My solutions to the ISI MMath 2026 (PMB) entrance exam problems.

These are my solutions to the problems in the subjective part (PMB) of the entrance examination to the Indian Statistical Institute's Master in Mathematics programme held in 2026.

1. Suppose \(A \in M_n(\mathbb{R})\) has \(n\) distinct real eigenvalues. Prove that there exists \(c \in \mathbb{R}\) such that all eigenvalues of the matrix \(A + cI\) are positive.
   
   Proof. \(A \in M_n(\mathbb R)\) has \(n\) distinct eigenvalues and is hence diagonalizable. Let \(P\) be a matrix such that \[ PAP^{-1}=D, \] where \(D\) is a diagonal matrix with eigenvalues \(\lambda_1,\ldots,\lambda_n\). Take \[ m:=\max\{|\lambda_1|,\ldots,|\lambda_n|\}, \qquad c:=m+1. \] \[ D+cI = PAP^{-1}+cI = P(A+cI)P^{-1}. \tag{1} \] In (1), \(D+cI\) is a diagonal matrix with only positive eigenvalues, and it is similar to \(A+cI\) and hence shares the same eigenvalues.
   \(\blacksquare\)



2. Let \(f : \mathbb{R} \to (0,\infty)\) be a differentiable function and \[ g(t)=\log_e f(t), \qquad t\in\mathbb{R}. \] Suppose \(f\!\left(\frac{\pi}{4}\right)=\sqrt{2}\,f(0).\) Show that there exists \(c>0\) such that \(g'(c)=\tan c.\)
   [Hint: Consider the function \( h(t)=f(t)\cos t. \)]
   
   Proof. We have, \(g(t)=\log_e f(t)\). Define \(h(t):=f(t)\cos t\). Now, \[ g'(t)=\frac{f'(t)}{f(t)} \quad \text{and} \quad h'(t)=f'(t)\cos t-\sin t\,f(t) \] \[ \Rightarrow g'(t) = \frac{f'(t)}{f(t)} = \frac{h'(t)}{h(t)} +\tan(t) \tag{1} \] Notice, \[ h(0)=f(0)=\frac{1}{\sqrt{2}}\,f\!\left(\frac{\pi}{4}\right) \quad \text{and} \quad h\!\left(\frac{\pi}{4}\right) = f\!\left(\frac{\pi}{4}\right)\frac{1}{\sqrt{2}} = f(0). \] Hence, \( h(0)=h\!\left(\frac{\pi}{4}\right) \) and by the mean value theorem \( \exists\,c\in\left(0,\frac{\pi}{4}\right) \) such that \( h'(c)=0. \) Evaluating \((1)\) at \(c\), \[ g'(c) = \frac{f'(c)}{f(c)} = \tan c. \]
   \(\blacksquare\)



3. Let \(W\) be a subspace of \(\mathbb{R}^3\) of dimension \(2\). Consider the vector space \[ V=\{T:\mathbb{R}^3\to\mathbb{R}^3 \mid T \text{ is linear and } T(W)\subseteq W\}. \] Find a basis of \(V\). Justify your answer.
   
   Proof. Let \(A_{ij}\) be the \(3\times 3\) matrix with \(0\) at all indices except the \(i\)-th row and \(j\)-th column entry where it is \(1\). A basis \(B\) for \(V\) is \[ B:=\{A_{ij}\mid 1\le i,j\le 3 \text{ but } (i,j)\ne(3,1) \text{ and } (3,2)\}. \] A linear combination of \(B\) is of the form, \[ \sum a_{ij}A_{ij} = \begin{pmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ 0 & 0 & a_{33} \end{pmatrix}, \qquad a_{ij} \in \mathbb R. \tag{1} \] Let \(\{v_1,v_2\}\) be a basis for \(W\) that is extended to \(\{v_1,v_2,v_3\}\), a basis for \(\mathbb R^3\). Hence every \(w\in W\) has the coordinate vector \[ \begin{pmatrix} a\\ b\\ 0 \end{pmatrix}, \qquad a,b\in\mathbb R. \] Since if \(T\in V\), \(T(W)\subseteq W\), \[ T \begin{pmatrix} a\\ b\\ 0 \end{pmatrix} = \begin{pmatrix} e\\ f\\ 0 \end{pmatrix}, \qquad a,b,e,f \in \mathbb R. \] Since this is for all vectors in \(W\), \(T\) must have form (1) and conversely every matrix of form (1) is invariant in \(W\). Hence \( \operatorname{span}(B)=V \) and since \(B\) is a subset of the standard basis in \(M_n(\mathbb R)\), it is linearly independent.
   \(\blacksquare\)



