intended polynomial-evaluation map is the $K$-algebra map just defined.

The kernel of a ring homomorphism is an ideal of the principal ideal domain $K[X]$. It is nonzero because finitely many powers of $A$ must become linearly dependent in the finite-dimensional matrix space. Its unique monic generator is precisely the minimal polynomial $m$: division by $m$ shows that a polynomial killing $A$ has zero remainder, by minimality. Hence $\ker \varphi =(m)$ and $R\cong K[X]/(m)$.

If $\gcd (f,m)=1$, Bézout's identity gives $uf+vm=1$; evaluation gives $u(A)f(A)=I$. Conversely an inverse in $R$ has form $g(A)$, so $fg-1\in (m)$, which forces $\gcd (f,m)=1$. Thus invertibility in this ring is exactly coprimality with the minimal polynomial.

The displayed matrix is block diagonal with a $1\times 1$ block $1$ and a $2\times 2$ block $B$ having characteristic polynomial $X^2-4X-5=(X-5)(X+1)$. The latter block is not scalar, so its minimal polynomial has degree two and is exactly that polynomial. Taking the least common multiple of the block minimal polynomials gives $m=(X-1)(X-5)(X+1)$ over $\mathbb{R}$.

Ideals of $K[X]/(m)$ correspond to ideals of $K[X]$ containing $(m)$, namely the principal ideals generated by the monic divisors of $m$. Every one of the three distinct linear factors may be included or omitted, soThis includes both the zero ideal and the whole ring. Equivalently the Chinese remainder theorem identifies this ring with ${\mathbb{R}}^3$.

Reduce the matrix modulo two. It is $A=1\oplus \left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, so $A\ne I$ but $(A-I{)}^2=0$. Thus the minimal polynomial is $\boxed{m=(X+1{)}^2}$. Its monic divisors are $1,X+1,(X+1{)}^2$, giving exactly $\boxed{3\text{ ideals}}$ in the quotient.

This ring has the four elements $0,I,A,I+A$. Coprimality with $X+1$ is equivalent to the representing polynomial taking value one at $1$. Thus precisely $I$ and $A$ are units: $\boxed{2\text{ invertible matrices}}$. Indeed $A^2=I$, whereas $(I+A{)}^2=0$.