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past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-1.bigb
= Paper 1
{scope}
https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2024/paperia_1_2024.pdf
= 1A
{parent=Paper 1}
{scope}
{title2=Vectors and Matrices}
= a
{parent=1a}
{scope}
= i
{parent=a}
{scope}
= Solution
{parent=i}
The fixed-point equation is
$$
\frac{az+b}{z+1}=z
\quad\Longleftrightarrow\quad
z^2+(1-a)z-b=0.
$$
A unique solution at $z=i$ means that this <polynomial> is
$$
(z-i)^2=z^2-2iz-1.
$$
Comparing coefficients gives
$$
\boxed{a=1+2i,\qquad b=1}.
$$
This is the repeated-root case of <prescribed fixed points of a Möbius transformation>.
Solved by gpt-5.6-sol high.
= ii
{parent=a}
{scope}
= Solution
{parent=ii}
For roots $3i$ and $1+i$, the fixed-point <polynomial> must be
$$
(z-3i)(z-1-i)
=z^2-(1+4i)z+(3-3i).
$$
Comparing this with $z^2+(1-a)z-b$ gives
$$
\boxed{a=2+4i,\qquad b=-3+3i}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=1a}
{scope}
= Solution
{parent=b}
Write $z=x+iy$. Rationalizing gives
$$
\frac{z-2}{z}
=\frac{x^2-2x+y^2+2iy}{x^2+y^2}.
$$
An argument of $\pi/4$ requires equal positive real and imaginary parts. Hence
$$
x^2-2x+y^2=2y,
\qquad y>0.
$$
Completing squares yields
$$
\boxed{(x-1)^2+(y-1)^2=2,\qquad y>0}.
$$
Thus the <constant-argument locus of a Möbius transformation> is the part above the real axis of the circle with centre $(1,1)$ and radius $\sqrt2$. Its endpoints $z=0$ and $z=2$ are excluded because the quotient is respectively undefined and zero.
Solved by gpt-5.6-sol high.
= 2C
{parent=Paper 1}
{scope}
{title2=Vectors and Matrices}
= a
{parent=2c}
{scope}
= i
{parent=a}
{scope}
= Solution
{parent=i}
The $j$th column is
$$
c_j=v+aj\mathbf1,
\qquad
v=(1+b,2+b,3+b)^T.
$$
All three columns therefore lie in the two-dimensional span of $v$ and $\mathbf1$, so
$$
\boxed{\det A=0}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=a}
{scope}
= Solution
{parent=ii}
For $x=(x_1,x_2,x_3)^T$,
$$
Ax=v(x_1+x_2+x_3)
+a\mathbf1(x_1+2x_2+3x_3).
$$
The <vectors> $v$ and $\mathbf1$ are linearly independent, and $a\ne0$. Thus $Ax=0$ exactly when
$$
x_1+x_2+x_3=0,
\qquad
x_1+2x_2+3x_3=0.
$$
Solving gives
$$
\boxed{\ker A=\operatorname{span}\{(1,-2,1)^T\}},
\qquad
\boxed{\operatorname{nullity}A=1}.
$$
Solved by gpt-5.6-sol high.
= iii
{parent=a}
{scope}
= Solution
{parent=iii}
Every column lies in $\operatorname{span}\{v,\mathbf1\}$. Conversely, differences of two columns are nonzero multiples of $\mathbf1$, while any one column together with $\mathbf1$ recovers $v$. Hence
$$
\boxed{\operatorname{im}A
=\operatorname{span}\{(1+b,2+b,3+b)^T,(1,1,1)^T\}},
$$
and
$$
\boxed{\operatorname{rank}A=2}.
$$
This is an instance of the <rank of a matrix with affine columns>.
Solved by gpt-5.6-sol high.
= iv
{parent=a}
{scope}
= Solution
{parent=iv}
The <rank-nullity theorem> states that for a <linear map> with finite-dimensional domain,
$$
\dim(\text{domain})
=\operatorname{rank}+\operatorname{nullity}.
$$
Here the domain is $\mathbb R^3$, and the preceding parts give
$$
3=2+1,
$$
which verifies the theorem.
Solved by gpt-5.6-sol high.
= b
{parent=2c}
{scope}
= Solution
{parent=b}
The $j$th column of $B$ is
$$
c_j=v+aj\mathbf1,
\qquad
v=(1^2+b,2^2+b,\ldots,n^2+b)^T.
$$
Thus the image is contained in $\operatorname{span}\{v,\mathbf1\}$. Since $n>1$, $v$ is not constant; since $m>1$ and $a\ne0$, two columns have a nonzero constant difference. The two spanning <vectors> therefore both lie in the image, so $\operatorname{rank}B=2$. By <rank-nullity theorem>,
$$
\boxed{\operatorname{nullity}B=m-2}.
$$
Solved by gpt-5.6-sol high.