4. For any function \(g:[0,1]\to\mathbb{R}\), let \[ Z(g)=\{x\in[0,1]:g(x)=0\}. \] Let \(\{f_n\}_{n=1}^{\infty}\) be a sequence of real-valued continuous functions on \([0,1]\) converging uniformly on \([0,1]\) to a function \(f\).
   If \(Z(f)=[a,b]\), then show that for any \(\varepsilon>0\), there exists \(N_\varepsilon\in\mathbb{N}\) such that \( Z(f_n)\subset (a-\varepsilon,b+\varepsilon) \) for all \(n\ge N_\varepsilon\).
   
   Proof. The condition \( Z(f_n)\nsubseteq (a-\varepsilon,b+\varepsilon) \) implies \(\exists\,x\notin (a-\varepsilon,b+\varepsilon)\) such that \( f_n(x)=0. \) (Note, \(Z(f_n)=\varnothing \subset (a-\varepsilon,b+\varepsilon)\).)
   Assume the statement does not hold. Take \(\varepsilon>0\), then there exists a sequence \((n_k)_{k=0}^{\infty}\) such that \( f_{n_k}(x_{n_k})=0 \) and \( x_{n_k}\notin (a-\varepsilon,b+\varepsilon). \)
   Now \( (x_{n_k})\subset [0,1] \) and hence has a convergent subsequence by Bolzano-Weierstrass Theorem. We rename \((x_{n_k})\) to mean this subsequence. Let \[ x_{n_k}\to x, \] now, by the uniform continuity of \((f_n)\), \[ f_{n_k}(x_{n_k})\to f(x). \] Since \( f_{n_k}(x_{n_k})=0 \; \forall k\in\mathbb N, \) \( f(x)=0. \) Now, assume \( x\in [a,b], \) this implies \(\exists\,K\in\mathbb N\) such that \[ |x_{n_k}-x|<\frac{\varepsilon}{2} \qquad \forall k\ge K. \] Hence, \( x_{n_k}\in (a-\varepsilon,b+\varepsilon). \) This contradicts our initial assumption and hence \(x \notin [a,b]\).
   \(\blacksquare\)



5. For each \(n\in\mathbb{N}\), let \( g_n:[0,1]\to[0,1] \) be a continuous function. Assume that the sequence \(\{g_n\}_{n=1}^{\infty}\) converges uniformly to \(0\) on \([0,c]\) for all \(c<1\). Let \( f:[0,1]\to\mathbb{R} \) be a continuous function with \(f(0)=0\). Prove that \[ \lim_{n\to\infty}\int_0^1 f(g_n(x))\,dx = 0. \]
   
   Proof. Define \( c_n=1-\frac1n, \forall n\in\mathbb N. \) \(\{g_n\}\) converges uniformly on \([0,c_n]\), \(\forall n\in\mathbb N\) by hypothesis. Now, \[ I_{n'}:=\lim_{n\to\infty} \int_0^{c_{n'}} f(g_n(x))\,dx. \] Since \(c_n<1\), \[ I_{n'} = \int_0^{c_{n'}} f\!\left(\lim_{n\to\infty}g_n(x)\right)\,dx = \int_0^{c_{n'}}f(0)\,dx = 0, \qquad \forall n\in\mathbb N. \] As \(n\to\infty\), \( c_n\to 1 \) and \( I_n\to 0. \) Since \[ \int_0^1 f(x)\,dx = \lim_{n\to\infty} \int_0^{c_n} f(x)\,dx, \] we have \[ \lim_{n\to\infty} \int_0^1 f(g_n(x))\,dx = 0. \]
   \(\blacksquare\)