= 3F
{parent=Paper 1}
{scope}
{title2=Analysis I P∞}
= a
{parent=3f}
{scope}
= Solution
{parent=a}
Convergence of $\sum a_n$ implies $a_n\to0$. Hence $0<a_n\leq1$ for all sufficiently large $n$, and then $a_n^2\leq a_n$. The comparison test, with the finitely many initial terms treated separately, proves that
$$
\boxed{\sum_{n=1}^{\infty}a_n^2<\infty}.
$$
This is the <squares of a summable positive sequence> lemma.
Solved by gpt-5.6-sol high.
= b
{parent=3f}
{scope}
= Solution
{parent=b}
For every $N$, the <Cauchy-Schwarz inequality> gives
$$
\sum_{n=1}^N\sqrt{a_nb_n}
\leq
\left(\sum_{n=1}^Na_n\right)^{1/2}
\left(\sum_{n=1}^Nb_n\right)^{1/2}.
$$
The right-hand side is bounded independently of $N$. Since the summands are positive, the partial sums increase to a finite <limit>, proving the <geometric means of two summable positive sequences> result.
Solved by gpt-5.6-sol high.
= c
{parent=3f}
{scope}
= Solution
{parent=c}
Another application of the <Cauchy-Schwarz inequality> gives
$$
\sum_{n=1}^N\sqrt{a_n}\,n^{-p}
\leq
\left(\sum_{n=1}^Na_n\right)^{1/2}
\left(\sum_{n=1}^Nn^{-2p}\right)^{1/2}.
$$
When $p>1/2$, the second factor is bounded because $2p>1$, so the <series> converges.
At the endpoint, take $a_1=1$ and, for $n\geq2$,
$$
a_n=\frac1{n(\log n)^2}.
$$
Then $\sum a_n$ converges, while
$$
\sqrt{a_n}\,n^{-1/2}=\frac1{n\log n},
$$
whose <series> diverges. This proves both parts of <weighted square roots of a summable sequence>.
Solved by gpt-5.6-sol high.
= 4D
{parent=Paper 1}
{scope}
{title2=AnalysisPI}
= a
{parent=4d}
{scope}
= Solution
{parent=a}
Suppose the <series> converges at $w\ne0$. Then its terms $a_nw^n$ are bounded, say by $M$. Whenever $|z|<|w|$,
$$
|a_nz^n|
\leq M\left|\frac zw\right|^n,
$$
so comparison with a geometric <series> proves absolute convergence at $z$.
Let
$$
R=\sup\{|w|:\sum_{n=0}^{\infty}a_nw^n
\text{ converges}\},
$$
with the natural values $0$ or $\infty$ allowed. If $|z|<R$, choose a convergent point $w$ with $|z|<|w|\leq R$; the preceding argument gives convergence at $z$. If $|z|>R$, convergence at $z$ would contradict the definition of $R$. Hence the <series> converges for $|z|<R$ and diverges for $|z|>R$. No universal assertion is possible on $|z|=R$.
Solved by gpt-5.6-sol high.
= b
{parent=4d}
{scope}
= Solution
{parent=b}
By the <Cauchy-Hadamard theorem>, put
$$
L=\limsup_{n\to\infty}a_n^{1/n},
\qquad R=\frac1L.
$$
The squared coefficients have root limsup $L^2$, so their radius is $R^2$. Equal finite nonzero radii therefore require $R=R^2$, hence $R=1$. This can happen: for $a_n=n+1$, both $\sum a_nz^n$ and $\sum a_n^2z^n$ have radius one.
For the coefficients $a_n^{a_n}$, the relevant root terms are
$$
\left(a_n^{a_n}\right)^{1/n}
=\exp\left(\frac{a_n\log a_n}{n}\right).
$$
Equality can again occur. For example, take
$$
a_n=\log(n+2).
$$
Then $a_n\to\infty$, $a_n^{1/n}\to1$, and
$$
\frac{a_n\log a_n}{n}\to0.
$$
Thus both $\sum a_nz^n$ and $\sum a_n^{a_n}z^n$ have radius one. These examples illustrate the <radius of convergence after powering coefficients>.
Solved by gpt-5.6-sol high.
= 5A
{parent=Paper 1}
{scope}
{title2=Vectors and Matrices}
= a
{parent=5a}
{scope}
= i
{parent=a}
{scope}
= Solution
{parent=i}
Both sides are antisymmetric in $j,k$ and in $p,q$. If either pair has equal indices, both sides vanish. It therefore suffices to take $j\ne k$ and $p\ne q$. The sum over $i$ is nonzero exactly when $(p,q)=(j,k)$ or $(k,j)$; it is respectively $1$ or $-1$. This is exactly
$$
\boxed{\epsilon_{ijk}\epsilon_{ipq}
=\delta_{jp}\delta_{kq}-\delta_{jq}\delta_{kp}},
$$
the <contraction of two Levi-Civita symbols>.
Solved by gpt-5.6-sol high.
= ii
{parent=a}
{scope}
= Solution
{parent=ii}
Expanding a sum of squares gives the $n$-dimensional <Lagrange identity for the cross product>:
$$
\frac12\sum_{i,j}(x_iy_j-x_jy_i)^2
=\sum_{i,j}x_i^2y_j^2-\sum_{i,j}x_ix_jy_iy_j
=|x|^2|y|^2-(x\cdot y)^2.
$$
The left-hand side is nonnegative, so
$$
\boxed{|x|^2|y|^2\geq|x\cdot y|^2}.
$$
Solved by gpt-5.6-sol high.
= iii
{parent=a}
{scope}
= Solution
{parent=iii}
In components, part (i) gives
$$
\begin{aligned}
[a\times(b\times c)]_i
&=\epsilon_{ijk}a_j\epsilon_{klm}b_lc_m\\
&=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})
a_jb_lc_m\\
&=b_i(a\cdot c)-c_i(a\cdot b).
\end{aligned}
$$
Therefore the <vector triple product> is
$$
\boxed{a\times(b\times c)
=b(a\cdot c)-c(a\cdot b)}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=5a}
{scope}
= Solution
{parent=b}
Put
$$
c=y\cdot z,\qquad p=x\cdot y,\qquad q=x\cdot z.
$$
Since $x,y,z$ are unit <vectors>,
$$
S=2-p^2-q^2+p+q
=\frac52-\left(p-\frac12\right)^2
-\left(q-\frac12\right)^2.
$$
The <feasible inner products with two unit vectors> satisfy
$$
\frac{p^2-2cpq+q^2}{1-c^2}\leq1.
$$
The desired pair $p=q=1/2$ is feasible exactly when $c\geq-1/2$. In that case every maximizer has these <inner products>. If $c<-1/2$, symmetry and strict convexity force $p=q$, and the nearest feasible diagonal point is
$$
p=q=\sqrt{\frac{1+c}{2}}.
$$
Using the <vector triple product>,
$$
x\times(y\times z)=y(x\cdot z)-z(x\cdot y)=qy-pz.
$$
Hence
$$
\boxed{
F(y,z)=
\begin{cases}
\dfrac12(y-z),&y\cdot z\geq-\dfrac12,\\[6pt]
\sqrt{\dfrac{1+y\cdot z}{2}}\,(y-z),
&y\cdot z<-\dfrac12.
\end{cases}}
$$
Rotations preserve dot products and cross-product lengths, while $z'=2z$. Thus
$$
S'=5-p^2-4q^2+p+2q
=\frac{11}{2}
-\left(p-\frac12\right)^2
-4\left(q-\frac14\right)^2.
$$
Let
$$
t=\begin{pmatrix}1/2\\1/4\end{pmatrix},
\qquad
K=\begin{pmatrix}1&c\\c&1\end{pmatrix},
\qquad
W=\begin{pmatrix}1&0\\0&4\end{pmatrix}.
$$
The target pair is feasible precisely when
$$
t^TK^{-1}t=\frac{5-4c}{16(1-c^2)}\leq1,
$$
or
$$
\frac{1-3\sqrt5}{8}\leq c\leq
\frac{1+3\sqrt5}{8}.
$$
On this interval,
$$
\boxed{G(y,z)=\frac14y-\frac12z}.
$$
Outside this interval, let $\lambda>0$ be the unique number for which
$$
r=(W+\lambda K^{-1})^{-1}Wt,
\qquad
r^TK^{-1}r=1,
$$
and write $r=(p,q)^T$. These equations give the unique weighted projection of $t$ onto the feasible ellipse, and therefore determine the maximizing <inner products> solely from $c=y\cdot z$. In the remaining cases,
$$
\boxed{G(y,z)=qy-pz}.
$$
Neither answer depends on $\theta_1$ or $\theta_2$.
Solved by gpt-5.6-sol high.
= 6C
{parent=Paper 1}
{scope}
{title2=Vectors and Matrices}
= a
{parent=6c}
{scope}
= Solution
{parent=a}
Since $P=P'=0$, it suffices to map the two edge <vectors>. Let the columns of the required <matrix> be $u_1,u_2,u_3$. The condition $MQ=Q'$ gives
$$
u_2=\left(-\frac12,0,\frac{\sqrt3}{2}\right)^T.
$$
The condition $MR=R'$ then gives
$$
\frac{\sqrt3}{2}u_1+\frac12u_2=R',
$$
and hence
$$
u_1=\left(\frac{\sqrt3}{2},0,\frac12\right)^T.
$$
These are orthonormal. Taking $u_3=u_1\times u_2=(0,-1,0)^T$ produces the orthogonal <matrix>
$$
\boxed{
M=
\begin{pmatrix}
\sqrt3/2&-1/2&0\\
0&0&-1\\
1/2&\sqrt3/2&0
\end{pmatrix}}.
$$
Its columns are orthonormal, and direct substitution verifies $MP=P'$, $MQ=Q'$, and $MR=R'$.