6. Let \((X,d)\) be a metric space. Let \(B\) be an uncountable subset of \(X\) with the property that \( d(x,y)=1 \) for all \(x,y\in B\) such that \(x\neq y\). Let \(A\) be a dense subset of \((X,d)\). For \(b\in B\), define \[ S_b=\{x\in A:d(b,x)<\tfrac14\}. \]
   1. Show that \(S_b\) is nonempty for all \(b\in B\).
   2. Show that \(S_b\cap S_{b'}=\varnothing\) if \(b,b'\in B\) and \(b\neq b'\).
   3. Show that \(A\) is uncountable.
   
   
   Proof. \(A\) is dense in \(X\), i.e., \( \forall x\in X,\ \forall \varepsilon>0, B(x,\varepsilon)\cap A\ne\varnothing. \)
   a. Given \(b\in B\), note \(b\in X\) and hence \( \exists y\in B\!\left(b,\frac14\right)\cap A \) such that \( d(y,b)<\frac14 \) and hence \( y\in S_b\ne\varnothing. \)
   b. If \( y\in S_b \cap S_{b'} \neq \varnothing, \) then \( d(b,b') \le d(b,y)+d(y,b') <\frac12, \) which contradicts our hypothesis and \( S_b \cap S_{b'}=\varnothing. \)
   c. Using the Axiom of Choice, pick a single \( x_b\in S_b, \forall b\in B. \) Since \(B\) is uncountable and each \(B\) only belongs to a single \(S_b\), \( \{S_b\mid b\in B\} \) is uncountable which in turn makes \( \{x_b\mid b\in B\} \) uncountable. Since \(A\) has an uncountable subset, it must itself be uncountable.
   \(\blacksquare\)



7. Let \(G\) be a group. Suppose \(H\) and \(K\) are normal subgroups of \(G\) such that \(G/H\) and \(G/K\) are abelian.
   1. Prove that \(H\cap K\) is a normal subgroup of \(G\) and that \( G/(H\cap K) \) is abelian.
   2. Suppose \(G/H\) and \(G/K\) are cyclic groups of orders \(m\) and \(n\) respectively with \( \gcd(m,n)=1. \) Prove that \( G/(H\cap K) \) is a cyclic group of order \(mn\).
   
   
   Proof. Given \(a\in H\cap K\), \(\forall\,b\in G\), \( bab^{-1}\in H \) and \( bab^{-1}\in K. \) Since \(H\) and \(K\) are normal, \( bab^{-1}\in H\cap K \) \( \Rightarrow b(H\cap K)b^{-1}\subseteq (H\cap K) \; \forall\,b\in G. \) Since \(H\cap K\) is also a subgroup, it is a normal subgroup.
   Define \( \phi:G/(H\cap K)\longrightarrow G/H\times G/K, \; g+(H\cap K)\longmapsto (g+H,g+K). \)
   Take \( a+(H\cap K),\; b+(H\cap K)\in G/(H\cap K). \) \[ \begin{align*} \phi\big((a+(H\cap K))+(b+(H\cap K))\big) &= \phi(a+b+(H\cap K)) \\ &= (a+b+H,\;a+b+K) \\ &= ((a+H)+(b+H),\;(a+K)+(b+K)) \\ &= (a+H,a+K)+(b+H,b+K) \\ &= \phi(a+(H\cap K))+\phi(b+(H\cap K)). \end{align*} \] Note that the identity in \(G/H\times G/K\) is \( (0+H,\;0+K), \) \( \phi(a+(H\cap K))=0 \) iff \( a\in H\cap K. \) Hence \[ G/(H\cap K)\cong G/H\times G/K. \] Further, \(G/(H\cap K)\) is abelian, has order \(mn\), and is a cyclic group.
   \(\blacksquare\)