Solved by gpt-5.6-sol high.
= b
{parent=6c}
{scope}
= Solution
{parent=b}
Write $a=(a_1,a_2,a_3,a_4)^T$. The <reflection in a hyperplane> orthogonal to $a$ sends
$$
x\longmapsto x-2\frac{a^Tx}{a^Ta}a,
$$
so
$$
\boxed{M=I_4-\frac{2aa^T}{a^Ta}}.
$$
It acts as $-1$ on $\operatorname{span}\{a\}$ and as $1$ on the three-dimensional hyperplane $a^\perp$. Therefore
$$
\boxed{\det M=-1}.
$$
Solved by gpt-5.6-sol high.
= c
{parent=6c}
{scope}
= Solution
{parent=c}
For $a=(a_1,a_2,a_3)^T$, the <cross-product matrix> is
$$
\boxed{
A=
\begin{pmatrix}
0&-a_3&a_2\\
a_3&0&-a_1\\
-a_2&a_1&0
\end{pmatrix}}.
$$
Its diagonal is zero, so $\operatorname{tr}A=0$. Also $Aa=a\times a=0$, and $a\ne0$, so $A$ has a nontrivial kernel. Hence
$$
\boxed{\operatorname{tr}A=0,\qquad\det A=0}.
$$
Solved by gpt-5.6-sol high.
= d
{parent=6c}
{scope}
= i
{parent=d}
{scope}
= Solution
{parent=i}
Expanding $\det A^{(n)}$ along its last row gives the last diagonal contribution $a_nd_{n-1}$. The only other nonzero entry is $c_{n-1}$; expanding its minor along the last column contributes
$$
-b_{n-1}c_{n-1}d_{n-2}.
$$
Thus the <tridiagonal determinant recurrence> is
$$
\boxed{d_n=a_nd_{n-1}-b_{n-1}c_{n-1}d_{n-2}},
$$
so
$$
\boxed{X_n=a_n,\qquad Y_n=-b_{n-1}c_{n-1}}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=d}
{scope}
= Solution
{parent=ii}
The recurrence becomes
$$
d_n=-2d_{n-1}-d_{n-2},
\qquad d_0=1,\quad d_1=-2.
$$
Its characteristic <polynomial> is
$$
r^2+2r+1=(r+1)^2.
$$
Therefore $d_n=(\alpha+\beta n)(-1)^n$. The initial conditions give $\alpha=1$ and $\beta=1$, so
$$
\boxed{d_n=(-1)^n(n+1)}.
$$
Solved by gpt-5.6-sol high.
= 7B
{parent=Paper 1}
{scope}
{title2=Vectors and Matrices}
= a
{parent=7b}
{scope}
= Solution
{parent=a}
The <matrix trace> is
$$
\operatorname{Tr}(A)=\sum_{i=1}^nA_{ii}.
$$
Using <matrix> multiplication and interchanging dummy indices,
$$
\operatorname{Tr}(AB)
=\sum_{i,j}A_{ij}B_{ji}
=\sum_{j,i}B_{ji}A_{ij}
=\operatorname{Tr}(BA).
$$
The proposed noncyclic interchange is false. With <matrix> units,
$$
A=E_{11},\qquad B=E_{12},\qquad C=E_{21},
$$
one has
$$
\operatorname{Tr}(ABC)=1,
\qquad
\operatorname{Tr}(ACB)=0.
$$
Trace is invariant under cyclic permutations, but an arbitrary swap need not be cyclic.
Solved by gpt-5.6-sol high.
= b
{parent=7b}
{scope}
= Solution
{parent=b}
The <characteristic polynomial> of $A$ is
$$
\chi_A(t)=\det(tI-A).
$$
A <scalar> $\lambda\in\mathbb C$ is an <eigenvalue> when there is a nonzero <vector> $v$ such that
$$
Av=\lambda v.
$$
Such a $v$ is an <eigenvector> belonging to $\lambda$. Equivalently, $\lambda$ is a root of $\chi_A$ and the corresponding <eigenvectors> are the nonzero elements of $\ker(A-\lambda I)$.
Solved by gpt-5.6-sol high.
= c
{parent=7b}
{scope}
= Solution
{parent=c}
Consider the real <polynomial>
$$
p(t)=\det(A+tB).
$$
The hypothesis says $p(i)\ne0$, so $p$ is not identically zero. A nonzero <polynomial> has only finitely many roots, whereas $\mathbb R$ is infinite. Hence there is a real $\lambda$ with
$$
p(\lambda)=\det(A+\lambda B)\ne0.
$$
Thus $A+\lambda B$ is invertible, as summarized by the <real parameter avoiding a singular matrix pencil>.
Solved by gpt-5.6-sol high.
= d
{parent=7b}
{scope}
= Solution
{parent=d}
Write the complex similarity <matrix> as
$$
S=X+iY,
$$
with $X,Y$ real. From $C=SDS^{-1}$ we obtain $CS=SD$, and equating real and imaginary parts gives
$$
CX=XD,\qquad CY=YD.
$$
Therefore
$$
C(X+tY)=(X+tY)D
$$
for every real $t$.
Now $p(t)=\det(X+tY)$ is a real <polynomial> and
$$
p(i)=\det(X+iY)=\det S\ne0.
$$
Thus some real $t$ has $p(t)\ne0$. Taking $T=X+tY$, we have a real invertible <matrix> satisfying $CT=TD$, or
$$
\boxed{C=TDT^{-1}}.
$$
This proves <real similarity from complex similarity>.
Solved by gpt-5.6-sol high.
= e
{parent=7b}
{scope}
= Solution
{parent=e}
<Eigenvectors> belonging to distinct <eigenvalues> are linearly independent. Indeed, suppose
$$
\sum_{j=1}^m c_jv_j=0
$$
is a shortest nontrivial relation, where $Av_j=\lambda_jv_j$ and the $\lambda_j$ are distinct. Applying $A-\lambda_mI$ gives
$$
\sum_{j=1}^{m-1}c_j(\lambda_j-\lambda_m)v_j=0,
$$
a shorter nontrivial relation, a contradiction.
Thus $n$ distinct <eigenvalues> provide a <basis> of $n$ <eigenvectors>, proving that $A$ is diagonalizable over $\mathbb C$. If $A$ is real and all its <eigenvalues> are real, each real singular <matrix> $A-\lambda I$ has a nonzero real null <vector>. Choosing one for each <eigenvalue> gives a real eigenbasis. Hence the real <matrix> is diagonalizable over $\mathbb R$. This is the <distinct eigenvalues imply diagonalizability> criterion.
Solved by gpt-5.6-sol high.
= 8B
{parent=Paper 1}
{scope}
{title2=Vectors and Matrices}
= a
{parent=8b}
{scope}
= Solution
{parent=a}
The <characteristic polynomial> $\chi_A(t)=\det(tI-A)$ has degree $n$. By the <fundamental theorem of algebra>, it has a complex root $\lambda$. Hence
$$
\det(\lambda I-A)=0,
$$
so $\lambda I-A$ has a nonzero kernel. Any nonzero $v$ in that kernel satisfies
$$
Av=\lambda v.
$$
Thus $\lambda$ is an <eigenvalue> and $v$ is a corresponding <eigenvector>.
Solved by gpt-5.6-sol high.
= b
{parent=8b}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
The <adjoint matrix> of $A$ is its conjugate transpose:
$$
\boxed{A^\dagger=\overline A^T},
\qquad
(A^\dagger)_{ij}=\overline{A_{ji}}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
A complex <matrix> $U$ is <unitary matrix> when
$$
U^\dagger U=UU^\dagger=I.
$$
If $Uv=\lambda v$ with $v\ne0$, unitarity preserves norms, so
$$
\|v\|=\|Uv\|=\|\lambda v\|=|\lambda|\,\|v\|.
$$
Therefore every <eigenvalue> satisfies
$$
\boxed{|\lambda|=1}.
$$
Solved by gpt-5.6-sol high.
= iii
{parent=b}
{scope}
= Solution
{parent=iii}
Suppose $A$ is diagonalized by a unitary <matrix>:
$$
A=UDU^\dagger,
$$
where $D$ is diagonal. Then
$$
A^\dagger=UD^\dagger U^\dagger.
$$
Since the diagonal <matrices> $D$ and $D^\dagger$ commute,
$$
AA^\dagger=UDD^\dagger U^\dagger
=UD^\dagger DU^\dagger=A^\dagger A.
$$
Thus $A$ is a <normal matrix>.
Solved by gpt-5.6-sol high.
= c
{parent=8b}
{scope}
= Solution
{parent=c}
First note that if $A$ is normal, then
$$
\|(A-\lambda I)v\|
=\|(A^\dagger-\overline\lambda I)v\|
$$
for every $v$. Hence an <eigenvector> $v$ with $Av=\lambda v$ also satisfies
$$
A^\dagger v=\overline\lambda v.
$$
Normalize $v$ and extend it to an orthonormal <basis>. Let $U$ have these <basis> <vectors> as columns. The first column of $U^\dagger AU$ is $(\lambda,0,\ldots,0)^T$. For every $w\perp v$,
$$
\langle v,Aw\rangle
=\langle A^\dagger v,w\rangle
=\langle\overline\lambda v,w\rangle=0,
$$
so the first row also has no off-diagonal entries. Therefore
$$
U^\dagger AU=
\begin{pmatrix}
\lambda&0\\
0&B
\end{pmatrix}.
$$
Comparing the two block products in the normality identity shows that $BB^\dagger=B^\dagger B$, so $B$ is normal.
The result is trivial in dimension one. Applying the induction hypothesis to $B$ and adjoining the <eigenvector> $v$ gives an orthonormal eigenbasis for $A$. This proves the <unitary diagonalization of a normal matrix>.