8. Consider the subrings \(R\) and \(\mathbb{Z}[i]\) of \(\mathbb{C}\) given by \[ R=\{a+2bi:a,b\in\mathbb{Z}\} \quad \text{and} \quad \mathbb{Z}[i]=\{a+bi:a,b\in\mathbb{Z}\}. \] Let \( I=\{2a+2bi:a,b\in\mathbb{Z}\}\subseteq R. \)
   1. Prove that \(I\) is an ideal of \(R\) as well as of \(\mathbb{Z}[i]\).
   2. Show that \(I\) is a prime ideal of \(R\) but not a prime ideal of \(\mathbb{Z}[i]\).
   3. Examine whether \(3R\) is a prime ideal of \(R\).
   
   
   Proof. Note, first that \(R\) is a subgroup of \(\mathbb{Z}[i]\). Let \(h = 2a+2bi \in I\).
   a. Take \(r\in\mathbb{Z}[i]\), \(r=c+di\). Then \[ rh=(c+di)(2a+i\,2b) =(2ac-2bd)+i(2bc+2ad) =2(ac-bd)+i\,2(bc+ad)\in I. \] This is true \(\forall r\in\mathbb{Z}[i]\) and \(h\in I\). Hence, \(rI=Ir=I\) for all \(r\in R\) and for all \(r\in\mathbb{Z}[i]\). Hence \(I\) is an ideal of \(\mathbb{Z}[i]\) and \(R\).
   b. Let \(r_1=c+2di,\ r_2=e+2fi\in R\) such that \(r_1r_2=2a+2bi\in I\), for some \(a,b\in\mathbb Z\). \[ r_1r_2=(ce-4df)+2(de+fc)i. \] \[ \Rightarrow ce-4df=2k,\qquad k\in\mathbb Z \] \[ \Rightarrow ce=2(k+2df). \] i.e. either \(c\) or \(e\) has \(2\) as a factor, wlog, \(c=2c'\). Then \(r_1=2c'+2di\in I\). \(I\) is prime in \(R\). \((1+i),\ (3+i)\notin I\) but \[ (1+i)(3+i) =(3-1)+(3+1)i =2+4i\in I. \] \(I\) is not prime in \(\mathbb{Z}[i].\)
   c. \(3R=\{\,3a+6bi \mid a,b\in\mathbb Z\,\}\). \[ (3a+6bi)(c+2di) = 3(a+2bi)(c+2di) = 3e+6fi \in 3R. \] (Since \((a+2bi),(c+2di)\in R\), so does their product.) Hence \(3R\) is an ideal in \(R\).
   
   (The remaining proof is due to my friend, Borishan.)
   To see that this is a prime ideal, take \(a+i2b, c+i2d \in R\) and \((a+i2b)(c+i2d) \in 3R\) implies, \[ 3\mid(ac-4bd), \qquad 3\mid(bc+ad). \] Thus \[ ac-4bd \equiv 0 \pmod 3 \implies ac-bd \equiv 0 \pmod 3, \] and \[ bc+ad \equiv 0 \pmod 3. \] Assume at least one of \(3\nmid a\) and \(3\nmid b\) is true. Then \[ (ac-bd)a+(bc+ad)b = a^2c-abd+b^2c+abd = (a^2+b^2)c \equiv 0 \pmod 3. \] Since \(a^2+b^2\not\equiv 0 \pmod 3\), \( c\equiv 0 \pmod 3. \)
   [Note, \((3k+1)^2 = 3k'+1\), \((3k+2)^2 = 3k'+1\) and \((3k)^2 = 3k'\). Hence the sum of a pair of perfect squares is not divisible by 3 if at least one of the squares is not divisible by 3.]
   Similarly, \[ (ac-bd)b-(bc+ad)a = abc-b^2d-abc-a^2d = -(a^2+b^2)d \equiv 0 \pmod 3. \] Hence \( d\equiv 0 \pmod 3. \) Therefore, \[ 3\nmid a \lor 3\nmid b \implies 3\mid c \text{ and } 3\mid d. \] Thus, without loss of generality, at least one of the divisors belongs to \(3R\) making it a prime ideal.
   \(\blacksquare\)