Solved by gpt-5.6-sol high.
= d
{parent=8b}
{scope}
= i
{parent=d}
{scope}
= Solution
{parent=i}
The proof is by induction on $n$. Choose a unit <eigenvector> $v$ of $A$, which exists by part (a), and extend it to an orthonormal <basis>. In that <basis>,
$$
U_1^\dagger AU_1=
\begin{pmatrix}
\lambda&*\\
0&C
\end{pmatrix}.
$$
By induction, a unitary change of <basis> on the last $n-1$ coordinates makes $C$ upper triangular. Extending that change by the identity on the first coordinate leaves the displayed zero block intact. Their product is unitary and makes all of $A$ upper triangular. This proves <Schur triangularization>.
Solved by gpt-5.6-sol high.
= ii
{parent=d}
{scope}
= Solution
{parent=ii}
Let
$$
U=(e_3,e_1,e_2)
=
\begin{pmatrix}
0&1&0\\
0&0&1\\
1&0&0
\end{pmatrix}.
$$
This permutation <matrix> is unitary. The actions
$$
Ae_3=e_3,\qquad
Ae_1=2e_3+4e_1,\qquad
Ae_2=3e_3+5e_1+6e_2
$$
give
$$
\boxed{
U^\dagger AU=
\begin{pmatrix}
1&2&3\\
0&4&5\\
0&0&6
\end{pmatrix}},
$$
which is upper triangular.
Solved by gpt-5.6-sol high.
= 9F
{parent=Paper 1}
{scope}
{title2=Analysis I}
= i
{parent=9f}
{scope}
= Solution
{parent=i}
The <ratio test> says that if
$$
\limsup_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L<1,
$$
then $\sum a_n$ converges absolutely. If the ratio has a <limit> $L>1$, the <series> diverges.
For the first assertion, choose $q$ with $L<q<1$. For all sufficiently large $n$,
$$
|a_{n+1}|\leq q|a_n|.
$$
Iteration bounds the tail by a constant multiple of the convergent geometric <series> $\sum q^n$. If $L>1$, choose $q$ with $1<q<L$. Eventually $|a_{n+1}|\geq q|a_n|$, so $a_n$ cannot tend to zero; the <series> therefore diverges.
Solved by gpt-5.6-sol high.
= ii
{parent=9f}
{scope}
= Solution
{parent=ii}
First suppose $L>0$. Taking logarithms,
$$
\log a_n
=\log a_1+\sum_{k=1}^{n-1}
\log\left(\frac{a_{k+1}}{a_k}\right).
$$
Divide by $n$. Since the summands tend to $\log L$, their Cesàro means also tend to $\log L$. Hence
$$
\frac1n\log a_n\to\log L,
$$
and exponentiation gives $a_n^{1/n}\to L$.
If $L=0$, then for every $\varepsilon>0$ the ratios are eventually at most $\varepsilon$, so $a_n\leq C\varepsilon^n$ for a suitable constant $C$. Thus
$$
\limsup a_n^{1/n}\leq\varepsilon.
$$
Letting $\varepsilon\downarrow0$ proves the <ratio limit implies root limit> in all cases.
Solved by gpt-5.6-sol high.
= iii
{parent=9f}
{scope}
= Solution
{parent=iii}
Set
$$
a_n=\frac{n!}{n^n}.
$$
Then
$$
\frac{a_{n+1}}{a_n}
=\frac{(n+1)!}{(n+1)^{n+1}}
\frac{n^n}{n!}
=\left(\frac n{n+1}\right)^n
\longrightarrow e^{-1}<1.
$$
The <ratio test> therefore proves
$$
\boxed{\sum_{n=1}^{\infty}\frac{n!}{n^n}<\infty}.
$$
Solved by gpt-5.6-sol high.
= iv
{parent=9f}
{scope}
= Solution
{parent=iv}
For the <sequence> in part (iii), part (ii) gives
$$
\left(\frac{n!}{n^n}\right)^{1/n}\longrightarrow e^{-1}.
$$
Taking reciprocals yields the <factorial-over-power series> <limit>
$$
\boxed{\lim_{n\to\infty}\frac{n}{(n!)^{1/n}}=e}.
$$
Solved by gpt-5.6-sol high.
= 10E
{parent=Paper 1}
{scope}
{title2=Analysis}
= a
{parent=10e}
{scope}
= Solution
{parent=a}
The <function> $f$ is <differentiable> at $a$ when there is a real number $L$ such that
$$
\lim_{h\to0}\frac{f(a+h)-f(a)-Lh}{h}=0.
$$
Equivalently, the difference quotient has a finite <limit>, and
$$
\boxed{f'(a)=L
=\lim_{h\to0}\frac{f(a+h)-f(a)}h}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=10e}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
Write
$$
\begin{aligned}
f(a+h)g(a+h)-f(a)g(a)
={}&[f(a+h)-f(a)]g(a+h)\\
&+f(a)[g(a+h)-g(a)].
\end{aligned}
$$
Divide by $h$ and let $h\to0$. Differentiability implies continuity, so $g(a+h)\to g(a)$. Therefore the <product rule> gives
$$
\boxed{(fg)'(a)=f'(a)g(a)+f(a)g'(a)}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
No. Take $a=0$ and
$$
f(x)=x,\qquad g(x)=1+|x|.
$$
Then $f$ is <differentiable> and
$$
(fg)(x)=x+x|x|.
$$
Since $x|x|/x=|x|\to0$, the product is <differentiable> at zero with
$$
(fg)'(0)=1\ne0.
$$
However, $g$ has one-sided <derivatives> $-1$ and $1$ and is not <differentiable> at zero. This is a <nondifferentiable factor with a differentiable nondegenerate product>.
Solved by gpt-5.6-sol high.
= c
{parent=10e}
{scope}
= i
{parent=c}
{scope}
= Solution
{parent=i}
Differentiability lets us write
$$
f(a+h)=f(a)+f'(a)h+r(h),
\qquad \frac{r(h)}h\to0,
$$
and, with $k\to0$,
$$
g(f(a)+k)=g(f(a))+g'(f(a))k+s(k),
\qquad \frac{s(k)}k\to0.
$$
Substitute $k=f(a+h)-f(a)=f'(a)h+r(h)$. Since $k=O(h)$, the remainder $s(k)$ is $o(h)$. Thus
$$
g(f(a+h))-g(f(a))
=g'(f(a))f'(a)h+o(h),
$$
and the <chain rule> is
$$
\boxed{(g\circ f)'(a)=g'(f(a))f'(a)}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=c}
{scope}
= Solution
{parent=ii}
No. At $a=0$, let
$$
f(x)=\sqrt[3]x,\qquad g(y)=y^3.
$$
The <function> $g$ is <differentiable> at $f(0)=0$, and
$$
(g\circ f)(x)=x,
$$
so the composite has <derivative> one at zero. But
$$
\frac{f(h)-f(0)}h=|h|^{-2/3}
$$
is unbounded, so $f$ is not <differentiable>. This is a <nondifferentiable inner function with a differentiable nondegenerate composite>.
Solved by gpt-5.6-sol high.
= iii
{parent=c}
{scope}
= Solution
{parent=iii}
To say that $f$ is twice <differentiable> at $a$ means that $f'$ is defined in a neighbourhood of $a$ and is <differentiable> at $a$; its <derivative> is $f''(a)$.
By the <chain rule>, near $a$,
$$
(g\circ f)'=(g'\circ f)f'.
$$
Since $f$ is <differentiable> and $g'$ is <differentiable> at $f(a)$, another use of the chain rule shows that $g'\circ f$ is <differentiable> at $a$. The <product rule> then proves that the displayed product is <differentiable>. Hence $g\circ f$ is twice <differentiable>, with the <second derivative chain rule>
$$
\boxed{(g\circ f)''(a)
=g''(f(a))[f'(a)]^2+g'(f(a))f''(a)}.
$$
Solved by gpt-5.6-sol high.
= 11E
{parent=Paper 1}
{scope}
{title2=Analysis}
= Solution
{parent=11e}
The <intermediate value theorem> states that if $f:[\alpha,\beta]\to\mathbb R$ is continuous and $y$ lies between $f(\alpha)$ and $f(\beta)$, then some $c\in[\alpha,\beta]$ satisfies $f(c)=y$.
It is enough to prove the case
$$
f(\alpha)<y<f(\beta).
$$
Let
$$
E=\{x\in[\alpha,\beta]:f(x)<y\},
\qquad c=\sup E.
$$
Choose $x_n\in E$ with $x_n\to c$. Continuity gives $f(c)\leq y$. If $f(c)<y$, continuity would make $f(x)<y$ for some $x>c$, contradicting that $c$ is an upper bound of $E$. Thus $f(c)=y$. The case with reversed endpoint inequalities follows by replacing $f$ with $-f$.
Solved by gpt-5.6-sol high.
= a
{parent=11e}
{scope}
= Solution
{parent=a}
Choose $z_+,z_-\in\mathbb C$ with
$$
f(z_+)>0,\qquad f(z_-)<0.
$$
Restrict $f$ to the <line segment> between them:
$$
h(t)=f\bigl((1-t)z_-+tz_+\bigr),
\qquad 0\leq t\leq1.
$$
This is continuous, with $h(0)<0<h(1)$. The <intermediate value theorem> gives a $t_0$ with $h(t_0)=0$. Therefore
$$
z=(1-t_0)z_-+t_0z_+
$$
is the required <zero on a line segment>.
Solved by gpt-5.6-sol high.
= b
{parent=11e}
{scope}
= Solution
{parent=b}
Define the <continuous function>
$$
h(x)=g(x+a)-g(x)-a,
\qquad 0\leq x\leq(n-1)a.
$$
At the mesh points its values telescope:
$$
\sum_{k=0}^{n-1}h(ka)
=g(na)-g(0)-na=0.
$$
If one term is zero, the result follows immediately. Otherwise the terms cannot all have the same sign, so there are mesh points at which $h$ is positive and negative. The <intermediate value theorem> between those points gives an $x\in[0,(n-1)a]$ with $h(x)=0$. Thus the <telescoping increment lemma> yields
$$
\boxed{g(x+a)=g(x)+a}.
$$
Solved by gpt-5.6-sol high.
= 12D
{parent=Paper 1}
{scope}
{title2=Analysis I}
= a
{parent=12d}
{scope}
= i
{parent=a}
{scope}
= Solution
{parent=i}
For a partition
$$
P:\quad 0=x_0<x_1<\cdots<x_m=1,
$$
put
$$
M_j=\sup_{[x_{j-1},x_j]}f,
\qquad
m_j=\inf_{[x_{j-1},x_j]}f.
$$
The upper and lower Darboux sums are
$$
U(f,P)=\sum_{j=1}^mM_j(x_j-x_{j-1}),
\qquad
L(f,P)=\sum_{j=1}^mm_j(x_j-x_{j-1}).
$$
The <upper and lower Darboux integrals> are
$$
\overline{\int_0^1}f=\inf_PU(f,P),
\qquad
\underline{\int_0^1}f=\sup_PL(f,P).
$$
The <bounded function> is Riemann integrable when these two numbers agree; their common value is its <integral>.
Solved by gpt-5.6-sol high.
= ii
{parent=a}
{scope}
= Solution
{parent=ii}
Let $f$ be continuous on $[0,1]$. Compactness makes it uniformly continuous. Given $\varepsilon>0$, choose $\delta>0$ such that
$$
|x-y|<\delta\quad\Longrightarrow\quad
|f(x)-f(y)|<\varepsilon.
$$
Choose a partition with mesh smaller than $\delta$. The oscillation $M_j-m_j$ on every subinterval is then at most $\varepsilon$, so
$$
U(f,P)-L(f,P)
=\sum_j(M_j-m_j)(x_j-x_{j-1})
\leq\varepsilon.
$$
The <Riemann integrability criterion> proves that every <continuous function> on $[0,1]$ is Riemann integrable.
Solved by gpt-5.6-sol high.
= b
{parent=12d}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
No. Define
$$
f(0)=0,\qquad f(x)=\frac1x\quad(0<x\leq1).
$$
It is continuous at every point of $(0,1)$ and is unbounded. Its truncation is Riemann integrable, but
$$
\int_0^1\min\left(\frac1x,r\right)\,dx
=\int_0^{1/r}r\,dx+\int_{1/r}^1\frac{dx}{x}
=1+\log r\longrightarrow\infty.
$$
Thus it is not improperly integrable with a finite <integral>.
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
No. Enumerate the rationals in $[0,1]$ as
$$
A=\{q_1,q_2,\ldots\},
$$
and define
$$
f(q_n)=n,\qquad f(x)=0\quad(x\notin A).
$$
The <function> is nonnegative and unbounded, while it is identically zero on $[0,1]\setminus A$. For any $r>0$, however, $f_r$ is positive on the dense set of rationals and zero on the dense set of irrationals. Every interval therefore has positive oscillation, and the upper and lower Darboux <integrals> differ. Thus even the truncations need not be Riemann integrable.
Solved by gpt-5.6-sol high.
= iii
{parent=b}
{scope}
= Solution
{parent=iii}
For $k\geq1$, let $E_k$ be the set of numbers whose first digit $5$ occurs in position $k$. The first $k-1$ digits each have nine allowed values, while the $k$th is fixed, so $E_k$ is a union of decimal intervals of total length
$$
|E_k|=\frac1{10}\left(\frac9{10}\right)^{k-1}.
$$
The set $E_\infty$ of numbers having no digit $5$ can, after $N$ places, be covered by intervals of total length $(9/10)^N$; hence it has length zero.
For each fixed truncation level, the <function> is constant on finitely many decimal cylinders apart from the remaining set, whose covering length can be made arbitrarily small. The <Riemann integrability criterion> therefore shows that every truncation is Riemann integrable.
For integer $N$,
$$
\int_0^1\min(f,N)\,dx
=\sum_{k=1}^{N-1}
\frac{k}{10}\left(\frac9{10}\right)^{k-1}
+N\left(\frac9{10}\right)^{N-1}.
$$
The final term tends to zero, and monotonicity handles noninteger truncation levels. Therefore the <first occurrence of a decimal digit> calculation gives
$$
\int_0^1f(x)\,dx
=\sum_{k=1}^{\infty}
\frac{k}{10}\left(\frac9{10}\right)^{k-1}
=\frac{1/10}{(1-9/10)^2}
=\boxed{10}.
$$
Solved by gpt-5.6-sol high.